Download Chapter 6 - Foothill College

Energy and Chemical Reactions - Thermochemistry
Chapter 6 and Chapter 9 Sections 9.2 and 9.4
Alkali Metals in Water: http://www.youtube.com/watch?v=uixxJtJPVXk&feature=related
Key Concepts-Our Goals are to:
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
develop an understanding of Potential and Kinetic Energy and the ability to describe the relationship
between the two in Thermochemistry.
be able to define the terms System and Surroundings and to identify a System and its Surroundings for
physical and chemical changes.
develop an understanding of the concept of Heat Transfer and the ability to describe the process.
develop an understanding of the concept of Pressure-Volume Work and the ability to recognize when a
system can do work or when a system has work done on it.
develop the ability to apply and use the First Law of Thermodynamics.
develop an understanding of internal energy change for a system, ∆Esys; be able to relate this to heat flow
and work (Learn the sign conventions!).
develop an understanding of the concept of a State Function; know what variables in chemistry are State
Functions.
develop an understanding of ENTHALPY Change, ∆H; know that this is heat transfer at constant pressure, a
State Function.
develop an understanding of Enthalpies of Reaction, ∆Hrxn; be able to apply Hess’s Law to determine ∆Hrxn
and to calculate ∆Hrxn from measured heat flow.
develop an understanding of Enthalpies of Formation, ∆H°f, and the ability to write reactions corresponding
to standard enthalpies of formation.
develop the ability to do stoichiometric calculations that include heats of reaction.
learn the basics of Calorimetry; be able to use calorimetry data to calculate ∆Hrxn, specific heat capacity,
temperature change, amount of heat flow, etc..(Covered in LAB only.)
13.
revisit the concept of lattice energy and learn how lattice energies are calculated using known
thermodynamic data. (Section 9.2)
14.
understand the relationship between covalent bond energies and reactions. Learn how to estimate ∆H for a
reaction using bond energies. (Section 9.4)
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1
Energy-Kinetic vs. Potential Energy
Energy - The capacity to do work. Common units are Joules and Calories
Joule (J): SI unit of energy (A very small amount of energy so we commonly use the kilojoule, kJ.)
1 J = 1 Newton•m = 1 kg•m2/s2 calorie (cal): 1 cal = 4.184 J (The heat required to raise the temperature of 1 g of water from 14.5°C to 1 5.5°C.)
Nutritional calorie (Cal) = 1 kcal = 1000 cal
Kinetic Energy: Energy due to motion of matter.
In chemistry we are often concerned with the thermal (heat) energy of atoms/molecules/ions. Heat is a
manifestation of random kinetic energy. Heat is directly proportional to temperature.
Heat and Temperature are not the same!
Heat is a measure of energy content.
Heat is an extensive property.
Temperature is a measure of AVERAGE kinetic energy.
Temperature is an intensive property.
Heat is always transferred from an object with a higher temperature
(higher average kinetic energy of the particles) to one of lower
temperature (lower average kinetic energy of the particles) until
THERMAL EQUILIBRIUM is reached.
Amount of Heat Lost = Amount of Heat Gained
Potential Energy: Stored energy in a system. Weakly temperature dependent.
Chemical Potential Energy-The kind of Potential Energy of interest in Thermochemistry.
Chemical potential energy results from attractive and repulsive forces between nuclei and electrons in a system.
Includes the bond energies between atoms bonded together within a molecule (intramolecular), forces between
ions, and forces between different molecules (intermolecular) in liquids and solids.
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Kinetic Energy
Kinetic Energy (KE)
To calculate the kinetic energy of a moving object in joules, what units must you use for the
object’s mass, m, and velocity, u?
1
Ek = mu 2
2
Question:
a) Calculate the kinetic energy in Joules of a nitrogen molecule that is moving at 535 m/s.
b)
Consider a oxygen molecule with the same kinetic energy as the above nitrogen
molecule. What can you conclude about the oxygen molecule’s velocity?
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Thermochemistry Basic Terms
When we study chemical systems we actually study ENERGY TRANSFER NOT ENERGY CONTENT.
Energy is TRANSFERRED between a SYSTEM and SURROUNDINGS as heat (q) and/or work (w).
System: Process under study (chemical reaction or physical change).
Surroundings: Everything else. May be isolated as in a calorimetry experiment.
Universe
Universe = System + Surroundings
System
heat
work
Surroundings
The Internal Energy (E) of the SYSTEM: The sum of all the kinetic and potential
energies of the matter in the system. CANNOT BE absolutely determined.
However, we can easily measure changes in internal energy (∆Esys) for a
system:
∆Esys = Efinal - Einitial = q + w
∆ Esys (+) SYSTEM internal energy INCREASES.
∆ Esys (−) SYSTEM internal energy DECREASES.
Sign conventions:
1.
2.
3.
4.
q (+) HEAT flows into the system from the surroundings. ENDOTHERMIC process
q (-) HEAT flows out of the system to the surroundings. EXOTHERMIC process
w(+) WORK is done ON the system by the surroundings.
w(-) WORK is done BY the system on the surroundings.
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Thermochemistry-First Law
First Law of Thermodynamics:
(Law of Conservation of Energy)
Energy can be transferred from one form to
another, but cannot be created or destroyed
in a physical or chemical change.
∆ Euniv = ∆ Esys + ∆ Esurr = 0
∆ Esys = −∆ Esurr
For which of the above is energy released
to the surroundings?
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Definitions and Sign Conventions for Work (w)
Work (w): Energy that moves an object
against a force is called work. Work is
quantified as force x distance:
w = F •d
Only pressure-volume work will be
considered here. For a gaseous system
expanding or contracting in volume
(distance) of ∆V = Vfinal – Vinital, against a
constant external pressure (force), P, the
work is defined as:
w = F • d = −P(∆ V )
wsys > 0, ∆Esys increases
Work is done on the system. This is the case when the volume of a gas system
decreases, ∆V is negative, so w = –P∆V is positive.
wsys < 0, ∆Esys decreases
Work is done by the system. This is the case when the volume of a gas system
increases, ∆V is positive, so w = –P∆V is negative.
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Example Work for a Physical Change: Solid —> Gas
Reaction systems that result in a change in moles of gas experience a volume change. An example
where work is done by the system is sublimation of CO2: CO2(s) —> CO2(g)
Think about it:
What are the signs of q
and w for the system
in this case?
What are the signs of q
and w for the
surroundings?
Based on this
information, can you
conclude the sign of
∆Esys?
Another example is the denotation of TNT. Here 15 moles of gas are created for every 2 moles (450 g) of TNT.
Assuming a temperature of 525°C, the hot expanding gasses can do ≈100 kJ of work.
Can you think of a simple example where wsys > 0?
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State Functions in Chemistry
State functions are functions whose CHANGE in value depends only
upon the the initial and final states of the system, not on the pathway
from initial to final.
∆(State function) = Final State - Initial State
By convention, all state functions are represented by UPPERCASE letters
in chemistry.
∆T (temperature), ∆P (pressure), ∆V (volume), ∆E (internal energy) are
all state functions.
Variables that are not state functions: w (work), d (distance), t (time),
q (heat), etc.
Note: the sum q + w = ∆E is a state function.
A Special State Function: ENTHALPY, H
At constant external pressure, ∆P = 0, the heat transferred
between the system and surroundings is called enthalpy (H). The
change in enthalpy is defined as:
∆ H sys = ∆ E p + P ∆V = ?
∆Hsys is the Enthalpy of Reaction
(Heat of reaction at constant P, a STATE FUNCTION.)
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The relative amounts of heat and work for a
given ∆E depends upon the pathway over
which the energy change occurs. Consider the
discharge of the potential energy stored in a
battery. We can vary the proportions of heat to
work by changing the circuitry attached to the
battery. The same amount of energy is used in
both cases, but the pathway of energy use is
different.
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Heat Transfer at Constant Pressure, ∆H
Note how changes in internal energy and changes in enthalpy (heat content) at constant pressure
are related, differing only in the quantity of energy transferred to or from the system as work:
∆E = ∆H – P∆V = ∆H + w
(Remember w = – P∆V)
For many processes ∆V is very small, or negligible, making ∆E ≈ ∆H.
What are some examples of processes with negligible ∆V?
(Hint: What processes DO NOT involve a change in the number of moles of gas, thus ∆V ≈ 0?)
In processes where ∆V is large, ∆E and ∆H will be different.
What are some examples of processes with significant ∆V?
(Hint: What processes DO involve a change in the number of moles of gas, thus ∆V ≠ 0?)
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Example Chemical System
Consider the case where this reaction is done at a
constant external pressure of 1.00 atm and T = 25°C.
For the complete reaction of 1 mole of zinc:
1.
The amount of heat absorbed by the surroundings
is 154 kJ.
2.
Zn(s) + 2 HCl(aq)
SYSTEM
Enthalpy, ∆Hsys
Consider the redox reaction of zinc metal
with hydrochloric acid:
Zn(s) + 2 HCl(aq) –> ZnCl2(aq) + H2(g)
Initial State of
SYSTEM
Enthalpy Diagram
–154 kJ
ZnCl2(aq) + H2(g)
Final State of
SYSTEM
The volume of H2 gas generated is 24.5 liters.
We symbolize and quantify this as follows:
1. qp = ∆Hsys = –154 kJ (Exothermic)
Work
2. ∆Hsurr = – ∆Hsys = +154 kJ
3. w = -P(∆V) = (-1.00 atm)(+24.5 L)
= -24.5 L*atm = -2.48 kJ
4. ∆Esys = ∆Hsys + w = (–154 + -2.48) kJ = -156 kJ
5. ∆Esurr = – ∆Esys = +156 kJ
Observations:
The chemical reaction is EXOTHERMIC. The surroundings will gain heat and should show an increase in T.
The SYSTEM loses 154 kJ of POTENTIAL energy (not thermal energy) to the surroundings.
The SYSTEM does 2.48 kJ of work on the surroundings.
In this chemical system, |∆Hsys| >> |w| making ∆Esys ≈ ∆Hsys.
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∆H of Reactions, ∆Hsys
CH4(g) + 2O2(g) —> CO2(g) + 2H2O(l) ∆Hrxn = –890 kJ
For the above reaction the value of ∆Hsys represents the
difference in chemical potential energy between the
products (final state) and the reactants (initial state)
∆Hsys = H(products) - H(reactants)
So if ∆Hsys is a difference in potential energy then why do we
refer to it as a heat? Because we measure the system’s
difference in potential energy (∆Hsys) by measuring the HEAT
change of the surroundings (qsurr) when the chemical
reaction takes place.
∆(potential energysys) = ∆(heatsurr)
Thus, ∆Hsys = -qsurr. In the surroundings we measure a heat
change by experimentally observing a change in temperature.
This is the basis for calorimetry (covered in lab).
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Review of Energy Changes
Universe
Potential Energy
Changes
Kinetic Energy
Changes
(Temperature constant)
heat
work
System
(Temperature changes)
q
w
Surroundings
Make sure that you understand the changes in the system versus the surroundings when we study chemical
reactions and physical state changes. The “system” undergoes potential energy changes when the reactants
are transformed into products or when a physical state change occurs. The temperature of the system is taken
as the same before and after reaction, no thermal energy change in the system. Heat flow to or from the
surroundings maintains the system’s temperature.
The surroundings are a reservoir of thermal energy where heat is absorbed or released. Any potential energy
change in the system is reflected by a corresponding thermal energy change in the surroundings. This is what
we can “see” (measure) in the laboratory. How do we know when a potential energy change takes place in a
system? We observe the effects of the system’s potential energy change on the surroundings because the
surroundings experience a change in temperature. This is the basis for calorimetry.
Daley/Larson
Energy of Reactions
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Heat of Reactions, ∆Hrxn
For chemical reactions we measure or calculate ∆Hsys for a balanced chemical reaction.
CH4(g) + 2O2(g) —> CO2(g) + 2 H2O(l) ∆Hrxn = –890 kJ = ∆Hsys
1.
∆Hrxn is specific to the reaction given with the coefficients given.
2.
The physical states (s, l, g, aq) are required and important!
3.
The coefficients in the balanced equation determine the extent of ∆Hrxn.
∆Hrxn therefore depends on the amount of reaction, and is an EXTENSIVE property.
If you multiply the coefficients of a balanced equation by a constant, ∆Hrxn is
multiplied by the same constant.
4.
If a chemical reaction is reversed, then the sign of ∆Hrxn changes but the magnitude
remains the same.
CO2(g) + 2H2O(l) —> CH4(g) + 2 O2(g) ∆Hrxn = +890 kJ
5.
For an exothermic system we think of heat as a product of the reaction:
reactants —> products + heat
(Thermodynamically favorable, but can still be non-spontaneous for certain systems under certain conditions.
This will be explored in more detail in chemistry 1B.)
6.
For an endothermic system we think of heat as a reactant of the reaction:
heat + reactants —> products
(Thermodynamically unfavorable, but can still occur spontaneously for certain systems under the right
conditions. This will be explored in more detail in chemistry 1B.)
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Stoichiometry and ∆Hrxn
For the reaction:
N2(g) + 3 H2(g) —> 2 NH3(g)
∆Hrxn = –91.80 kJ (so heat is a product, exothermic)
We can interpret the magnitude of ∆Hrxn in many ways:
1. 91.80 kJ of heat is produced when 1 mol of N2 reacts or
2. 91.80 kJ of heat is produced when 3 mol of H2 reacts or
3. 91.80 kJ of heat is produced when 2 mol of NH3 is formed
(a) What is ∆H/mol N2(g), ∆H/mol H2(g), ∆H/mol NH3(g)?
(b) How much heat is produced if 23.6 g of hydrogen reacts?
(c) How many grams of ammonia are produced when 750 kJ of heat is given off?
(d) For the system, is the work term + or –?
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Finding ∆Hrxn from a Measured Heat Transfer
Given a measured heat transfer for a reaction (qsurr) we can determine the ∆Hrxn.
Experiment shows 1.14 kJ of heat is needed to decompose 1.50 g of NaHCO3(s).
2 NaHCO3(s) —> Na2CO3(s) + H2O(g) + CO2(g)
∆Hrxn = ?
From the data, find ∆Hrxn.
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Hess’s Law of Heat Summation
Hess’s Law states that if a reaction is the sum of two or more other reactions, then the net ∆Hrxn is
the sum of the individual ∆Hrxn values.
A —> B + C
B —> D
A —> C + D
∆H1
∆H2
∆H1 + ∆H2 = ∆Hrxn
In other words, the overall change in enthalpy (∆H) for a process is independent of the path by
which the change occurs. Why is this true?
Consider the following reactions:
1. 2 NaHCO3(s) —> Na2CO3(s) + H2O(g) + CO2(g);
∆H1 = 128 kJ
2. Na2CO3(s) —> Na2O(s) + CO2(g);
∆H2 = 138 kJ
3. 2 NaHCO3(s) —> Na2O(s) + H2O(g) + 2 CO2(g);
∆H3 = 266 kJ = ∆H1 + ∆H2
Reaction (3) is the sum of reactions (1) and (2). If reaction 1 is carried out, followed by (2), the overall enthalpy
change will be equal to the enthalpy change for reaction (3) alone.
Bottom-line: For any reaction that can be written as the sum of simpler reactions, ∆Hrxn is the sum of the heats from
the individual reactions.
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Using Hess’ Law
Use Hess’ Law to find the unknown ∆Hrxn.
1. 2 Al(s) + 3/2 O2(g) —> Al2O3(s)
∆Hrxn = –1669.8 kJ
2. Al(s) + 3/2 Cl2(g) —> AlCl3(s)
∆Hrxn = –705.6 kJ
3. H2(g) + 1/2 O2(g) —> H2O(g)
∆Hrxn = –241.82 kJ
4. 1/2 N2(g) + 2 H2(g) + 1/2 Cl2(g) + 2 O2(g)—> NH4ClO4(s)
∆Hrxn = –295.3 kJ
5. 6 NH4ClO4(s) + 10 Al(s) —> 3 N2(g) + 12 H2O(g) + 4 Al2O3(s) + 2 AlCl3(s)
∆Hrxn = ?
1. Work with only the reactants and products given in the target reaction. (All other species should cancel when summed.)
2. Reverse equations where reactants or products are on the incorrect side. Change the sign of ∆H.
3. Get the correct stoichiometric amounts (correct coefficients) of reactants and products on each side by multiplying each equation by the
appropriate factor. Multiply ∆H by the same factor. (Sometimes fractions are needed.)
4. Add the reactions together, canceling species with the same formula and phase that are on opposite sides of the reaction.
5. Check that you have obtained the target reaction.
6. Sum the ∆H’s.
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Standard Enthalpies of Formation, ∆H°f
Standard Enthalpy of Formation (∆Hf°), or Heat of Formation.
1. The enthalpy change for the formation of one mole of a compound (the product) directly from its component
elements (the reactants) in their standard states.
2.
3.
The standard state of an element is its most stable form in the physical state it exists at a pressure of 1bar
(about 1atm) and at a specified temperature.
Data for the heat of formation of almost all compounds has been measured and/or calculated under Standard
Thermochemical Conditions.
3.1. Standard State Thermochemical Conditions:
1 bar pressure ≈ 1 atm
25° C = 298.15 K
Aqueous Solutions: 1M
Pure elements: most stable form at 25° C and 1 bar.
Example: C(s) is more stable as C(s,graphite) not C(s,diamond).
C(s,graphite) is used for a formation reaction involving carbon.
Examples:
H2(g) + 1/2 O2(g) → H2O(l)
∆Hf˚ = –285.8 kJ/mol
H2(g) + 1/2 O2(g) → H2O(g)
∆Hf˚ = –241.8 kJ/mol
2 C(graphite) + 3 H2(g) + 1/2 O2(g) → CH3CH2OH(l)
∆Hf˚ = –277.7 kJ/mol
Write a standard enthalpy of formation reaction for Na2CO3(s).
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Notes on Standard Enthalpies of Formation, ∆H°f
1. The ∆H°f for all elements in their standard states are zero. Elements are not formed in nature, they already exist.
2. Most ∆H°f values are negative indicating most compounds are more “stable” than their elements.
3. ∆H°f can be used to compare stabilities of compounds. Generally, the more negative ∆H°f the more “stable”.
4. For aqueous compounds, the values refer to 1 M solutions being formed. This includes any energy changes that
occur during hydration of ions.
5. For some ionic compounds dissolved in water, you may only be able to find ∆Hf˚ values for the separate aqueous
ions. For example, for FeCl3(aq) you may not find the ∆Hf˚ value for FeCl3(aq), but may find the ions separately:
Fe3+(aq) and Cl–(aq). (In other words, you may need to use net ionic equations in some cases when you use
∆Hf˚ values.)
Why is H2O(s) not listed?
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Hess’ Law and ∆H°rxn
We can use tabulated values of ∆H°f to calculate the standard heat of reaction, ∆H°rxn, for any possible reaction.
Use Appendix B in the textbook to find ∆H°f values.
We apply Hess’ Law:
o
∆ H rxn
= ∑ m ∆ H of ( products ) − ∑ n ∆ H of ( reactants )
Example: Find ∆H°rxn for the following:
6 NH4ClO4(s) + 10 Al(s) —> 3 N2(g) + 12 H2O(g) + 4 Al2O3(s) + 2 AlCl3(s)
∆H°f (kJ/mol) –295.3
0
0
–241.82
–1669.8
–705.6 Look-up values using reference source.
∆H°f (kJ) –1771.8
∑∆H°f (kJ) .
0
–1771.8 (reactants)
0
∆H°rxn (kJ) = –10992.2 – (–1771.8) = -9220.4 kJ Larson-Foothill College
–2901.84
–6679.2
–10992.2 (products)
–1411.2 Mult. by the molar stoichiometric coeff.
Sum products and reactants separately
Subtract reactant sum from product sum.
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Using ∆H°f to Calculate ∆H°rxn
Use Appendix B and Hess’ Law to find ∆H°rxn for each of the following:
1.
H2O(l) —> H2O(g)
2.
Fe2O3(s) + 2 Al(s) —> 2 Fe(s) + Al2O3(s)
3.
2NaCl(aq) + Pb(NO3)2(aq) —> PbCl2(s) + 2NaNO3(aq)
Thermite Welding
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Practice Problems
Imagine a container placed in a tub of water, as depicted in the following diagram:
(a) If the contents of the container are the system and heat is able to flow through the container
walls, what qualitative changes will occur in the temperatures of the system and in its
surroundings?
What is the sign of q associated with each change?
From the system’s perspective, is the process exothermic or endothermic?
(b) If neither the volume nor the pressure of the system changes during the process, how is the
change in internal energy related to the change in enthalpy?
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Practice Problems
A gas (the system) is confined to a cylinder fitted with a piston and an electrical heater, as
shown in the accompanying illustration:
Suppose that current is supplied to the heater so that 100 J of energy is added. Consider
two different situations. In case (1) the piston is allowed to move as the energy is added.
In case (2) the piston is fixed so that it cannot move.
(a) What can you say about the values of q and w for the system in each of these
cases?
(b)
In which case does the gas have the higher temperature after addition of the
electrical energy? Explain.
(c)
What can you say about the relative values of ∆E for the system in the two cases?
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Practice Problems
Calculate the standard enthalpy of formation of gaseous diborane (B2H6) using the
following thermochemical equations
4B(s) + 3O2(g)
—> 2B2O3(s)
∆H° = –2509.1 kJ
2H2(g) + O2(g) —> 2H2O(l)
∆H° = –571.7 kJ
B2H6(g) + 3O2(g) —> B2O3(s) + 3H2O(l) ∆H° = –2147.5 kJ
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Practice Problems
When 1.0 mol of KBr(s) decomposes to its elements, 394 kJ of heat is absorbed.
(a) Given that elemental bromine is a liquid at 25°C and 1 atm, write a balanced
thermochemical equation for the process.
(b)
What is ∆H when 10.0 kg of KBr(s) forms from its elements?
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Practice Problems
Consider the following unbalanced oxidation-reduction reaction in aqueous solution:
(a)
(b)
(c)
Balance the reaction.
By using data in Appendix B, calculate ∆H° for the reaction.
Based on the value you obtain for would you expect the reaction to be
thermodynamically favorable?
(d)
Use the activity series to predict if this reaction should occur. Is this prediction
consistent with your conclusion in part (c) of this problem?
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Practice Problems
The heat of combustion of ethanol, C2H5OH(l), is −1367 kJ/mol.
(a)
Write the balanced equation and sketch an enthalpy diagram for the reaction.
(b)
A batch of Sauvignon Blanc wine contains 10.6% ethanol by mass. Assuming the
density of the wine to be 1.0 g/mL, what caloric content does the alcohol (ethanol) in a
6-oz glass of wine (177 mL) have?
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Practice Problems
Meals-Ready-to-Eat (MREs) are military meals that can by heated on a flameless
heater. The heat is produced by the following reaction:
Mg(s) + 2H2O(l) —> Mg(OH)2(s) + H2(g)
Calculate the number of grams of Mg needed for this reaction to release enough
energy to increase the temperature of 75 g of water from 21°C to 79°C.
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Revisiting Energetics of Ionic Bond Formation – Lattice Energy of Ionic Crystals
Section 9.2
The lattice energy is the energy needed to break apart the solid crystal lattice and form ions in
the gas phase. This is always ENDOTHERMIC!
NaCl(s) —> Na+(g) + Cl–(g)
∆Hlattice = +788 kJ/mol
The lattice energy IS NOT the opposite of ∆H°f.
For NaCl(s) ∆H°f is –410.9 kJ/mol
The lattice energy is an indication of the force
of attraction between the ions. This attractive
force depends on the charge of the ions and
distance between ions in the solid phase
(Coulomb’s Law); charge having the greater
affect.
(n +e)(n - e)
∆ H Lattice ∝ attractive force = k
d2
•
k is a constant
•
n+ and n- are the magnitude of the (+)
and (–) charges
•
e is the charge of an electron
•
d is the distance between the ions
(directly related to ionic radius)
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Lattice Energy cannot be easily measured directly. It is calculated
using the Born-Haber cycle and Hess’ Law.
Steps in the cycle:
1. Na(s) —> Na(g)
2. 1/2 Cl2(g) —> Cl(g) 3. Na(g) —> Na+(g) + e– 4. Cl(g) + e– —> Cl–(g) 5. Na+(g) + Cl–(g) —> NaCl(s)
4.
3.
∆H°f Na(g) ∆H°f Cl(g)
IE1(Na)
EA(Cl)
–∆Hlattice
= 108 kJ (sublimation)
= 122 kJ (1/2 bond energy)
= 496 kJ
= -349 kJ
=?
Add these 5 steps together: to get step 6, ∆H°f NaCl(s):
5.
6. Na(s)+ 1/2 Cl2(g) —> NaCl(s)
∆H°f NaCl(s)= -411 kJ
2.
1.
Start at standard states.
6.
Using Hess’ Law and solving for the energy of step 5:
5 = 6 - {1 + 2 + 3 + 4}
So we have:
–∆Hlattice = 5 = -411 kJ - {108 + 122 +496 + -349} kJ = –788 kJ
∆Hlattice = +788 kJ
NaCl(s) —> Na+(g) + Cl–(g)
End at standard state.
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Practice (Review and Check of Skills and Knowledge)
•
Consider the gas-phase transfer of an electron from a sodium atom to a chlorine atom in the gas
phase:
(a) Write this reaction as the sum of two reactions, one that relates to an ionization energy and one
that relates to an electron affinity.
(b) Use the result from part (a) and Hess’s law to calculate the enthalpy of the above reaction. (You
will need to look up some thermochemical data.) Is the reaction exothermic or endothermic?
(c) The reaction between sodium metal and chlorine gas to produce NaCl(s) is highly exothermic.
Why is this reaction exothermic given your answer to part (b)?
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Lattice Energy Calculations Using the Born-Haber Cycle
Write the appropriate equations and perform the calculations to determine the lattice energy
for the crystal CaCl2. Data are as follows:
Ca(s) to Ca(g) ∆Hsublimation = 178 kJ/mol
IE1 Ca(g) = 590 kJ/mol
IE2 Ca+(g) = 1145 kJ/mol
Cl2(g) bond energy = 242 kJ/mol
EA Cl(g) = -349 kJ/mol
∆H°f CaCl2(s) = -796 kJ/mol
Answer 2253 kJ
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Revisiting bond enthalpy (energy, strength) (Section 9.4): The bond enthalpy, BE, is
the enthalpy change (∆H) per mole required to break a covalent bond in the gas
phase. Table 9.2 Average Bond Energies (kJ/mol) and Bond Lengths (pm)
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Using Bond Enthalpies (Energies) to Estimate ∆Hrxn
The heat released or absorbed during a chemical change is due to differences between
the bond energies of reactants and products.Therefore, the heat of reaction between
covalent molecules can be estimated from bond energies, BE.
∆Hrxn = ∑ (bond energies broken) - ∑ (bond energies formed)
∆Hrxn= ∑ BEreactants) - ∑ (BEproducts)
1
2
= [BE(C-H) + BE(Cl-Cl)]-[BE(C-Cl)+BE(H-Cl)]
= [(413+243)-[339+427] kJ = –110 kJ
The actual value is -101.1 kJ, so our estimate
has a 9% error.
Why is this an estimate?
CH4 (g) + Cl2 (g) —> CH3Cl (g) + HCl (g)
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Using Bond Energies to Estimate ∆Hrxn
Ammonia is produced directly from nitrogen and hydrogen by using the Haber process. The
chemical reaction is
3 H2(g) + N2(g) —> 2 NH3(g)
(a) Use bond energies to estimate the enthalpy change for the reaction. Is it exothermic
or endothermic.
(b) Compare the enthalpy change you calculated in (a) to the true enthalpy change as
obtained using ∆H°f values.
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Using Bond Enthalpies to Calculate Heat of Reaction
An important reaction for the conversion of natural gas to other useful hydrocarbons is
the conversion of methane to ethane.
In practice, this reaction is carried out in the presence of oxygen, which converts the
hydrogen produced to water.
Use bond energies to estimate ΔH for these two reactions. Why is the conversion of
methane to ethane more favorable when oxygen is used?
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Calorimetry and Heat Transfer (Lab Notes)
Calorimetry is the technique used to measure heat exchange between the system and
the surroundings. First some definitions:
Heat Capacity: The amount of heat required to raise the temperature of an object by 1 K or
1 °C. Units are J/°C or J/K: The value in J/°C is numerically equivalent to the value in J/
K since a ∆T in °C is the same as a ∆T in kelvin units.
Specific Heat Capacity or Specific Heat (c): The amount of heat required to raise the
temperature of a fixed amount of a substance by 1 K or 1 °C. Specific heat capacities are
often quoted on a per gram or per mole basis. We will use per gram:
c=
(quantity of heat transferred)
q
=
(grams of substance) × (temperature change) m × ∆ T
Units are: J/(g•°C) = J/(g•K)
• What is the heat capacity of 100.0 g of water?
• How much heat is required to raise the temperature of 100.0 g
of water from 25.0°C to 100.0°C?
• If this same amount of heat is absorbed by 100.0 g of iron at
25.0°C, what will be the final temperature of the iron?
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Constant Pressure Calorimetry
We use constant pressure calorimetry to find ∆Hrxn for a system.
In constant pressure calorimetry ∆P = 0. Thus qsurr = qp = -qsys.
(Remember, the heat absorbed or released by the surroundings is a result of a potential energy change in the system.)
The calorimeter and its contents become our surroundings and as heat flows into or out of
the surroundings we observe a temperature change. Heat flows to or from two sources:
1.
the contents (solution) in the calorimeter and
2.
the physical calorimeter itself (cup and thermometer in General Chemistry lab).
We can represent this in equation form as follows:
qsurroundings = qcontents + qcalorimeter
qcontents = c ( contents ) x mcontents x ∆ Tcontents
Note:
For our laboratory experiment, we will
include qcalorimeter in our calculations.
However, for text book examples and
problems qcalorimeter is considered negligible.
qcalorimeter = heat capacity ( calorimeter ) x ∆ Tcalorimeter
We relate the heat change of the surroundings to the potential energy
change of the system: q = −q
rxn
surroundings
∆Hrxn is determined by considering the limiting reactant (molrxn) for the
qrxn
system:
∆ H rxn =
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mol of rxn
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Constant Pressure Calorimetry: Lab Lecture Demonstration
Magnesium reacts with hydrochloric acid in a single replacement reaction.
Write the balanced chemical equation for the reaction:
Data:
Mass Mg:
Molarity HCl(aq):
Volume HCl(aq):
Mass calorimeter (with stir bar):
Mass calorimeter plus HCl(aq):
Initial temperature of solution in calorimeter:
Final (maximum) temperature of solution in calorimeter:
Based on the data, is this an exothermic or endothermic reaction?
Using the data, determine the heat of reaction, ∆H. Assume that the heat capacity of the Calorimeter is 25 J/°C.
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