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Arnold & Marie Schwartz College of Pharmacy and Health Sciences, Long Island University
PH931
Instructor: M. Savva, Ph.D.
CHAPTER I
STATES OF MATTER
INTRODUCTION
Advanced Physical pharmacy is a required three credit-hour course offered to the MS
students of the Pharmaceutics & Industrial Pharmacy program. The course discusses
states of matter, ideal and real gases, enthalpy and thermochemistry, introduction to
thermodynamics, intermolecular forces in liquids and solids, chemical equilibria and
entropy, Gibbs free energy, kinetics, solution theory, diffusion and dissolution principles.
The application of these subject areas to the preparation of solid and liquid dosage forms,
aerosol and other rate-controlled and targeted drug delivery systems is discussed in
subsequent courses. The material presented in this chapter aims to help the students:
1. Learn about and distinguish between the different forms and the three different states
of matter.
2. Understand that conversion of a drug molecule into a different state is due to physical
changes that are intimately related to intermolecular forces. Physical changes are
reversible. Chemical changes are usually related to the spatial arrangement of atoms
within the molecule (interatomic or intramolecular forces) and they always result in the
creation of a new substance.
3. Develop critical thinking of how the physicochemical properties of a formulated drug
product can be affected by the “inert” excipients and how one can go about detecting the
drug in a particular dosage form.
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Arnold & Marie Schwartz College of Pharmacy and Health Sciences, Long Island University
PH931
Instructor: M. Savva, Ph.D.
4. Understand the interplay between molecular structure, physical properties and
pharmacological action of a drug.
STATES OF MATTER
Matter is the material of the universe and it can be defined as anything that has mass
and occupies space. Based on its composition and properties, matter can be classified as
mixtures, pure substances, pure compounds and elements.
Matter
Mixtures of
substances
Heterogeneous
mixtures
Pure
substances
Homogeneous
mixtures
Compounds
Elements
A substance is a form of matter that has a constant composition. The physicochemical
properties of a substance are dependent on the way its atoms are organized. For example,
n-butane has exactly the same chemical formula as iso-butane, C4H10. Their physical
properties, e.g., boiling and melting point as shown in Table I, vapor pressure at a given
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Arnold & Marie Schwartz College of Pharmacy and Health Sciences, Long Island University
PH931
Instructor: M. Savva, Ph.D.
temperature, and their chemical properties, e.g., reactivity to a carbocation or a free
radical, differ due to a different organization of the same atoms in each molecule, that is,
they have different structural formulas (n-butane: CH3-CH2-CH2-CH3; iso-butane: CH3CH(CH3)-CH3.
Table I. Physical Constants for n-butane and isobutene*
n-butane
isobutane
Boiling point
0 ˚C
-12 ˚C
Melting point
-138 ˚C
-159 ˚C
Relative density at -20 ˚C
0.622
0.604
* Values are adopted from: Morrison, R.T., and Boyd, R.N., Organic Chemistry, 5th Edition, Allyn and
Bacon Inc.: Massachusets, 1987.
Nitrogen (gas), water (liquid), glucose (solid) are examples of three different
substances existing in different physical states under normal conditions (1 atmosphere, 22
˚C). Ice water, liquid water and vapor water, are examples of a substance in the three
different states. Reversible changes of the physical states of a substance are physical
changes. Physical changes are due to reorganization of the molecules in a substance.
Contrary to that, chemical changes are due to the way the substance’s atoms are
organized. Chemical changes may be irreversible, fully or not fully reversible (the
majority of chemical reactions are reversible only to some extent) and they always result
in a change of a substance to a new one having different properties. An example of an
irreversible chemical change is decomposition of water causing the molecules to break
apart and form hydrogen and oxygen, two new substances. The esterification of salicylic
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Arnold & Marie Schwartz College of Pharmacy and Health Sciences, Long Island University
PH931
Instructor: M. Savva, Ph.D.
acid with malonic anhydride to form aspirin (Fig. 1) is a reversible chemical change. The
product to reactant’s ratio of a reversible chemical reaction under a given set of
conditions at equilibrium, is always the same and is expressed by the equilibrium
constant, K of the reaction.
O
O
O
OH
OH
O
+
OH
O
O
O
CH3
Fig. 1. Synthesis of aspirin from salicylic acid and malonic anhydride.
A compound is a form of substance in which two or more atoms (elements) are
chemically linked. Molecular compounds can be broken down to pure elements only by
chemical means.
An element is a substance that cannot be further divided by chemical means. It is
defined by its atomic number. Elements have isotopes. For example, the radioactive
that is frequently used in thyroid cancer treatment is an isotope of the stable
127
125
I
I. All
isotopes have the same atomic number but they have a different mass number (different
number of neutrons). Pharmaceutical scientists frequently use radioisotopes as a means to
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Arnold & Marie Schwartz College of Pharmacy and Health Sciences, Long Island University
PH931
Instructor: M. Savva, Ph.D.
follow the in vivo fate of (tagged) biologically active macromolecules and synthetic drug
compounds.
A mixture is a combination of two or more substances in which the substances may or
may not retain their physicochemical properties intact. Mixtures are classified as
homogeneous and heterogeneous mixtures.
In homogenous mixtures of solids and liquids, the chemical and physical properties of
the individual substances cannot be detected (intact) by any method of instrumental
analysis. Fig. 2 shows the melting point of solid crystalline aspirin centered around 135
˚C.
120
aspirin, crystals
100
aspirin, solution
80
60
40
20
0
20
0
17
0
15
5
13
2
6
4.
13
4
4.
13
13
0
12
80
0
40
% of crystals remaining
Melting-point curve of aspirin
Temperature
Fig. 2. Melting of aspirin crystals as determined by a scanning calorimeter that measures
the heat of fusion. No melting of aspirin can be detected in the aspirin solution since
the forces that hold the aspirin crystal have been destroyed by the solvent, during a
process called dissolution. Notice that the temperature scale is not linear.
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Arnold & Marie Schwartz College of Pharmacy and Health Sciences, Long Island University
PH931
Instructor: M. Savva, Ph.D.
Naturally in the aspirin solution the aspirin crystal is dissolved in water. The solvent
has destroyed the intermolecular forces that hold the aspirin molecules in a crystalline
arrangement, during the process of dissolution. Formation of a molecular dispersion
requires mutual interaction between solute and solvent. As a result, the properties of the
individual components of the mixture are altered. We are rather dealing with the unique
properties of homogenous mixtures that have resulted from the (spontaneous) mixing of
the individual substances. All the physical properties of aspirin are altered because of
interaction with the aqueous solvent. Similarly the properties of water are affected by the
presence of the solute. Absorption of electromagnetic radiation is another physical
property that is altered as a result of homogeneous mixing. Consider for example the
inhalation anesthetic halothane. The absorption of light in the visible and ultraviolet
region of halothane as pure liquid and as a solution in organic solvents is not the same.
It is important to note that contrary to the homogeneous mixtures of liquids and
solids, mixtures of gases are always homogeneous. The composition of a homogeneous
mixture is always the same throughout. Examples of gas, liquid and solid pharmaceutical
homogeneous mixtures, respectively, are: 1) nitrous oxide gas with oxygen at a ratio
80:20 by volume used for general anesthesia. 2) medicated simple syrup (85 % w/w), in
which sucrose is dissolved in water forming a molecular dispersion. 3) suppositories
composed of a mixture of PEG (polyethylene glycol) 8000 (40 %) and PEG 400 (60 %)
prepared by the melting method and allowed to congeal to the solid state at room
temperature.
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Arnold & Marie Schwartz College of Pharmacy and Health Sciences, Long Island University
PH931
Instructor: M. Savva, Ph.D.
Contrary to the above, a heterogeneous mixture is one in which the individual
components that make up the mixture retain their physicochemical properties intact. The
composition of a heterogeneous mixture may or may not be (statistically) uniform
throughout. The components of homogeneous and heterogeneous mixtures can be
separated and recovered as pure substances by means of physical methods. However, in
the case of homogenous mixtures one has to be very careful with the recovery of pure
solid substances. Consider for example the case of a simple syrup. Water can be removed
by boiling the solution and condensing the vapor to pure water with the aid of a
distillation apparatus, leaving behind the pure dry sugar powder. The compound is
successfully recovered in a pure, but not necessarily in the original, crystalline state.
Different crystalline states of a drug, called polymorphs, may present distinctly different
solubility, dissolution, bioavailability and pharmacological profiles.
A tablet prepared by direct compression of a drug, lactose, Actisol® and magnesium
stearate is an example of a heterogeneous solid mixture. Lactose grains remain separate
from the magnesium stearate and the drug grains. In order for the excipients to play their
role in the tablet, they have to retain their distinct identity along with their
physicochemical properties within the powder mixture. Actisol® is the disintegrant. Its
swelling properties facilitate tablet disintegration in aqueous media, a process that greatly
accelerates drug dissolution and absorption. Interaction of Actisol® with the drug or with
any of the other excipients would change or even neutralize the disintegration properties
of the excipient. Similarly, interaction of magnesium stearate (lubricant) with the other
excipients, could eliminate its lubricant properties. The tablet would stick to the punches
during compression resulting in a damaged or complete removal of the tablet surface; a
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Arnold & Marie Schwartz College of Pharmacy and Health Sciences, Long Island University
PH931
Instructor: M. Savva, Ph.D.
phenomenon known as “capping”. More importantly, active drug-excipient interaction
not previously anticipated by the pharmaceutical scientist could result in product
instability, inefficient therapy or toxicity. The presence of a basic excipient, like
carbonate salts commonly used in effervescent tablets, may cause hydrolysis of an ester
drug in the presence of moisture. Reduction of the quantity of the active drug in the
dosage form would result in lower blood concentration of the drug and inefficient
therapy. Similarly, interaction of the drug with excipients may result in a complex of
reduced solubility, thus, reduced absorption and inefficient therapy, again. A completely
different scenario arises when the crystalline state of a drug changes to less stable (higher
energy) crystalline state or to the least stable amorphous state, due to drug-excipient
interactions. The solubility of an amorphous solid is always higher that that of the
crystalline solid. Faster or increased solubility of the drug may result in increased levels
of drug in the blood, which in turn can cause toxicity.
Pharmaceutical suspensions are examples of heterogeneous liquid mixtures. They are
liquids in which the insoluble drug, present as fine particles, is somewhat uniformly
dispersed in aqueous media. Brownian motion due to the forces exerted by the water
molecules on the suspended drug molecules is primarily responsible for the suspension of
the particles. The larger the particles are, the more difficult it is to keep them uniformly
suspended in the water. Because the drug solubility is so small, the physicochemical
properties of drug and water in pharmaceutical suspensions remain practically intact.
Since drug and dispersion medium exist as two discrete phases, one cannot talk about
colligative properties of pharmaceutical suspensions.
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Arnold & Marie Schwartz College of Pharmacy and Health Sciences, Long Island University
PH931
Instructor: M. Savva, Ph.D.
As previously discussed, matter exists in three distinct physical states: solid, liquid
and gas.
Gas
Liquid
Solid
Molecules in a solid are held close together in an orderly fashion with very little
freedom of motion. Solids are characterized by: 1) shape 2) strong interatomic or
intermolecular interactions; high density 3) very little or no compressibility.
On the other hand, in a gas the component molecules are far apart. They are in
random rapid motion and they exert very small forces on each other. They are therefore
characterized, by: 1) no shape 2) weak or no intermolecular forces; low density 3) high
compressibility.
Liquids also do not have a shape. Their properties lie somewhere between those of
solids and gases. Intermolecular attractive forces in liquids are closer to those in solids
although they are significantly weaker. The molecules are close together but not as
rigidly as in solids and they can move past each other. Liquids are in general not
compressible.
Lastly, the physicochemical properties of matter are further classified into extensive
and intensive properties.
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Arnold & Marie Schwartz College of Pharmacy and Health Sciences, Long Island University
PH931
Instructor: M. Savva, Ph.D.
Extensive properties are additive; i.e. the value of an extensive property is
proportional to the quantity of the substance in the system. Mass and volume are
extensive properties. For example, mixing 25 grams Petrolatum Alba with 5 grams of 1
% hydrocortisone ointment will yield a total mass of 30 grams of hydrocortisone
hydrophilic ointment.
In contrast, the value of an intensive property such as temperature and density is not
dependent to the amount of a substance. For example, mixing 1 L of water 22 ºC with 1 L
of water 30 ºC will make a 2 L water of temperature somewhere in between 22 ºC and 30
ºC, but definitely not 52 ºC.
PROBLEMS
1. Define
a) matter b) substance c) element
2. Give one pharmaceutical example of:
a) gas homogeneous mixture
b) liquid homogeneous mixture
c) solid homogeneous mixture
d) solid heterogeneous mixture
e) liquid heterogeneous mixture
3. How is a chemical change different from a physical change?
4. What is the difference between a homogeneous and a heterogeneous pharmaceutical
mixture?
5. Explain the difference between intensive and extensive properties.
6. Classify the following as intensive and extensive properties.
a) Volume
b) temperature
d) mass
e) length
c) density
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Arnold & Marie Schwartz College of Pharmacy and Health Sciences, Long Island University
PH931
Instructor: M. Savva, Ph.D.
CHAPTER II
GASES
PROBLEMS
1. Define pressure in mathematical terms. Express it in mm Hg, atmospheres, torrs, psi
and pascals.
2. Write down the difference between a gas and a vapor.
3. What are the standard temperature and pressure conditions (STP conditions) in gases?
4. Where is atmospheric pressure greater, at Mt. Everest or at sea level?
5. Calculate the volume occupied by 2 moles of carbon dioxide at STP conditions.
(44.82 L)
6. Calculate the number of moles of a gas present in a container of 0.0432 m3 volume at
temperature and pressure 21 ºC and 15.4 atmospheres.
(27.6 moles)
7. A 40 L cylinder contains 30.5 g of Nitrogen gas (N2) at 21 ºC. What is the pressure
inside the cylinder expressed in psi units?
(9.66 psi)
8. Calculate the volume occupied by 60 g of oxygen gas (O2) at a temperature and
pressure of 25 ºC and 24 atm, respectively.
(1.91 L)
9. A 5.7 x 10-4 m3 aerosol can is filled with 0.0875 moles of compressed gas. Calculate
the pressure inside the aerosol at: a) 20 ºC and b) 60 ºC. What storage conditions would
you recommend?
(a) 3.69; b) 4.19)
10. Express the universal gas constant, R, in three different energy units.
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Arnold & Marie Schwartz College of Pharmacy and Health Sciences, Long Island University
PH931
Instructor: M. Savva, Ph.D.
11. The manufacturing of a parenteral drug solution involves dissolution of the drug at
temperature and pressure of 40 ºC and 1.2 atm, respectively. In an effort to prevent
oxidation reactions, the pharmaceutical scientist, is bubbling N2 (gas) from beneath. A
small bubble with a 0.100 mL volume rises from the bottom of the container to the
surface of the solution where the temperature is 35 ºC and the pressure is 744.8 torrs.
What is the volume of the bubble when it reaches the surface of the solution?
(0.12 mL)
12. A gas exerts a pressure of 2.3 atmospheres at 45 ºC. Calculate the pressure in torrs
when the volume of the gas is reduced to 5 % of its original value, at constant
temperature.
(34960 torrs)
13. Calculate the pressure of a gas inside a balloon that was inflated with 1.3 L
compressed gas at 15.4 atmospheres to a final volume of 20 L. Assume that the
temperature remains constant during the inflation.
(1atm)
14. 10 L of oxygen gas at 70 ºC is cooled down at constant pressure to a final volume of 8
L. Calculate the final temperature of the gas in centigrade.
(1 ºC)
15. How much N2 (gas) is required to fill up a 450 mL aerosol can to 26 psi at 21ºC.
What is the molar concentration of N2 (gas)?
(924 mg; 0.073 M)
16. Calculate the concentration of 30 g of carbon dioxide (CO2) at a 25°C temperature
and pressure of 750 torrs. Atomic weights: C=12; O=16.
(0.04 M)
17. Calculate the volume of carbon dioxide generated from compete decomposition of 1.5
moles of MgCO3 at 25 ºC and pressure equal to 1 atmosphere.
(36.7 L)
18. a) Calculate the molar mass of 118.8 g of an anesthetic gas that occupies 13.46 L at
temperature and pressure of 22°C and 1.08 atm, respectively.
b) Chemical analysis showed that the gas contained C=12.1%, F=28.8%, Cl=18.2%,
Br=40.4% and H=0.5%, by mass. Determine the chemical formula of the gas. Atomic
weights: C=12, F=19, Cl=35.5, Br=80, H=1.
(a) 198 g/mol; b) C2HBrClF3)
19. Calculate the molar mass of 6.7 g of N2 (gas) – O2 (gas) mixture that occupies a
volume of 5.6 L at 22°C temperature and pressure of 1 atm. Determine the ratio of N2 to
O2 in the gas mixture. Atomic weights: N=14, O=16.
(a) 28.9 g/mol; b) 77.5:22.5)
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Arnold & Marie Schwartz College of Pharmacy and Health Sciences, Long Island University
PH931
Instructor: M. Savva, Ph.D.
20. Calculate the density of 0.5 moles oxygen gas at 25°C and 783 torrs.
(1.3 g/L)
21. What is the density of Xe at 2368 kPa pressure and temperature a) 22 °C b) 50 °C.
Atomic weights: Xe=131.
(a) 126.5 g/L; b) 115.5 g/L)
22. Calculate the root-mean-square speed of oxygen molecules at 25°C. Compare it with
that of hydrogen molecules. Atomic weights: O=16, H=1.
(√ū2(O2) = 482 m/s; √ū2(H2) = 1928 m/s)
23. Which of the following is true with regard to the kinetic energy and speed of gas
molecules?
a) the kinetic energy of all the molecules of 1 mole of oxygen is the same
b) the average kinetic energy of gas molecules increases when the volume of the gas
increases.
c) the average kinetic energy of one mole of N2 and one mole of O2 under same
temperature and pressure conditions is the same.
d) the average kinetic energy of a gas increases with increasing temperature
e) b, c and d are true
24. Calculate the ratio of the effusion rate of N2 (gas) to H2 (gas) through small holes of
10 nm diameter at 67 °C.
(1:3.74)
25. A mixture of 0.2 moles of oxygen and 0.8 moles of nitrogen is placed in a 0.043 m3
container.
a) Calculate the partial pressure of each gas and the total pressure, at 0°C.
b) The total pressure at 120°C.
(a) Pox = 0.10 atm, PNit = 0.42 atm, Ptotal = 0.52 atm; b) Ptotal = 0.75 atm)
26. A 900 mL chamber contains 2.184 g of N2 and 1.45 g of O2 at 22 °C. Calculate:
a) the mole fractions of each gas
b) the total gas pressure exerted by the mixture
c) the partial pressures of each gas
(a) XNit = 0.634, Xox= 0.366; b) Ptotal = 3.31 atm c) Pox = 1.21 atm, PNit = 2.1 atm)
27. Calculate the pressure in atmospheres of 60 moles of oxygen gas as supplied in
0.0432 m3 cylinders at 22°C using:
a) the ideal gas equation
b) the van der Waals equation
a = 1.360 L2 atm/mol2; b = 0.03183 L/mol
(a) P = 33.62; b) P = 32.56)
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Arnold & Marie Schwartz College of Pharmacy and Health Sciences, Long Island University
PH931
Instructor: M. Savva, Ph.D.
CHAPTER III
THERMOCHEMISTRY
PROBLEMS
1. Write down the two fundamental laws that chemical reactions obey.
All chemical reactions obey the law of conservation of energy and the law of
conservation of mass.
2. What is energy? Write down two different units of energy.
Energy is defined as the capacity to do work. Units of energy are Joules and Calories.
3. Give an example of conversion of a) Solar energy to thermal energy and b) thermal
energy to chemical energy c) chemical energy to kinetic energy.
a) solar energy can be converted to thermal energy on our skin when we get exposed
to the sun.
b) heating up a reaction will result in formation of products
c) when we work out, chemical energy stored as fat is converted to kinetic energy
4. Explain the difference between thermal energy and heat.
Heat is the transfer of thermal energy between two bodies that are at different
temperature. We can thus talk of ‘heat flow’.
5. What is the objective of thermochemistry?
Thermochemistry studies heat changes in chemical reactions.
6. Define a) the system b) the surroundings and c) the universe, in a titration experiment
of a base with an inorganic acid in aqueous media.
a) the base, the acid and the reaction participating water molecules is the system
b) the (bulk) water that does not participate in the neutralization reaction is the
surroundings
c) the system (reactants) and the surroundings comprise the universe (of the reacting
system)
7. Describe the three general types of systems.
a) open system: mass and energy can be exchanged usually in the form of heat with
the surroundings.
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Arnold & Marie Schwartz College of Pharmacy and Health Sciences, Long Island University
PH931
Instructor: M. Savva, Ph.D.
b) closed system: energy can be exchanged with the surroundings but not mass.
c) isolated system: no transfer of energy or mass.
8. Explain what is an endothermic and what is an exothermic process.
During an endothermic process there is a net gain of heat from the surroundings to the
system.
During an exothermic process the system gives off heat
9. Write down an expression of Enthalpy. Describe its uses and factors that affect its
magnitude.
H = E + PV
Enthalpy can be used to quantify heat flow for physical or chemical processes that
take place under conditions of constant pressure. It is an extensive property and its
magnitude depends on the amount of the substance present.
10. a) Write an expression for the enthalpy of a reaction in terms of the enthalpies of
reactants and products.
b) The equation was derived based on what property of the enthalpy?
c) Under what conditions is the heat of reaction equal to the enthalpy change of the
same reaction?
d) Based on c), what is the sign of ∆H for and endothermic and what is the sign for an
exothermic process, respectively.
a) ∆H = H (products) – H (reactants)
Although the absolute values of enthalpies cannot be measured, we can express the
enthalpies of products and reactants through the Standard Enthalpy of Formation,
∆Hf°, which is defined as the heat change that results when 1 mole of a compound is
formed from its elements at a pressure of 1 atmosphere (Although standard state in
solids, liquids and real gases does not specify temperature, a temperature of 25 °C is
assumed). By convention the ∆Hf° of any element in its most stable form is zero.
Once we know the values of ∆Hf°, we can calculate the standard enthalpy of reaction,
∆H°rxn, defined as the enthalpy of reaction carried out at 1 atmosphere.
b) The above equation is true because enthalpy is a state function. In other words, the
magnitude of the enthalpy change (∆H) depends only on the initial and final states
of the system and not on how the change is accomplished.
c) Under conditions of constant pressure.
d) Since ∆H is the heat of reaction, for an endothermic reaction it is positive whereas
for an exothermic reaction it is negative.
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Arnold & Marie Schwartz College of Pharmacy and Health Sciences, Long Island University
PH931
Instructor: M. Savva, Ph.D.
11. Methanol can be converted to water and carbon dioxide according to the following
reaction:
2 CH3OH (l) + 3 O2 (g)
4 H20 (l) + 2 CO2 (g)
∆H = -1453 kJ
Atomic weights: C=12; H=1; O=16. Standard Enthalpies of Formation, ∆Hf° (CH3OH
(l)) = -239 kJ/mol; ∆Hf° (H2O (l)) = -286 kJ/mol; ∆Hf° (CO2 (g)) = -393.5 kJ/mol; ∆Hf°
(H2O (v)) = -242 kJ/mol.
a) Calculate the heat evolved when 50 g of CH3OH (l) is converted to CO2 according to
the reaction above.
b) Use the ∆Hf° of reactants and products to verify that the ∆H°rxn is equal to -1453 kJ.
c) What is the value of ∆H if the equation is multiplied throughout by 2?
d) What is the value of ∆H if the direction of the reaction is reversed?
e) calculate the ∆H if water vapor is formed.
f) Write down guidelines that can be useful for the correct interpretation of
thermochemical equations.
a) –1135.1 kJ
c) – 2906 kJ
d) 1453 kJ
e) –1277 kJ
f) 1. The stoichiometric coefficients always refer to the number of moles of a
substance
2. When we reverse an equation we change the roles of reactants and products
accordingly.
3. If we multiply both sides of a thermochemical equation by a factor n, then ∆H
must also be multiplied by n.
4. When writing thermochemical equations we must always specify the physical
states of all reactants and products.
12. Calculate the ∆H° for the following reactions:
a) 2 NH3(g) + 3 O2(g) + 2 CH4(g)
b) Ca3(PO4)2(s) + 3 H2SO4(l)
2 HCN(g) + 6 H2O(g)
3 CaSO4(s) + 2 H3PO4(l)
c) NH3(g) + HCl(g)
NH4CL(s)
d) MgO(s) + H2O(l)
Mg(OH)2(s)
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Arnold & Marie Schwartz College of Pharmacy and Health Sciences, Long Island University
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Instructor: M. Savva, Ph.D.
Standard Enthalpies of Formation: ∆Hf°(NH3(g)) = - 46 kJ/mol; ∆Hf°(O2(g)) = 0;
∆Hf°(CH4(g)) = - 75 kJ/mol; ∆Hf°(HCN(g)) = 135 kJ/mol; ∆Hf°(H2O (g)) = -242
kJ/mol; ∆Hf°(Ca3(PO4)2(s)) = -4126 kJ/mol; ∆Hf°(CaSO4(s)) = - 1433
kJ/mol; ∆Hf°(H3PO4(l)) = -1267 kJ/mol; ∆Hf°(H2SO4(l))= - 814 kJ/mol; ∆Hf°(HCl(g))
= - 92 kJ/mol; ∆Hf°( NH4CL(s))= - 314 kJ/mol; ∆Hf°( MgO (s)) = -602 kJ/mol;
∆Hf°(H2O (l)) = -286 kJ/mol; ∆Hf°( Mg(OH)2 (s)) = - 925 kJ/mol.
a) – 940 kJ
b) – 265 kJ
c) – 176 kJ
d) – 37 kJ
13. a) State the Hess’s law and explain its usefulness in thermochemistry.
The Hess’s law states that the change in enthalpy, ∆H, of a chemical reaction that
involves product formation from reactants is the same whether the reaction takes place
in one step or in a series of steps.
The usefulness of Hess’s law lies in the fact that it allows us to calculate enthalpies or
heats of reactions for compounds that cannot be directly synthesized from their
elements.
14. Calculate the ∆Hrxn for the reaction below,
FeO(s) + CO(g)
Fe(s) + CO2(g)
Given the following data:
Fe2O3(s) + 3 CO(g)
2 Fe(s) + 3 CO2(g)
∆H°= -23 kJ
3 Fe2O3(s) + CO(g)
2 Fe3O4(s) + CO2(g)
∆H°= -39 kJ
Fe3O4(s) + CO(g)
3 FeO(s) + CO2(g)
∆H°= +18 kJ
(∆H°rxn = -11 kJ)
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Arnold & Marie Schwartz College of Pharmacy and Health Sciences, Long Island University
PH931
Instructor: M. Savva, Ph.D.
15. Calculate the ∆H°rxn for the reaction below
2 C (gr) + H2(g)
C2H2(g)
Given the following data:
C (gr) + O2(g)
∆H°= -393.5 kJ
CO2(g)
H2(g) + ½ O2(g)
H2O(l)
∆H°= -286 kJ
2 C2H2(g)+ 5 O2(g)
4 CO2(g) + 2 H2O(l)
∆H°= - 2599 kJ
(∆H°rxn = +226.5 kJ)
16. Give the definitions of
a) heat capacity (C)
b) specific heat capacity (s)
a) heat capacity, C of a substance is defined as the amount of heat required raising the
temperature of a given quantity of the substance by one degree Celsius. The units of
heat capacity is J/°C.
b) specific heat capacity, s, of a substance is defined as the amount of heat required to
raise the temperature of 1 gram of the substance by one degree Celsius. The specific
heat capacity is measured in J/g °C.
17. Give a mathematical expression for the heat capacity of a substance.
Heat absorbed
C=
q
=
Increase in temperature
∆t
18. What is the sign of q for an endothermic and exothermic reaction?
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Arnold & Marie Schwartz College of Pharmacy and Health Sciences, Long Island University
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Instructor: M. Savva, Ph.D.
The sign convention for q is the same as for enthalpy changes. In other words, q is
positive and q is negative for endothermic and for exothermic reactions, respectively.
19. Write a mathematical expression that relates the specific heat capacity with heat
capacity.
C=ms
20. A 0.5 L of water is heated from 10°C to 25°C. Calculate the amount of heat absorbed
by the water. s(H2O) = 4.184 J/g °C; d(H2O) = 1.0 g/cm3.
(31.38 kJ)
21. What is the specific heat capacity of silver if the heat capacity of 362g of silver has a
heat capacity of 85.7 J/°C.
(0.237 J/g °C)
22. A 1.5 g of copper metal is heated from 222.5 °C to 230 °C. Calculate the heat
absorbed by the metal. s(Cu) = 0.385 J/g °C.
(4.33 J)
23. Calculate the amount of heat liberated from 366 g of mercury when it cools from 77
°C to 25 °C. s(Hg) = 0.139 J/g °C.
(-2.645 kJ)
24. Consider two metals A and B, each having a mass of 100 g and an initial temperature
of 20 °C. The specific heat capacity of A is larger than that of B. Under the same
heating conditions, which metal would take longer to reach a temperature of 21 °C?
(metal A)
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25. How are enthalpy changes determined experimentally?
Enthalpy changes can experimentally be determined with a device known as constantpressure calorimeter.
26. Define:
a) enthalpy of solution
b) enthalpy of dilution
a) Enthalpy of solution or Heat of solution, ∆Hsln is the heat generated or absorbed when
a certain amount of solute dissolves in a certain amount of solvent under conditions
of constant pressure.
b) Enthalpy of dilution or Heat of dilution, ∆Hdln is the heat generated or absorbed when
a solution is diluted with more solvent under conditions of constant pressure.
27. 100 mL of 0.5 M HCl is mixed with 100 mL of 0.5 M NaOH in a constant-pressure
calorimeter that has a heat capacity of 0.23 kJ/°C, as shown below:
NaOH (aq) + HCl (aq)
NaCl (aq) + H2O (l)
Calculate the heat change of neutralization if the temperature after the reaction
increases by 1.5 °C. s(solution) = 4.184 J/g °C; d(solution) = 1.0 g/cm3. (NOTE: the
heat of neutralization is usually given per mole of reactants).
(-32 kJ/mol)
28. 400 mL of 0.6 M HCl is mixed with 400 mL of 0.6 M NaOH in a constant-pressure
calorimeter that has a heat capacity of 0.387 kJ/°C, as shown below:
NaOH (aq) + HCl (aq)
NaCl (aq) + H2O (l)
Calculate the final temperature of the solution after the reaction if the initial
temperature of both solutions is 18.88 °C. The enthalpy of neutralization is –56.2
kJ/mol. s(solution) = 4.184 J/g °C; d(solution) = 1.0 g/cm3.
(22.5 °C)
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Arnold & Marie Schwartz College of Pharmacy and Health Sciences, Long Island University
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Instructor: M. Savva, Ph.D.
29. 400 mL of 0.6 M HCl is mixed with 400 mL of 0.3 M Ba(OH)2 in a constant-pressure
calorimeter that has a heat capacity of 0.387 kJ/°C, as shown below:
Ba(OH)2 (aq) + 2 HCl (aq)
Ba(Cl)2 (aq) + 2 H2O (l)
Calculate the final temperature of the solution after the reaction if the initial
temperature of both solutions is 18.88 °C. The enthalpy of neutralization is –56.2
kJ/mol. s(solution) = 4.184 J/g °C; d(solution) = 1.0 g/cm3.
(22.5 °C)
30. 150 mL of 0.15 M AgNO3 is mixed with 150 mL of 0.15 M HCl in a constantpressure calorimeter that has a heat capacity of 0.23 kJ/°C, as shown below:
AgNO3 (aq) + HCl (aq)
AgCl (s) + HNO3 (l)
Calculate the heat of neutralization if the temperature after the reaction increases by
0.75 °C. s(solution) = 4.184 J/g °C; d(solution) = 1.0 g/cm3.
(-49.51 kJ/mol)
31. 150 mL of 0.15 M AgNO3 is mixed with 150 mL of 0.15 M HCl in a constantpressure calorimeter, as shown below:
AgNO3 (aq) + HCl (aq)
AgCl (s) + HNO3 (l)
Calculate the heat of neutralization if the temperature after the reaction increases by
0.75 °C. s(solution) = 4.184 J/g °C; d(solution) = 1.0 g/cm3. Assume that there is no
heat loss to the calorimeter.
(-41.84 kJ/mol)
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Arnold & Marie Schwartz College of Pharmacy and Health Sciences, Long Island University
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Instructor: M. Savva, Ph.D.
32. 0.5 L of 0.5 N HNO3 is mixed with 0.5 L of 0.5 N Mg(OH)2 in a constant-pressure
calorimeter that has a heat capacity of 350 J/°C. Calculate the final temperature of
the solution if the initial temperature of both solutions was 20 °C. The enthalpy of
neutralization is –56 kJ/mol.
(23.1°C)
33. Calcium Chloride is dissolved in a 100 g of water at 20.5 °C to a final concentration
of 1.1 % by weight. Enthalpy of dissolution was calculated in a constant-pressure
calorimeter (Ccal = 345 J/°C) to be – 78 kJ/mol. Calculate the final temperature of the
reaction. Atomic weights: Ca = 40; Cl = 35.5.
(21.5°C)
34. 100 mL of 0.5 M HCl is mixed with 100 mL of 0.25 M Ba(OH)2 in a constantpressure calorimeter that has a heat capacity of 235 J/°C, as shown below:
Ba(OH)2 (aq) + 2 HCl (aq)
BaCl2 (aq) + 2 H2O (l)
Calculate the final temperature of the solution if the heat change of neutralization is –
55 kJ/mol and the initial temperature of the reaction is 20 °C. s(solution) = 4.184 J/g
°C; d(solution) = 1.0 g/cm3.
(22.6°C)
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Arnold & Marie Schwartz College of Pharmacy and Health Sciences, Long Island University
PH931
Instructor: M. Savva, Ph.D.
CHAPTER IV
INTRODUCTION TO THERMODYNAMICS
THE 1ST LAW OF THEMODYNAMICS
PROBLEMS
1. What is thermodynamics?
Thermodynamics is the scientific discipline that deals with the study of laws that
govern the conversion of energy from one form to another, the direction of the flow
of heat and the availability of energy to do work.
2) How is the state of a system defined?
The state of a system is defined by the values of all relevant macroscopic properties,
for example composition, energy, pressure, volume and temperature. These properties
are said to be state functions.
3) What is a state function?
A state function is a property of the present state of the system, which is independent
of the way the state was prepared. The change of any state function depends only on
the value of the function in the initial and final states and not on how the change is
accomplished.
4) State the 1st law of thermodynamics.
The 1st law of thermodynamics, which is based on the law of conservation of energy,
states that energy can be converted from one form to another, but cannot be created or
destroyed.
5) Calculate the work done (in joules) by a gas that expands in volume from 1.5 L to 3.2
L against (1 L atm = 101.3 J):
a) an external pressure of 1.12 atm
b) a vacuum
a) - 192.9 J
b) 0
6) Is work a state function?
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Instructor: M. Savva, Ph.D.
No. Although the initial and final states in a) and b) are the same the amount of work
done is different because the external pressures are different. In general, the work
done depends not only on the initial and final states, but also on how the process is
carried out.
7) Is heat a state function?
No. In summary, heat (q) and work (w) are not state functions because they are
not properties of the system. They manifest themselves only during a process and
therefore, their values depend on the path of the process.
8) Is energy a state function?
Yes.
∆E = Ef - Ei
9) Calculate the work done (in joules) by a gas that expands in volume from 500 cm3 to
0.0044 m3 against:
a) an external pressure of 790 torrs
b) a vacuum
a) – 410.9 J
b) 0
10) The work when nitrogen gas is compressed in a cylinder is 340 J. Calculate the
energy change for the process if during the compression 111 J of heat is released to the
surroundings. Explain your answer.
(+ 229 J)
11) The work on the surroundings during an expansion of a gas equals 1.32 kJ. The
process of the gas expansion is accompanied by 1.12 kJ heat transfer from the
surroundings to the system. What is the change in energy of the system? Explain your
answer.
(-200 J)
12) A gas expands and does work on the surroundings equal to 0.234 kJ. During the
process of expansion 0.217 kJ of heat is absorbed by the system. Calculate the energy
change of the system after the expansion. Explain your answer.
(-17 J)
13) To compress oxygen, 567 J of work is done by the surroundings. Calculate the energy
changes of the system if during the compression 245 J of heat is given off to the
surroundings. Explain your answer.
(+332 J)
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Arnold & Marie Schwartz College of Pharmacy and Health Sciences, Long Island University
PH931
Instructor: M. Savva, Ph.D.
14) The heat of the reaction of sodium metal with water as shown below, was measured
in a constant-pressure calorimeter (P = 1 atm) and found to be equal to –378 kJ.
2 Na (s) + 2 H2O (l)
2 NaOH (aq) + H2 (g)
∆H = - 378 kJ
In order for the H2 (g) to enter to the atmosphere some of the energy produced by the
reaction is used to do the work of pushing back a volume of air (∆V) against atmospheric
pressure. Calculate the energy change of the system at the end of the reaction. Assume a
constant temperature of 25 ºC and neglect the small changes in the volume of the
solution. Explain your answer.
(-380.48 kJ; The ∆H value is a little smaller (absolute value) than ∆E because some of the
energy released is used to do gas expansion work, so less heat is evolved.)
15) Calculate the change in internal energy, ∆Eº, of the following reaction under normal
temperature and pressure conditions (25 ºC, 1 atm): Explain your answer.
2 CO (g) + O2 (g)
2 CO2 (g)
∆Hº = - 566 kJ
(-563.52 kJ)
16) What is the change in internal energy, ∆Eº, of the following reactions under normal
temperature and pressure conditions (25 ºC, 1 atm):
a)
NO (g) + O (g)
NO 2 (g)
∆Hº = - 233 kJ
b)
N2 (g) + 2 O2 (g)
2 NO2 (g)
∆Hº = + 67.7 kJ
c)
C (gr) + 1/2 O2 (g)
CO (g)
∆Hº = -111.74 kJ
d)
C (gr) + O2 (g)
CO2 (g)
∆Hº = -394 kJ
(a) –230.52; b) +70.18 kJ; c)-111.74 kJ; d) –394 kJ)
17) Calculate the ∆Eº if 2 moles of nitrogen gas react with 6 moles of hydrogen gas at
normal temperature and pressure conditions:
a)
N2 (g) + 3 H2 (g)
2 NH3 (g)
∆Hºrxn = - 93 kJ
(-176.09 kJ)
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Arnold & Marie Schwartz College of Pharmacy and Health Sciences, Long Island University
PH931
Instructor: M. Savva, Ph.D.
18) Calculate the ∆Eº if 1/2 moles of Cl2 (g) react with 1/2 moles of H2 (g) at normal
temperature and pressure conditions:
a)
H2 (g) + Cl2 (g)
2 HCl (g)
∆Hºrxn = - 185 kJ
(-92.5 kJ)
19) Calculate the energy change of the system at the end of the hydroquinone oxidation
reaction at atmospheric pressures (760 torrs) as shown below. Assume a constant
temperature of 25 ºC and neglect the small changes in the volume of the solution
C6H4(OH)2 (aq)
C6H4O2 (aq) + H2 (g)
∆Hºrxn = +177 kJ
(+174.52 kJ)
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Arnold & Marie Schwartz College of Pharmacy and Health Sciences, Long Island University
PH931
Instructor: M. Savva, Ph.D.
CHAPTER V
INTERMOLECULAR FORCES IN LIQUIDS AND SOLIDS
Table 1:
State of
matter
Summary of properties of the states of matter
Description of properties
Gases
Negligible intermolecular (attractive) forces. As a result gases are
characterized by:
1. random (free) motion
2. large interparticle distances
3. large empty spaces
4. low density
5. high compressibility
6. no shape
7. no definite volume
Solids
Very strong intermolecular forces. As a result solids are characterized
by:
1. no translational motion. There is only vibration of the
consisting molecules about fixed (equilibrium) positions.
2. small interparticle distances
3. small empty spaces
4. high density
5. essentially no compressibility
6. definite shape
7. definite volume
Liquids
Intermolecular forces in liquids are not as strong as in solids but much
stronger than those in gases. As a result liquids are characterized by:
1. ‘cohesive’ translational motions, i.e. liquids can move as a whole.
The molecules can slide past one another but they cannot break
away from the intermolecular attractive forces while remaining
in the liquid state.
2. small empty spaces
solids < liquids <<< gases
3. high density
solids < liquids <<< gases
4. slightly compressible
solids < liquids <<< gases
5. no shape
6. definite volume
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Arnold & Marie Schwartz College of Pharmacy and Health Sciences, Long Island University
PH931
Instructor: M. Savva, Ph.D.
PROBLEMS
1. Give a description of intermolecular versus intramolecular forces.
As we’ve discussed previously, the attractive and repulsive forces between the
molecules of a substance are responsible for the physical properties of the substance,
e.g. melting point and boiling point. On the other hand, the intramolecular forces are
dealing with the organization of atoms within a molecule and are related to the
(physico)chemical properties of the substance. In general intramolecular forces are
much stronger than intermolecular forces.
The magnitude of the intermolecular forces of a substance is reflected through the
boiling point and melting point of the substance. Compare for example the m.p. of
water and chloroform. Since the m.p. of water (0 ºC) is higher than the m.p. of
chloroform (-64 ºC), the intermolecular forces in ice water are stronger than in
chloroform. Similarly, the intermolecular forces of liquid water (b.p. = 100 ºC) are
much stronger than the intermolecular forces in liquid ether (b.p. = 35 ºC). In other
words, more energy is required to overcome the attractive intermolecular forces in
liquid water for the water molecules to enter the vapor phase (vaporization).
2. Describe the various types of intermolecular forces.
Intermolecular forces are mainly the dipole-dipole (Keesom), dipole-induced dipole
(Debye) and dispersion forces (London), commonly refer to as van der Waals forces
and the ion-dipole forces.
a) dipole-dipole forces:
They are attractive forces between molecules that possess permanent
dipole moments. The larger the dipole moment the greater the attractive force. In
polar solids the molecules that have a permanent dipole moment tend to align
with opposite polarities. Such solids are characterized by a long-range order, that
is, the molecules are arranged in regular configurations. Solids that possess rigid
and long-range order are called crystalline solids. Solids that lack a well-defined
arrangement and a long-range molecular order, are called amorphous solids.
The hydrogen bond is a special type of dipole-dipole interaction between
the hydrogen atom in a polar bond such as N-H, O-H or F-H, and an
electronegative atom e.g. O, N, F. Because the average energy of a H-bond is
quite large for a dipole-dipole interaction, hydrogen bonds have quite a powerful
effect on the structure and properties of many compounds. The strength of
hydrogen bonding is determined by the coulombic interaction between the lone
pair of electrons of the electronegative atom and the hydrogen nucleus. The
magnitude of the intermolecular attractions due to hydrogen bonding is dependent
by: 1) the strength of hydrogen bonds, as stated above and 2) the number of
hydrogen bonds per molecule.
b) ion-induced dipole interactions:
This kind of interaction involves induction of a dipole to an apolar
molecule by an ion. Briefly, the electron distribution of an apolar molecule is
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Arnold & Marie Schwartz College of Pharmacy and Health Sciences, Long Island University
PH931
Instructor: M. Savva, Ph.D.
distorted by the close proximity of an ion, resulting in a dipole formation and an
effective separation of positive and negative charges in the non-polar molecule.
An attractive interaction is developed between the ion and the dipole.
c) dipole-induced dipole (London dispersion forces):
The close approach of a polarizable molecule that possess a dipole
moment, induces a dipole formation to another polarizable molecule. The
probability of a dipole moment being induced depends:
1) on the charge of the ions
2) on the strength of the dipole
3) on the polarizability of the atom or molecule
Polarizability is defined as the ease with which the electron distribution
can be distorted. In general, the larger the number of electrons and the more
diffuse the electron cloud in the molecule, the greater its polarizability. It is
important to note that London dispersion forces exist among both polar and apolar
molecules. This is because the origin of the London dispersion forces lies in the
distribution o electrons around the nucleus. As the electrons move about the
nucleus, a momentary asymmetrical electron distribution can develop that
produces a temporary dipolar arrangement of charge. The formation of this
instantaneous dipole can, in turn, affect the electron distribution of a neighboring
atom and so on. Dispersion forces are weak but additive in nature. In other words,
they increase in magnitude with molar mass because larger MW molecules tend to
have more electrons and are more polarizable.
d) ion-dipole forces:
They usually hold together an ion (anion or cation) with a polar molecule
that possess permanent dipole moment. The strength of the ion-dipole forces
depends on the size and charge of the ion and on the magnitude and size of the
molecule. Cations are usually smaller than ions and thus their charges are more
concentrated. An example of an ion-dipole interaction is the hydration of NaCl. In
a sodium chloride solution, the Na+ and Cl- ions are surrounded by water
molecules, which have a dipole moment of 1.87 D. Water is a good solvent for
NaCl because it keeps the Na+ and Cl- ions apart through ion-dipole interactions
with them. In contrast, hexanes (C6H14) is a poor solvent for salts because it lacks
the ability to participate in ion-dipole interactions.
3. What type of intermolecular forces exist between the following pairs?
a) HBr and H2S
b) Br2 and CCl4
c) I2 and NO3d) NH3 and C6H6
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Arnold & Marie Schwartz College of Pharmacy and Health Sciences, Long Island University
PH931
Instructor: M. Savva, Ph.D.
4. Which of the following species can form hydrogen bonds with water?
a) CH3OCH3
b) CH4
c) Fd) CH3COOH
e) Na+
5. Normally the boiling point of a series of similar compounds containing elements of the
same periodic group increases with increasing MW (WHY?).
a) Rationalize the higher b.p. of hydrogen iodide compared to the other hydrogen
halides as given HI > HBr > HCl.
b) Explain why the b.p. of HI but it is much lower than the b.p. of HF.
6. Why is so hard to compress liquids and solids?
The empty space in liquids and solids is quite small. In addition to that, we need to
remember that all molecules exert repulsive forces too, to one another. As two
molecules approach one another, repulsion between the electrons and their nuclei
come into play. In solids and liquids, the molecules are already close to each other.
Any attempt to bring them closer together, beyond a certain equilibrium distance is
energetically unfavorable.
7. Describe: a) crystalline solids and b) amorphous solids.
a) Crystalline solids possess long-range order; its atoms , molecules, or ions occupy
specific positions. The arrangement of such particles in a crystalline solid is such
that maximizes their attractive intermolecular forces. The forces responsible for
the stability of a crystal can be ionic, covalent, van der waals, hydrogen bonds or
a combination of these forces.
b) Amorphous solids lack a well-defined arrangement and long-range molecular
order. They are usually formed under conditions where the molecules do not have
time to align themselves and become locked in positions other than those of a
c) regular crystal.
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Arnold & Marie Schwartz College of Pharmacy and Health Sciences, Long Island University
PH931
Instructor: M. Savva, Ph.D.
8. Explain the differences in boiling points for each of the following pairs:
isobutane
Structure
CH3-CH(CH3)-CH3
Boiling point
-12 ºC
density
0.604
n-butane
CH3-CH2-CH2-CH3
0 ºC
0.622
b) Hydrogen fluoride
HF
20 ºC
Hydrogen chloride
c) n-butane
HCl
CH3-CH2-CH2-CH3
-85 ºC
0 ºC
n-propane
d) Hydrogen chloride
CH3-CH2-CH3
HCl
-42 ºC
-85 ºC
Lithium chloride
e) ethanol
LiCl
CH3-CH2-OH
1360 ºC
79 ºC
acetic acid
f) n-propane
CH3-COOH
CH3-CH2-CH3
118 ºC
-42 ºC
dimethyl ether
g) water
CH3-O-CH3
H2O
-25 ºC
100 ºC
Hydrogen fluoride
e) ethanol
HF
CH3-CH2-OH
20 ºC
79 ºC
CH3-O-CH3
-25 ºC
a)
dimethyl ether
9. What is a phase?
Phase is a homogeneous, physically distinct and mechanically separable part of a
system. If a system is composed of a single phase, then the phase represents the state of
the system. When a system is composed of many phases, then each phase is in contact
with the other phases but separated from them by a well-defined physical boundary.
10. Why phase changes take place with addition or removal of energy?
11. Define:
a) evaporation
b) condensation
c) vapor pressure
a) the process in which a liquid is transformed into a gas.
b) the process in which a gas is converted into a liquid.
c) The pressure at which liquid (solid) and vapor can coexist in
equilibrium at a given temperature.
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Arnold & Marie Schwartz College of Pharmacy and Health Sciences, Long Island University
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Instructor: M. Savva, Ph.D.
12. What is the difference between a vapor and a gas?
Vapor is the gaseous form of a substance that is a liquid or a solid under conditions of
normal temperature and pressure.
13. Give a definition for the molar heat of evaporation, ∆Hvap
The molar heat of evaporation, ∆Hvap, is defined as the energy required to vaporize 1
mole of liquid.
14. What is the relationship between molar heat of evaporation and vapor pressure in a
liquid.
The molar heat of vaporization is a measure of the strength of intermolecular forces
in a liquid. The higher the ∆Hvap of a liquid the stronger the intermolecular forces and
the lower is the vapor pressure (at normal temperatures) of the liquid.
15. Use the Clausius-Clapeyron equation to compute the vapor pressure of water at 55 ºC.
The vapor pressure of water at 25 ºC is 24 mm Hg and the ∆Hvap is 40.7 kJ/mol.
(107.8 mm Hg)
16. The vapor pressure of ethanol at 35 ºC is 100 torrs. Calculate the vapor pressure of
ethanol at 52.9 ºC if the ∆Hvap for ethanol is 38.6 kJ/mol.
(229 torr)
17. The vapor pressure of ether is 400 torrs at 18 ºC. What is the vapor pressure of diethyl
ether at 24.5 ºC. The ∆Hvap for ether is 26.0 kJ/mol at 25 ºC.
(505.9 torr)
18. The vapor pressure of ethanol is 23.6 torr at 10 ºC and 78.8 torr at 30 ºC. Calculate
the heat of vaporization of ethanol in cal/mol within the temperature range?
(43 kJ/mol)
19. Give a definition for the boiling point. Based on the definition you gave, where do
you think it will take longer to cook an egg, in high mountains or at sea level?
Boiling point is defined as the temperature at which the vapor pressure of a liquid is
equal to the external pressure. At this temperature the liquid begins to boil.
20. How is the boiling point and vapor pressure in a liquid related? Explain your answer.
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Arnold & Marie Schwartz College of Pharmacy and Health Sciences, Long Island University
PH931
Instructor: M. Savva, Ph.D.
21. Define critical temperature, Tc and critical pressure, Pc.
Critical temperature, Tc is defined as the temperature above which the gas cannot be
made to liquefy no matter how much pressure is exerted on the gas. Critical pressure,
Pc, is the minimum pressure that must be applied to liquefy the gas at its critical
temperature. Together the critical temperature and the critical pressure define the
critical point.
22. Can the Critical temperature, Tc be used as an indication of the intermolecular forces
in a liquid?
23. Define melting point.
The melting point of a solid or the freezing point of a liquid is the temperature at
which solid and liquid phases coexist at equilibrium.
24. What is the molar heat of evaporation, ∆Hfus.
Molar heat of evaporation, ∆Hfus is the energy required to melt 1 mole of a solid.
Note that, for the same substance ∆Hvap > ∆Hfus.
25. What is a supercooled liquid?
A supercooled liquid is a liquid that is cooled below its freezing point so rapidly that
the molecules didn’t have the time to assume the ordered structure of a solid and
thus, it continues to exist as liquid below its freezing point.
26. What is sublimation? Name a substance that is known to sublime at room temperature
and is commonly used as disinfectant.
Sublimation is defined as the process in which molecules go directly from the solid
into the vapor phase. The reverse process is called deposition.
27. What is the molar heat of sublimation, ∆Hsub.
Molar heat of sublimation, ∆Hsub is the energy required to sublime 1 mole of a solid.
Note that, for the same substance ∆Hsub > ∆Hvap > ∆Hfus. In fact,
∆Hsub = ∆Hfus + ∆Hvap
under the condition that all the phase changes occur at the same temperature. The
equation above is in accord with the Hess’s law, which states that the change in the
enthalpy or heat is the same, whether the reaction takes place in one or multiple steps.
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Arnold & Marie Schwartz College of Pharmacy and Health Sciences, Long Island University
PH931
Instructor: M. Savva, Ph.D.
28. How much energy is required to convert 270 grams of water at 7 ºC to steam at 167
ºC? s (H20 (l))= 4.18 J/g ºC; s ((H20 (v))= 2.0 J/g ºC; ∆Hvap = 40.8 kJ/mol.
(753.14 kJ)
29. Compute the heat released when 234 g of water steam at 174 ºC is converted to liquid
water at 25 ºC. s (H20 (l))= 4.18 J/g ºC; s ((H20 (v))= 2.0 J/g ºC; ∆Hvap = 40.8 kJ/mol.
(638.4 kJ)
30. How much energy is required to convert 0.324 kg of water at -7 ºC to steam at 167
ºC? s (H20 (l))= 4.18 J/g ºC; s (H20 (ice)) = s ((H20 (v))= 2.0 J/g ºC; ∆Hfus = 6.02
kJ/mol; ∆Hvap = 40.8 kJ/mol.
(1029.4 kJ)
31. Calculate the amount of heat required to convert 1 kg of ice at –20 ºC to liquid at 25
ºC. s (H20 (l))= 4.18 J/g ºC; s (H20 (ice)) = s ((H20 (v))= 2.0 J/g ºC; ∆Hfus = 6.02
kJ/mol; ∆Hvap = 40.8 kJ/mol.
(478.9 kJ)
32. How much energy is required to convert 162 g of water to liquid at 0 ºC? s (H20 (l))=
4.18 J/g ºC; s (H20 (ice)) = s ((H20 (v))= 2.0 J/g ºC; ∆Hfus = 6.02 kJ/mol; ∆Hvap = 40.8
kJ/mol.
(54.18 kJ)
33. How much energy is released when 154 g of water are converted to liquid at 100 ºC?
s (H20 (l))= 4.18 J/g ºC; s (H20 (ice)) = s ((H20 (v))= 2.0 J/g ºC; ∆Hfus = 6.02
kJ/mol; ∆Hvap = 40.8 kJ/mol.
(734.4 kJ)
34. What is the vapor pressure of mercury at 25 ºC. ∆Hvap = 59.1 kJ/mol; normal boiling
point = 357 ºC.
(0.0026 torr)
35. Calculate the molar heat of vaporization of liquid iodine. The molar heat of fusion
and sublimation of molecular iodine is 15.3 kJ/mol and 62.3 kJ/mol, respectively.
(47 kJ/mol)
36. Calculate the vapor pressure of absolute ethyl alcohol at 25 ºC. ∆Hvap = 39.3 kJ/mol;
normal boiling point = 78.3 ºC.
(68.5 torr)
37. Compute the molar heat of vaporization of a liquid whose vapor pressure doubles
when the temperature is raised from 50 ºC to 70ºC.
(31.92 kJ/mol)
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Arnold & Marie Schwartz College of Pharmacy and Health Sciences, Long Island University
PH931
Instructor: M. Savva, Ph.D.
CHAPTER VI
CHEMICAL EQUILIBRIA
Introduction
In principle, every substance in a closed system can coexist either as a pure phase or as
a mixture of the three phases. Water for example, at 0.01 ºC and 0.006 atm coexist as a
solid, liquid and vapor mixture. This is an example of a physical equilibrium. Increasing
the temperature and pressure of the system to 100 ºC and 1 atm, respectively, results in
physical changes so that liquid and vapor water coexist at equilibrium. In other words,
equal number of liquid molecules pass into the vapor phase as vapor molecules become
trapped into the liquid phase, establishing a dynamic equilibrium between the two phases.
Chemical equilibrium is also dynamic in nature. However, unlike the physical
equilibrium which involves changes in the phases of the same substance, chemical
equilibrium comprises changes in the concentration of reactants and products under the
conditions set during the chemical reaction. The predominant of this kind of chemical
reactions are reversible. Changes in the reactant and product concentration will take place
until the system reaches lowest energy state. At the lowest energy state the rates of
forward and reverse reaction become equal and the system is at a dynamic equilibrium,
that is, the ratio of the concentration of reactants to the concentration of products remains
constant. Exactly this ratio is the equilibrium constant, K, of the reaction.
PROBLEMS
1. Give the units of the equilibrium constant, K of a chemical reaction. Explain the
physical significance of K > 1 and K < 1.
2. What does the equilibrium constant, K tell us?
The equilibrium constant, K, helps us to predict the direction in which a reaction
mixture will proceed to achieve equilibrium and to calculate the concentrations of
reactants and products once equilibrium has been reached.
Applications of the equilibrium constant.
A) Predicting the direction of a reaction
The direction of a reaction can be determined at a given temperature by comparing the
value of the equilibrium constant, K with the value of the reaction quotient (Q).
The reaction quotient, Q, is defined as the ratio of the initial concentrations of products to
the initial concentration of reactants of a reaction that has not reached equilibrium.
Q<K
The reaction proceeds to the right until it reaches equilibrium
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Arnold & Marie Schwartz College of Pharmacy and Health Sciences, Long Island University
PH931
Q=K
Q>K
Instructor: M. Savva, Ph.D.
The reaction is at equilibrium
The reverse reaction is favored.
3. The equilibrium constant for the dissociation of acetic acid as given below is Ka = 1.8
x 10-5.
CH3COOH + H2O
CH3COO- + H+
At the beginning of the experiment, there are 0.25 M of acetic acid, and 0.0012 M of
acetate ion and 0.0012 M hydrogen ion. Predict the direction the reaction will proceed to
reach equilibrium.
(Q= 5.76 x 10-6 < Ka ; thus the reaction will proceed to the right, formation of more
products)
4. The equilibrium constant for the dissociation of formic acid as given below is Ka = 1.8
x 10-4.
HCOOH + H2O
HCOO- + H+
At the beginning of the experiment, there are 0.25 moles of formic acid, and 1.5 mmoles
of formic ion and 1.5 mmoles hydrogen ion in 1 L of solution. Predict the direction the
reaction will proceed to reach equilibrium.
(Q= 9 x 10-6 < Ka ; thus the reaction will proceed to the right, formation of more
products)
5. Ethyl acetate is synthesized from acetic acid and ethanol in an anhydrous organic
solvent according to the following reaction:
CH3COOH + C2H5OH
CH3COOC2H5 + H2O
K = 2.2
Predict the direction of the reaction for the mixtures below:
a) [CH3COOH] = 0.010 M; [C2H5OH] = 0.010 M; [CH3COOC2H5] = 0.22 M; [H2O] =
0.10 M
b) [CH3COOH] = 0.0020 M; [C2H5OH] = 0.1 M; [CH3COOC2H5] = 0.22 M; [H2O] =
0.002 M
c) [CH3COOH] = 0.044 M; [C2H5OH] = 6.0 M; [CH3COOC2H5] = 0.88 M; [H2O] = 0.12
M
d) [CH3COOH] = 0.88 M; [C2H5OH] = 10.0 M; [CH3COOC2H5] = 4.4 M; [H2O] = 4.4 M
e) What must the concentration of water be for a mixture with [CH3COOH] = 0.10 M;
[C2H5OH] = 5.0 M; [CH3COOC2H5] = 2.0 M to reach equilibrium?
f) Why is water included in the equilibrium expression for this reaction?
(a) Q = 220 > K; The reaction will proceed to the left.
(b) Q = 2.2 = K; Equilibrium conditions.
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Arnold & Marie Schwartz College of Pharmacy and Health Sciences, Long Island University
PH931
(c)
(d)
(e)
(f)
Instructor: M. Savva, Ph.D.
Q = 0.4 < K; The reaction will proceed to the right.
Q=K
[H2O] = 0.055 M
Water does not have a role of solvent here. It is a product of the reaction.
6. The synthesis of hydrogen fluoride as shown below, has an equilibrium constant, K, of
1.1 x 103.
H2 (g) + F2 (g)
2 HF (g)
Calculate the concentration of hydrogen fluoride if the concentration of [H2 (g)] = [F2 (g)]
= 0.0011 M when the system is analyzed at equilibrium conditions.
(0.0365 M)
B) Calculating Equilibrium Concentrations
Knowledge of the equilibrium constant and the initial concentration of reactants allow
one to calculate the percent yield of a reaction under a given set of conditions.
7. The dissociation of molecular iodine into iodine atoms is shown below:
I2 (g)
2 I (g)
K = 4.0 x 10-4 at 600 ºC
What are the concentrations of gases at equilibrium if the initial [I2 (g)]o = 0.0200 M
([I2 (g)] = 0.01885 M; [I (g)] = 0.0023 M)
8. A mixture of 0.5 M H2 and 0.5 M of I2 was placed in a flask at 450 ºC as shown below.
I2 (g) + H2 (g)
2 HI (g)
K = 50.0 at 450 ºC
Calculate the concentrations of [I2 (g)], [H2 (g)], and [HI (g)] at equilibrium.
([I2] = [H2] = 0.11 M; [HI] = 0.78 M)
9. The equilibrium constant K for the reaction
CO2 (g) + H2 (g)
H2O (g) + CO (g)
is 4.2 at 1600 ºC. The initial concentrations [CO2]o = [H2]o = 0.8 M. Calculate the
concentration of each species at equilibrium.
([CO2] = [H2] = 0.262 M; [H2O] = [CO] = 0.538 M)
10. Sodium bicarbonate, a component in effervescent tablets, undergoes thermal
decomposition as follows:
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Arnold & Marie Schwartz College of Pharmacy and Health Sciences, Long Island University
PH931
Instructor: M. Savva, Ph.D.
2 NaHCO3 (s)
Na2CO3 (s) + CO2 (g) + H2O (g)
In which case would we obtain more CO2 (g) and H2O (g) a) in a closed vessel b) in an
open vessel. Use the equilibrium expression to explain your answer.
11. What do you think is going to happen if we add citric acid (C6H8O7) to the above
mixture?
a) Na2CO3 (s) will increase
b) Na2CO3 (s) will decrease
c) CO2 (g) will increase
d) CO2 (g) will decrease
e) NaHCO3 (s) will increase
f) Write down the equation that describes the reaction.
12. The equilibrium constant for the following reaction is K = 3.5 at 100 ºC
H
H
H
H
cis-stilbene
trans-stilbene
What are the concentrations of cis- and trans-stilbene if initially only cis-stilbene were
present at a concentration of 0.7 M, at 100 ºC?
([cis-] = 0.155 M; [trans-] = 0.544 M)
13. A 0.1 M acetic acid solution was found by conductivity analysis to dissociate under
equilibrium conditions into 0.00132 M each of hydrogen and acetate ion at 25 ºC. What is
the equilibrium constant (Ka) for the dissociation of acetic acid at 25 ºC? (In this case, Ka
is the dissociation constant or acidity constant of the acetic acid).
CH3COOH
CH3COO- + H+
(Ka = 1.74 x 10-5)
38
Arnold & Marie Schwartz College of Pharmacy and Health Sciences, Long Island University
PH931
Instructor: M. Savva, Ph.D.
CHAPTER VII
ENTROPY, FREE ENERGY AND EQUILIBRIUM
Introduction
In the previous chapters we have discussed the 1st Law of thermodynamics which
states that the energy of universe is constant; it cannot be created or destroyed. Use of the
1st Law allows us 1) to determine whether energy in a certain form is released to or
absorbed by a system and, 2) to measure the exact amounts of energy changes involved in
a process. The 1st Law cannot predict however, whether a particular process will occur
under certain conditions or not. Often, pharmaceutical scientists need to predict whether a
particular reaction, for example synthesis of drugs or dissolution of drugs in a buffer for
parenteral administration is feasible before they actually do it.
The 2nd Law of thermodynamics can help us determine 1) whether a process is
feasible and 2) the probability of occurrence of a certain process.
A process that occurs under a given set of conditions is called a spontaneous
process. If a process does not occur under specified conditions, it is said to be nonspontaneous. Examples of spontaneous physical and chemical processes are:
a) Heat is transferred always from hot to cold. The reverse does not happen
spontaneously.
b) A ball always rolls down from a higher to lower altitude.
c) Crystalline sugar dissolves in water spontaneously. Dissolved sugar cannot
reappear in the same crystalline form spontaneously under the same conditions.
d) Water freezes spontaneously at temperatures below ºC and ice melts spontaneously
above ºC at atmospheric pressures.
Note that a spontaneous process does not necessarily have to be a fast process.
Thermodynamics cannot assess the speed of a reaction. Knowledge of the rate of a
reaction is the subject of chemical kinetics. For now, the important question we need to
ask ourselves is: what are the requirements for a spontaneous process to occur (always in
one direction)?
From examples a) and b), it appears that a spontaneous process occurs when the
energy of the system is decreased. This would imply that all exothermic reactions are
spontaneous, which is not true. Consider for example the spontaneous process of fusion
of ice to liquid water above ºC (∆Hfus = 6.01 kJ/mole, an endothermic process). Since the
above endothermic process is spontaneous, the reverse exothermic process of freezing,
under the same set of conditions is not spontaneous.
It turns out that all spontaneous endothermic processes involve an increase in a
property in the system called, ENTROPY (S). In general, for a spontaneous reaction to
occur the entropy of the universe has to be greater than zero.
Entropy can be viewed as a measure of randomness or disorder of a system. Highly
disordered systems have high entropy values and vice-versa. In general
S (gas) >> S (liquid) > S (solid)
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Arnold & Marie Schwartz College of Pharmacy and Health Sciences, Long Island University
PH931
Instructor: M. Savva, Ph.D.
Entropy, like Enthalpy and Energy is also a state function. That is, its value is determined
only by the initial and final states of the system:
∆S = Sf - Si
However, certain differences exist between Entropy and Enthalpy as summarized below:
Units
Entropy and
Enthalpy values
Entropy (S)
Enthalpy (H)
Absolute Entropy of a substance
J/(mole K)
Much smaller
N/A
J/mole
Much bigger
Entropies of elements and ∆Hºf for elements at their stable
compounds are all positive (Sº > form is zero. For compounds ∆Hºf
0).
may be > 0 or < 0.
PROBLEMS
1. Which of the following processes involves a change to a more disordered state?
a) melting
b) evaporation
c) dissolution of crystalline sugar in water
d) preparation of alcohol USP from absolute alcohol and water
a-b) In both processes Energy is given to the system in the form of heat. Specifically, the
thermal energy is absorbed by the molecules of the substance and is converted to
internal energy, that is, kinetic and potential molecular energy. Increased kinetic
energy is responsible for an increase in the average translational motion of
molecules. An increase in the potential energy increases the rotational and
vibrational motion of molecules. Clearly, the increase in the molecular motion
brought about by the increased kinetic and potential energy will increase the
randomness (Entropy) of the system.
c) When crystalline sugar dissolves in water the highly ordered crystalline structure of the
solid breaks down. In addition, the solute molecules perturb the ordered structure of
the water. The solution is more disordered than the pure solvent and pure solute.
However, another process that takes place during dissolution is hydration. Hydration
involves orientation of water molecules around the solute molecules thus decreasing
the entropy of the solution. In the great majority of the cases, the dissolution process
leads to an increase in Entropy.
d) Alcohol USP is 95 % Ethyl alcohol. Its preparation requires mixing 95 parts of pure
ethanol with 5 parts of water. An increase in entropy change is expected primarily
because of the increased volume available for every particle in the solution. In other
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Arnold & Marie Schwartz College of Pharmacy and Health Sciences, Long Island University
PH931
Instructor: M. Savva, Ph.D.
words, the molecules of each particle have more volume to occupy, thus the
randomness of the system is increased.
2. State the 2nd Law of thermodynamics.
‘The Entropy of the Universe increases in a spontaneous reaction involving a system
and its surroundings and remains unchanged when the system reaches equilibrium.’ Or
‘Spontaneous reactions always proceed in the direction in which the entropy of the
universe increases’.
3. Define:
a) Standard free energy of reaction, ∆Gºrxn
b) Standard free energy of formation, ∆Gºf
a) Standard free energy of reaction, ∆Gºrxn, is the free energy change of a reaction
that occurs under standard state conditions; P = 1atm.
b) Standard free energy of formation, ∆Gºf, of a compound is the free-energy change
that occurs when 1 mole of the compound is synthesized from its elements in the
standard states.
Note that, similarly to ∆Hºf, ∆Gºf for any element in its most stable form is zero.
4. State the 2nd Law of thermodynamics in terms of Free Energy.
‘A reaction at constant temperature and pressure will proceed spontaneously in the
direction that lowers its Free Energy.’
Thus, ∆Gºrxn < 0
∆Gºrxn = 0
favored
∆Gºrxn > 0
spontaneous reaction
reaction is at equilibrium; products and reactants are equally
spontaneous reverse reaction
5. Find out whether synthesis of methanol from carbon monoxide and hydrogen gas, as
shown below, is spontaneous at standard conditions (T = 25 ºC, P = 1 atm). ∆Hºf (CH3OH
(l)) = -239 kJ/mol; ∆Hºf (CO (g)) = -110 kJ/mol; ∆Hºf (H2 (g)) = ∆Gºf (H2 (g)) = 0
kJ/mol; ∆Gºf (CH3OH (l)) = -166 kJ/mol; ∆Gºf (CO (g)) = -137 kJ/mol; Sº (CH3OH (l)) =
127 J/(mol.K); Sº (CO (g)) = 198 J/(mol K); Sº (H2 (g)) = 65.5 J/(mol K).
CO (g) + 2 H2 (g)
CH3OH (l)
Use the Gibbs equation to determine ∆Gºrxn and then verify your answer taking advantage
of the state properties of the Free Energy.
( ∆Gºrxn = -29 kJ)
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Arnold & Marie Schwartz College of Pharmacy and Health Sciences, Long Island University
PH931
Instructor: M. Savva, Ph.D.
6. Calculate the standard free-energy changes for the following reactions at 25 ºC. Are
both reactions spontaneous? ∆Gºf (Mg (s)) = ∆Gºf (O2 (g)) = ∆Gºf (H2 (g)) = 0 kJ/mol;
∆Gºf (CH4 (g)) = -50.8 kJ/mol; ∆Gºf (CO2 (g)) = -394.4 kJ/mol; ∆Gºf (H2O (l)) = -237.2
kJ/mol; ∆Gºf (MgO (s)) = -596.6 kJ/mol.
a) CH4 (g) + 2 O2 (g)
CO2 (g) + 2 H2O (l)
b) 2 MgO (s)
2 Mg (s) + O2 (g)
(a) ∆Gºrxn = -818.0 kJ; b) ∆Gºrxn = -1139 kJ)
7. The molar heats of fusion and vaporization of benzene are 10.9 kJ/mol and 31.0
kJ/mol, respectively. Calculate the entropy changes for the solid- liquid and liquid-vapor
transitions for benzene. At 1 atm pressure, benzene melts at 5.5 ºC and boils at 80.1 ºC.
∆Hfus = 10.9 kJ/mole; ∆Hvap = 31.0 kJ/mole.
(∆Sfus= 39.1 J/(K mole); ∆Svap = 87.8 J/(K mole)
8. a) Calculate the entropy changes associated with the liquid-vapor water phase
transition under a pressure of 1 atm. Assume that the temperature throughout the phase
transition remains constant at T = 100 ºC. ∆Hvap = 40.79 kJ/mole.
b) Calculate the entropy of liquefaction at the above conditions?
(∆Svap= 109.3 J/(K mole); ∆Scond = -109.3 J/(K mole)
9. Calculate the ∆Gº (per mole) of the solubilization of silver chloride using the solubility
product of silver chloride at 25 ºC (Ksp = 1.6 x 10-10). Use the ∆Gº value to predict the
direction the reaction will follow. Comment on the solubility of the AgCl.
Ag+ (aq) + Cl- (aq)
AgCl (s)
(∆Gº = +55.88 kJ/mole)
10. Calculate ∆Gº for the following process at 25 ºC. (Ksp = 1.7 x 10-6). Analyze your
answer with regard to solubility and equilibrium direction. Compare ∆Gº and Ksp with
those of the previous problem.
BaF2 (s)
Ba2+ (aq) + 2 F- (aq)
(∆Gº = +32.9 kJ/mole)
11. Use the Hess’s law to calculate the ∆Gº for the C (graphite)
transition.
C (diamond)
C (diam) + O2 (g)
CO2 (g)
∆Gºrxn = -397 kJ/mole
C (gr)
CO2 (g)
∆Gºrxn = -394 kJ/mole
+ O2 (g)
( ∆Gºgr-diam = +3 kJ/mole)
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Arnold & Marie Schwartz College of Pharmacy and Health Sciences, Long Island University
PH931
Instructor: M. Savva, Ph.D.
12. Calculate the ∆G (for one round) of the following reaction at 50 ºC. ∆Gºf (CO (g)) = 137 kJ/mol; ∆Gºf (H2 (g)) = 0; ∆Gºf (CH3OH (l)) = -166 kJ/mol. Is the reaction more
favorable at 50 ºC? Why?
CO (g) + 2 H2 (g)
CH3OH (l)
(∆G = -30.86 kJ/mole)
13. Cells use the hydrolysis of adenosine triphosphate (ATP) as a source of energy
according to the following reaction:
ATP (aq) + H2O (l)
ADP (aq) + H2PO4- (aq)
Calculate K at 25 ºC.
∆Gº = -30.5 kJ/mol
(K = 2.22 x 105)
14. Calculate the ∆Gº for the ionization of acetic acid (Ka = 1.74 x 10-5). Which direction
is favored?
CH3COOH
CH3COO- + H+
(∆Gº = + 27.15 kJ/mol)
43