Unit 10 - Thermodynamics
Section 2 – Entropy and Free Energy
1.
2 PCl3(g) + O2(g) → 2 POCl3(g)
Thermodynamic values related to the above reaction are given in the table below.
a. Determine the enthalpy change for the above reaction.
H°rxn = [2(–542.2)] – [2(–287) + 1(0)] = –510.kJ
b. Estimate the PO bond energy.
The answer from part a equals the bonds broken minus the bonds formed. Both phosphorus molecules have
three P–Cl bonds, and O2 has an O=O bond.
[(2 x 3 x 331) + (498)] – [(2 PO) + (2 x 3 x 331)] = –510 kJ
PO = 504 kJ
c.
Is the PO bond a single or a double bond? Justify your answer.
It is a double bond. The value from part (b) is much higher than the single-bond values from the table.
d. Calculate the entropy change for the reaction.
S°rxn = [2(325)] – [2(312) + 1(205.0)] = –179 J/K
e.
Is this reaction spontaneous or nonspontaneous at 25°C? Justify your prediction.
The free-energy change must be calculated.
G°rxn = H°rxn – TS°rxn = –510.kJ – (298 K)(1 kJ/1000 J)(–179J/K) = –457 kJ
The negative value means the reaction is spontaneous.
2.
Answer the following question in terms of thermodynamic principles and concepts of kinetic molecular theory.
a. Consider the reaction represented below, which is spontaneous at 298 K.
CO2(g) + 2 NH3(g)  CO(NH2)2(s) + H2O(l) ∆H˚298 = – 134 kJ
i. For the reaction, indicate whether the standard entropy change, ∆S˚ 298 is positive, negative, or
zero. Justify your answer.
∆S˚ is negative because three molecules of reactant make two molecules of product (a decrease
in entropy) and two different gases make a solid and a liquid (both with smaller entropies).
ii. Which factor, the change in enthalpy, ∆H˚298, or the change in entropy, ∆S˚298, provides the
principle driving force for the reaction at 298 K? Explain.
natural tendency to maximize entropy and since this reaction decreases entropy and is
spontaneous (-∆G˚), then ∆H˚ must be negative to overcome the entropy change and drive this
reaction.
iii. For the reaction, how is the value of the standard free energy change, ∆G˚, affected by an
increase in temperature? Explain.
∆G˚ = ∆H˚ –T∆S˚ A T increase the value of –T∆S° increases and the value of –∆G˚
becomes a smaller negative number (i.e., moves toward zero).
b. Some reactions that are predicted by their sign of ∆G˚ to be spontaneous at room temperature do not
proceed at a measurable rate at room temperature.
i. Account for this apparent contradiction.
The sign of ∆G˚ (thermodynamics) does not account for activation energy (kinetics); a
large activation energy would effectively prevent a reaction even though there is a
favorable free energy change.
ii. A suitable catalyst increases the rate of such a reaction. What effect does the catalyst have on
∆G˚ for the reaction? Explain.
A catalyst changes neither the ∆H˚ nor the ∆S˚ for a reaction, therefore, it will have no
effect on the ∆G˚.
3.
2 NO(g) + O2(g)  2 NO2(g)
H°= -114.1 kJ, S°= -146.5 J K-1
The reaction represented above is one that contributes significantly to the formation of photochemical smog.
a. Calculate the quantity of heat released when 73.1 g of NO(g) is converted to NO 2(g).
73.1 g NO 1 mol NO

 2.44 mol NO
1
30 g NO
b.
For the reaction at 25C, the value of the standard free-energy change, G, is -70.4 kJ.
i. Calculate the value of the equilibrium constant, Keq, for the reaction at 25C.
ΔG° = - RT ln K
K = 2.19 x1012
ii.
so
- 70,400 J/(8.314)(298 K) = - ln K = 28.41
Indicate whether the value of G would become more negative, less negative, or remain
unchanged as the temperature is increased. Justify your answer.
When temperature goes up the ΔH° is negative and ΔS° is also negative.
c.
Use the data in the table below to calculate the value of the standard molar entropy, S, for O2(g) at
25C.
Standard Molar Entropy, S
(J K-1 mol-1)
NO(g)
210.8
NO2(g)
240.1
2 NO + O2  2 NO2
210.8
240.1
- 146.5 J = [480.2] - [421.6 + x]
ΔS = - 146.5 J/K
x = S°O2 = 205.1 J
d. Use the data in the table below to calculate the bond energy, in kJ mol -1, of the nitrogen-oxygen bond
in NO2 . Assume that the bonds in the NO2 molecule are equivalent (i.e., they have the same energy).
Bond Energy
(kJ mol-1)
Nitrogen-oxygen bond in NO
607
Oxygen-oxygen bond in O2
495
Nitrogen-oxygen bond in NO2
?
2 NO + O2 => 2 NO2
[2 (607) + (495)] [- 2X] = - 114.1 kJ
2 X = 1823.1 kJ
X = 911.55 kJ
ΔH° = - 114.1 kJ
2 equal bonds so each bond is 456 kJ