Unit 12
Thermodynamics – Free Response Review
PSI Chemistry
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
(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.
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-1at 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.
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
(a) (ii) ∆G° = ∆H° - T∆S°1 point
As the temperature increases, T∆S° will decrease (∆S° is negative).1 point
-T∆S° will increase, therefore ∆G° will become less negative.
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 temperature-dependent. (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°fproducts - ∆S°freactants = 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°fproducts - ∆S°freactantswith 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.
(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.
(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°298for O2 (g).
1 point is earned for the answer (with sign).
(b)∆G°rxn = ∆H°rxn - T∆S°rxn
= -283.0kJ mol-1– (298 K)(-0.0865kJ mol-1 K-1)
∆G°rxn = -257.2kJ 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))