IV. Thermodynamics (pg 5)
AP Chemistry
Second Law of Thermodynamics-Entropy
1) Predict the sign of ΔS for each of these changes.
a) O2 (g)  O2 (aq)
b) C6H6 (s)  C6H6 (l)
c) C (s) + CO2 (g)  2 CO (g)
d) C12H22O11 (s)  C12H22O11 (aq)
e) 2 NO2 (g)  N2O4 (g)
2) For each of the following pairs, choose the substance with the higher entropy per mole at a given temperature:
a) Ar (l) or Ar (g); b) He (g) at 3 atm pressure or He (g) at 1.5 atm pressure; c) 1 mole of Ne (g) in 15.0 L or 1 mole of Ne (g) in
1.50 L; d) CO2 (g) or CO2 (s).
Determining Reaction Entropy
3) Using S0 values from Appendix C, calculate ΔS0 values for the following reactions. In each case account for the sign of ΔS 0:
a) C2H4 (g) + H2 (g)  C2H6 (g)
b) N2O4 (g)  2 NO2 (g)
c) Be(OH)2 (s)  BeO (s) + H2O (g)
d) 2 CH3OH (g) + 3 O2 (g)  2 CO2 (g) + 4 H2O (g)
Determining Free Energy of a Reaction
4) Which of the following processes are spontaneous and which are non-spontaneous: a) the melting of ice cubes at –50C and 1
atm pressure; b) dissolution of sugar in a cup of hot coffee; c) the reaction of nitrogen atoms to form N 2 molecules at 250C and 1
atm; d) alignment of iron filings in a magnetic field; e) formation of CH4 and O2 molecules from CO2 and H2O at room
temperature and 1 atm of pressure?
5) Consider the vaporization of liquid water to steam at a pressure of 1 atm. a) Is this process endothermic or exothermic? b) In
what temperature range is it a spontaneous process? c) In what temperature range is it a non-spontaneous process? d) At what
temperature are the two phases in equilibrium?
6) Calculate ΔG for each process, and state if the process is spontaneous or non-spontaneous.
a) ΔH = 147 kJ, T = 422 K and ΔS = -67 J/K
b) ΔH = -43 kJ, T = 21 oC, and ΔS = - 118 J/K
c) ΔH = 227 kJ, T = 574 K, and ΔS = 349 J/K
7) For a certain chemical reaction, ΔH0 = - 35.4 kJ and ΔS0 = - 85.5 J/K. a) Is the reaction exothermic or endothermic? b) Does
the reaction lead to an increase or decrease in the disorder of the system? c) Calculate ΔG 0 for the reaction at 298 K. d) Is the
reaction spontaneous at 298 K?
8) Calculate the free energy change, ΔG, for the combustion of hydrogen sulfide according to the following chemical equation.
Assume reactants and products are at 298 K.
H2S (g) + O2 (g)  H2O (l) + SO2 (g) where ΔH = - 562.1 kJ/mole and ΔS = - 0.09278 kJ/mole K
9) Calculate the free energy change for the decomposition of sodium chlorate. Assume reactants and products are at 298 K.
NaClO3 (s)  NaCl (s) + O2 (g)
where ΔH = -19.1 kJ/mole and ΔS = 0.1768 kJ/mole K
10) Calculate the free energy change for the combustion of 1 mole of ethane. Assume reactants and products are at 298 K.
C2H6 (g) + O2 (g)  2 CO2 (g) + 3 H2O (l)
where ΔH = - 1561 kJ/mole and ΔS = 408.4 J/mole K
11) Using data in Appendix C, calculate ΔH0, ΔS0, and ΔG0 at 298 K for each of the following reaction. In each case show that
ΔG0 = ΔH0 - T ΔS0.
a) H2 (g) + F2 (g)  2 HF (g)
b) C (s, graphite) + 2 Cl2 (g)  CCl4 (g)
c) 2 PCl3 (g) + O2 (g)  2 POCl3 (g)
d) 2 CH3OH (g) + H2 (g)  C2H6 (g) + 2 H2O (g)
12) Explain qualitatively how ΔG changes for each of the following reactions as the partial pressure of O2 is increased:
a) 2 CO (g) + O2 (g)  2 CO2 (g)
b) 2 H2O2 (l)  2 H2O (l) + O2 (g)
c) 2 KClO3 (s)  2 KCl (s) + 3 O2 (g)
Relationship between Free Energy (∆G) and Keq
13) Use data from Appendix C to calculate the standard free energy change, ∆Go, and the equilibrium constant, Keq, at 298 K for
the following reaction: H2 (g) + Br2 (g) ↔ 2 HBr (g).
14) If the standard free energy of formation for HI from H 2 and I2 at 490 oC is –12.1 kJ/mole of HI, what is the equilibrium
constant for this reaction?
15) Phosgene is formed at 25 oC according to the equation
CO (g) + Cl2 (g)  COCl2 (g)
Calculate the free energy change ∆Go and the equilibrium constant, Keq, for this reaction.
Answers
1) a) Positive
b) positive
c) positive
2) a) Ar (g)
b) He (g) at 1.5 atm
c) 1 mole of Ne (g) in 15.0 L
3) a) -120. J/K
b) 177 J/K
c) 152 J/K
4) a) non-spontaneous
b) spontaneous
5) a) endothermic
b) above 100 oC
6) a) 175 kJ (non-spontaneous)
d) positive
c) spontaneous
e) negative
d) CO2 (g)
d) 92.3 J/K
d) spontaneous
c) below 100oC
e) non-spontaneous
d) at 100oC
b) -8.31 kJ (spontaneous) c) 26.6 kJ (non-spontaneous)
7) a) exothermic b) decrease in disorder
c) ∆G = -9920 J d) spontaneous
8) ∆G = -534 kJ
9) ∆G = -71.8 kJ
10) ∆G = -1680 kJ
11)
A
B
C
D
∆H (kJ)
-537
-107
-508
-166
∆S (J/K)
13.7
-142
-178
1.38
∆G (kJ)
-541
-64.0
-466
-166
∆G = ∆H - T∆S (kJ)
-541
-64.0
-455
-169
12) Use Le Chatelier’s Principle a) shifts to the right so ∆G negative b) shifts to the left, so ∆G positive
c) shifts to the left, so ∆G is positive
13) ∆H = -103 kJ, ∆S = 21.1 J/K, ∆G = -109 kJ, Keq = 1.54 x 1019
14) Keq = 45.4 because you need to multiply ∆G by 2 since you have 2 moles of HI
15) ΔGo = COCl2 = -73.1 kJ/mole; Keq= 6.51 x 1012