Download Chapter 5 Thermochemistry

Chapter 5 Thermochemistry
Transferring Energy
Energy is the ability to do work or transfer heat.
– Energy used to cause an object that has mass to move is called work. (W = F x d)
– Energy used to cause the temperature of an object to rise is called heat.
Conversion of Energy
Potential energy is stored energy due to position or chemical composition.
Kinetic energy is the energy of motion.
Units of Energy
The SI unit of energy is the joule (J).
An older, non-SI unit is still in widespread use: the calorie (cal).
1 cal = 4.184 J
1kcal = 1000 cal = 1 Cal (nutritional calorie)
Definitions: System and Surroundings
The system: part we want to study (here, the hydrogen and oxygen molecules).
The surroundings: everything else (here, the cylinder and piston).
5.2 The First Law of Thermodynamics
Energy is neither created nor destroyed.
The total energy of the universe is a constant. (If the system loses energy, it must be gained by the surroundings, and vice
versa.)
Internal Energy
The internal energy of a system is the sum of all kinetic and potential energies of all components of the system; we call it E.
We cannot calculate it! =(
But we can measure how it changes!! =)
E = Efinal − Einitial
Changes in Internal Energy
If E > 0
(+)
Efinal > Einitial
This energy change is called endergonic. (The system absorbed energy)
If E < 0,
(-)
Efinal < Einitial
This energy change is called exergonic. (The system released energy)
When energy is exchanged between the system and the surroundings, it is exchanged as either heat (q) or work (w).
That is: E = q + w.
E, q, w, and Their Signs (Table 5.1)
Notice that the sign is always positive when the system gains energy, and negative when the system loses it.
Problem 1
Calculate the change in internal energy of the system if the system absorbs 150 kJ of heat from the surroundings and does 30
kJ of work.
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State and Path Functions
State function: its value is determined by the state of the system.
Path function: its value depends on the path that was taken.
Internal Energy (E) is a state function.
Heat (q) and Work (w) are path functions.
Interestingly:
∆E = q + w
Example:
Whether the battery is shorted out by a wire or is discharged by running the fan, its E is the same.
– But q and w are different in the two cases.
Defining the Work done by a Gas
W = F x d (definition of work)
F = P x A (from definition of pressure)
d = ∆h (from the physics of the problem)
W = P x A x ∆h
∆V = A ∆h (from the physics of the problem)
W = P ∆V
Since the gas (system) is doing work (and loosing energy), but pressure is always positive and ∆V is positive too, we need to
introduce a negative sign, so we get:
W = -P∆V
Chemical reactions involve heat and work
P-V work (expansion or compression of gases).
w = -P∆V = - ∆ngasRT
We can use the balanced equation to get ∆ngas.
N2(g) + 3 H2(g) → 2 NH3(g) at 25°C
∆ngas = (final - initial) = 2 – 4 = – 2 mol
w = - ∆ngasRT = -(-2 mol)(8.314 J/mol-K)(298 K)
w = + 4955 J = + 4.96 kJ
5.3 Enthalpy (H)
Enthalpy is defined as:
H = E + PV
A change in Enthalpy would imply:
∆H = ∆E + ∆(PV) = ∆E + P∆V + V∆P
Since: ∆E = q + w; and W = - P∆V; we find:
∆H = q - P∆V + P∆V + V∆P
From this it follows that at constant P (process open to the atmosphere):
∆H = qp (equals heat at constant P)
Why Define Enthalpy?
1. We can measure or readily calculate qp. (i.e. we know the change in enthalpy)
2. Many physical and chemical changes occur at constant pressure.
3. For most reactions ∆H is similar to ∆E since the product P∆V is small.
4. It is a State Function!
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The Sign of Enthalpy
∆H < 0 (Surroundings feel hot.)
∆H > 0 (Surroundings feel cold.)
5.4 Enthalpies of Reaction
The change in enthalpy, H, is the enthalpy of the products minus the enthalpy of the reactants:
H = Hproducts − Hreactants
Notice the absolute value is independent of the direction because it is a state function.
This quantity, H, is called the enthalpy of reaction, or the heat of reaction.
The Truth about Enthalpy
1. Enthalpy is an extensive property.
2. H for a reaction in the forward direction is equal in size, but opposite in sign, to H for the reverse reaction.
3. H for a reaction depends on the state of the products and the state of the reactants.
Problem 2
When propane is burned:
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l) (ΔH = -2220 kJ)
How much heat is involved in burning 1.00 kg C3H8?
Problem 3
When 500. g O2 reacts with propane, how much heat is given off?
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l) (ΔH = -2220 kJ)
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5.5 Calorimetry
We measure H through calorimetry, the measurement of heat flow via temperature changes.
Heat Capacity (C)
The amount of heat required to raise the temperature of a substance by 1 K (1C) is its heat capacity. (An extensive property.)
J/°C or J/K or cal/°C or cal/K (etc)
It depends on the amount of substance.
Specific Heat (Cs)
We define specific heat capacity (or simply specific heat) as the amount of energy required to raise the temperature of 1 g of
a substance by 1 K. (An intensive property.) Cs
Specific heat, then, is
Problem 4
24.1 kJ is used to warm a 250. g piece of Al (Cs = 0.0902 J/g-K). What is the final temperature if the initial T = 5.0 °C?
Specific Heat Values (Table 5.2)
Notice the value for water…
Water has a very high specific heat.
Large bodies of water have a huge effect on climate.
Molar Heat Capacity (Cm)
The heat capacity of one mole of a substance.
Problem 5/Example
A 15.5 g piece of Cr is heated to 100.0 °C and then dropped into 55.5 g of water at 16.5 °C.
The final T of metal and water is 18.9 °C. Determine the specific heat of Cr.
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Constant Pressure Calorimetry
A reaction in aqueous solution. (Hrxn= qrxn)
The solution will either absorb from or give energy to the reaction. (qsol)
qrxn + qsol = 0
qsol = - qrxn
msol  Tsol  Cs sol = - qrxn
m  T  Cs = - qrxn
For water: Cs = 4.184 J/g-K
d = 1.00 g/mL
(Assume these values for dilute solutions if not provided.)
We can measure m.
Measuring Ti and Tf we calculate T.
Heat Lost = Heat Gained
Problem 6/Example
Mix 200.mL of 0.400M HCl with 200.mL 0.400M NaOH in a coffee-cup calorimeter. Ti = 25.10°C, Tf = 26.60°C. Determine Hrxn
for:
NaOH + HCl → H2O + NaCl
(Assume Cw = 4.20 J/g-°C and dw = 1.00 g/mL)
Problem 7
5.00g NaOH dissolves in 100.0g of water in a styrofoam cup. The T increases from 24.2°C to 36.9°C. What is Hrxn for:
NaOH(s) → Na+(aq) + OH–(aq)
Bomb Calorimeter
Closed, sealed, constant volume; used for Combustion Rxns.
Now: qcal = - qrxn = E (not H)
qcal = qbomb + qwater
We need to calibrate the calorimeter to know qcal.
-qrxn = Ccal  T
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Bomb Calorimeter
Since V = k (not P) we actually measure E, not H.
For most reactions, the difference is very small.
Measure T, calculate qcal, get qrxn , divide by the number of moles, adjust for coefficients.
Problem 8/Example
4.00g methylhydrazyne (CH6N2) is combusted in a bomb calorimeter. The T of the calorimeter (including water) increases
from 25.00°C to 39.50°C. The heat capacity of the entire calorimeter is 7.794 kJ/°C. Calcualte: a. the heat of combustion per
gram CH6N2.
b. the heat of combustion per mole CH6N2.
c. E of the following reaction as written:
2 CH6N2(l) + 5 O2(g) → 2N2(g) + CO2(g) + 6H2O(l)
5.6 Hess’s Law
It is a consequence of H being a state function.
“if a reaction is carried out in a series of steps, H for the overall reaction will be equal to the sum of the enthalpy changes
for the individual steps.”
Hess’s Law Problems
We use known values to calculate the enthalpy of other reactions. The rules are:
If you reverse a rxn, change the sign of ∆H.
If you multiply a rxn by a factor n, multiply ∆H by the same factor.
- If you add different rxns, add their ∆H values too.
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Example
Calculate ∆H for the reaction:
4 NH3 (g) + 5 O2 (g) → 4 NO (g) + 6 H2O (g)
from the following data.
a.
N2 (g) + O2 (g) → 2 NO (g)
∆H = -180.5 kJ
b.
N2 (g) + 3 H2 (g) → 2 NH3 (g)
∆H = -91.8 kJ
c.
2 H2 (g) + O2 (g) → 2 H2O (g)
∆H = -483.6 kJ
a.
2 N2 (g) + 2 O2 (g) → 4 NO (g)
∆H = 2 (-180.5 kJ)
b.
4 NH3 (g) → 2 N2 (g) + 6 H2 (g)
∆H = 2 (+91.8 kJ) (-1 and x2)
c.
6 H2 (g) + 3 O2 (g) → 6 H2O (g)
4 NH3 (g) + 5 O2 (g) → 4 NO (g) + 6 H2O (g)
∆H = 3 (-483.6 kJ)
∆ H = -1628. kJ
(x2)
(x3)
Problem 9
What is ∆H for: 2C(s) + 3H2(g) → C2H6(g)? Given:
1. C2H6(g) + 7/2O2(g) → 2CO2(g) + 3H2O(l)
∆H = – 1559.7 kJ
2. C(s) + O2(g) → CO2(g)
∆H = – 393.5 kJ
3. H2(g) + 1/2O2(g) → H2O(l)
∆H = – 285.8 kJ
5.7 Enthalpies of Formation
This is a special case of what we just did…
An enthalpy of formation or heat of formation, Hf, is defined as the enthalpy change for the reaction in which a compound is
made from its constituent elements in their elemental forms.
Are these formation reactions? (Why?)
CaO(s) + CO2(g) → CaCO3(s)
P4(s) + 6 Cl2(g) → 4 PCl3(l)
Example
Write the equation for the standard enthalpy of formation for 1 mol of thalidomide, which has a formula of C13H10O4N2 (s).
Include the states of all substances involved.
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Standard Enthalpies of Formation
Standard enthalpies of formation, Hf°, are measured under normal conditions (25 °C and 1.00 atm pressure). See Appendix
C (p. 1059-1061).
By definition, the standard enthalpy of formation, Hf°, for elements (in their standard states) is zero.
Hf° = 0 (elements in standard form)
Oxygen: O2(g)
Carbon: C(s, graphite)
Phosphorus: P4(s, white)
Calculation of H
C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)
•
Imagine this as occurring in three steps:
C3H8 (g)  3 C (graphite) + 4 H2 (g)
3 C (graphite) + 3 O2 (g)  3 CO2 (g)
4 H2 (g) + 2 O2 (g)  4 H2O (l)
C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)
We can use Hess’s law in this way:
H =  n Hf°products –  m Hf° reactants
where n and m are the stoichiometric coefficients.
C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)
H = [3(-393.5 kJ) + 4(-285.8 kJ)] – [1(-103.85 kJ) + 5(0 kJ)]
= [(-1180.5 kJ) + (-1143.2 kJ)] – [(-103.85 kJ) + (0 kJ)]
= (-2323.7 kJ) – (-103.85 kJ) = -2219.9 kJ
Problem 10
Calculate Hrxn° from Hf° values:
2 C2H6 (g) + 7 O2 (g)  4 CO2 (g) + 6 H2O (l)
-84.68kJ
0kJ
-393.5kJ
-285.8 kJ Hf°’s
5.8 Energy in Foods and Fuels
Most of the fuel in the food we eat comes from carbohydrates and fats. (Table 5.4)
Energy in Fuels
The vast majority of the energy consumed in this country comes from fossil fuels. (Table 5.5)
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Addendum Why Do Chemical Reactions Occur? Free Energy
Spontaneous Processes
• Processes that, once started, they proceed without any outside intervention.
• Processes that are spontaneous in one direction are nonspontaneous in the reverse direction.
Entropy (S)
• Entropy can be thought of as a measure of the randomness of a system.
• It has the unit of cal/mole·K.
Entropy Changes
• Entropy increases when
– There is a change of state
– S(g) > S(l) > S(s)
•
Entropy increases when
– Solid is dissolved.
– The number of gas molecules increases.
– The number of moles increases.
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•
•
Free energy change (DG): Free energy change is used to describe spontaneity of a process.
Exergonic: A spontaneous reaction or process (has a negative G).
Endergonic: A nonspontaneous reaction or process (has a positive G).
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