Unit 12
Thermodynamics – Free Response Review
PSI Chemistry
1.
2NO (g) + O2 (g) → 2NO2 (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) For the reaction at 25°C, the value of the standard free-energy change, ∆G°, is -70.4
kJ.
(ii) Indicate whether the value of ∆G° would become more negative, less negative, or
remain unchanged as the temperature is increased. Justify your answer.
(b) Use the data in the table below to calculate the value of 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.
(a) The reaction between nitrogen and hydrogen to form ammonia is represented below.
N2 (g) + 3H2 (g) → 2NH3 (g)
∆H° = -92.2 kJ
Predict the sign of the standard entropy change, ∆S°, for the reaction. Justify your
answer.
(b) The value of ∆G° for the reaction represented in part (b) is negative at low
temperatures but positive at high temperatures. Explain.
3.
2Fe (s) +
O2 (g) → Fe2O3 (s)
∆H°f = -824 kJ mol-1
Iron reacts with oxygen to produce iron (III) oxide, as represented by the equation
above.
(a) The standard free energy of formation, ∆G°f of Fe2O3 (s) is -740 kJ mol-1 at 298 K.
(i) Calculate the standard entropy of formation, ∆S°f, of Fe2O3 (s) at 298 K. Include
units with your answer.
(ii) Which is more responsible for the spontaneity of the formation reaction at 298 K,
the standard enthalpy of formation, ∆H°f, or the standard entropy of formation,
∆S°f ? Justify your answer.
4.
CO (g) +
O2 (g) → CO2 (g)
∆H°rxn = -283.0 kJ mol-1
The combustion of carbon monoxide is represented by the equation above.
(a) Determine the value of the standard entropy change, ∆S°rxn, for the combustion of
CO (g) at 298 K using the information in the following table.
Substance
S°298
(J mol-1 K-1)
CO (g)
197.7
CO2 (g)
213.7
O2 (g)
205.1
(b) Determine the standard free energy change, ∆G°rxn, for the reaction at 298 K.
Include units with your answer.
(c) Is the reaction spontaneous under standard conditions at 298 K? Justify your
answer.
Free Response Answers
1.
(a) (ii) ∆G° = ∆H° - T∆S°
As the temperature increases, T∆S° will decrease (∆S° is negative).
-T∆S° will increase, therefore ∆G° will become less negative.
1 point
1 point
Note: The first point is earned for using ∆G° = ∆H° - T∆S°, and the second point is
earned for the discussion and the correct conclusion. Students should focus on the sigh
of the ∆S° term. One point can be earned for a response discussing spontaneity with
reference to ∆H°, although this is not entirely correct because K is temperaturedependent. (Since ∆H° is negative, increasing the temperature sends the reaction to the
left, therefore the reaction Is less spontaneous, therefore ∆G° must be getting more
positive.)
(b)
∆S°rxn = ∆S°f products - ∆S°f reactants = 2 (SNO2) – [2(SNO) + SO2) ] 1 point
-146.5 J K-1 = 2(240.1) J K-1 – [2(210.8] J K-1 + SO2 ]
SO2 = 205.1 J K-1 = standard molar entropy of O2
1 point
Note: The first point is earned for correctly using the equation
∆S°rxn = ∆S°f products - ∆S°f reactants with the correct coefficients from the reaction
equation. (All four terms must be in the correct position in the equation.) The second
point is earned for calculating the correct answer.
2.
(a)
∆S° is negative. There are fewer moles of product gas (2 mol) compared
to reactant gases (4 mol), so the reaction is becoming more ordered.
1 point for correct sign
1 point for indicating fewer moles of products compared to reactants (in the gas
phase)
(b)
∆G° = ∆H° - T∆S°
∆H° and ∆S° are negative. At low temperatures, the T∆S° term is smaller than ∆H°, and
∆G° is negative. At high temperatures, the T∆S° term is higher than ∆H°, and ∆G° is
positive.
1 point each for using ∆G° = ∆H° - T∆S° to explain the sign of ∆G° at high and low
temperatures.
3.
(a) (i) ∆G°f = ∆H°f - T∆S°f
-740 kJ mol-1 = -824 kJ mol-1 – (298 K) ∆S°f + 84 kJ mol-1 = -(298 K) ∆S°f
∆S°f =
= -0.28 kJ mol-1 K-1
1 point for calculation of ∆S°f
1 point for correct units
(ii) ∆H°f is the more important factor. The reaction is exothermic, which favors
spontaneity. ∆S°f is negative, which means the system becomes more ordered as the
reaction proceeds. Greater order will not increase the spontaneity of the reaction.
1 point for indicating that ∆H°f is responsible and for an explanation that
addresses the signs of ∆H° and ∆S°
4.
(a)
∆S°rxn = 213.7 J mol-1 K-1 – (197.7 J mol-1 K-1 +
(205.1 J mol-1 K-1))
= -86.5 J mol-1 K-1
1 point is earned for taking one-half of S°298 for O2 (g).
1 point is earned for the answer (with sign).
(b)
∆G°rxn = ∆H°rxn - T∆S°rxn
= -283.0 kJ mol-1 – (298 K)(-0.0865 kJ mol-1 K-1)
∆G°rxn = -257.2 kJ mol-1
1 point is earned for substituting the values from parts (a) and (b) into the
equation.
1 point is earned for the answer (with sign and units)
(c)
Yes, the reaction is spontaneous because the value of ∆G°rxn for the
reaction is negative (-257.2 kJ mol-1)
1 point is earned for an answer with justification (consistent with the answer in
part (c))