Ch. 5 Thermochemistry (7 problems)
5.14) (a) Esys = -Esurr
(b) We do not need to measure the internal energy, only the change in internal
energy.
(c) The value of q will be negative if heat is transferred from system to
surroundings. The value of w will be negative if work is done by the system on
the surroundings.
5.16) (a) E = q + w = 670 J – 332 J = +338 J
endothermic change
5.18) (a)
w
q
system
(b) Given that E = q + w, if the work done on the system is greater in magnitude
than the heat lost by the system, then E will be positive.
(c) Given that E = q + w, if the heat lost by the system is greater in magnitude
than the work done on the system, then E will be negative.
5.28) (a) Since H is positive, the reaction is endothermic.
(b) 45.0 g CH3OH x (1 mol CH3OH/32.042 g CH3OH) x (90.7 kJ/1 mol CH3OH)
= 127 kJ
(c) 16.5 kJ x (2 mol H2/90.7 kJ) x (2.016 g H2/1 mol H2) = 0.733 g H2
(d) 10.0 g CO x (1 mol CO/28.01 g CO) x (-90.7 kJ/1 mol CO) = -32.4 kJ
5.40) q = mCT = (40.0 g)(1.13 J/gC)(28.0 C – 10.4 C) = 796 J
5.54) flip #2, divide by 2 NO2(g)  NO(g) + 1/2O2(g)
divide #3 by 2
N2O(g)  N2(g) + 1/2O2(g)
keep #1 the same + N2(g) + O2(g)  2NO(g)
N2O(g) + NO2(g)  3NO(g)
H = +56.55 kJ
H = -81.6 kJ
+ H = +180.7 kJ
H = 155.7 kJ
5.60) 2C4H10(l) + 13O2(g)  8CO2(g) + 10H2O(l)
H = [(8 mol)(-393.5 kJ/mol) + (10 mol)(-285.83 kJ/mol)] – [(2 mol)(-147.6
kJ/mol) + 0] = -5711.1 kJ
1.0 g C4H10 x (1 mol C4H10/58.12 g C4H10) x (-5711.1 kJ/2 mol C4H10) = -49 kJ
Ch. 19 Chemical Thermodynamics (11 problems)
Checked 19.66 for work.
19.6) (a) Freezing is an exothermic process.
(b) The freezing of hexane is spontaneous below -95C.
(c) The freezing of hexane is nonspontaneous above -95C.
(d) At -95C hexane liquid and solid are in equilibrium (no preferred direction).
19.8) (a) An irreversible process is one that cannot be returned to its original state by
the same path that the forward process took place. All spontaneous processes are
irreversible.
(b) Because the process is irreversible, the pathway by which the system returned
to its original state must have different values of q and w than the forward
process. As a result, the surrounding will have undergone a net change.
19.22) (a) Suniv  0
(b) If Ssystem < 0, then Ssurr > 0 to satisfy equation (a) for an irreversible
process.
(c) Suniv = 0 for a reversible process.
Suniv = Ssystem + Ssurr = Ssystem + 34 J/K = 0
Ssystem = -34 J/K
19.28) (a) Ssystem is negative, since freezing occurs.
(b) Ssystem is negative, since there are fewer molecules of gas in the products.
(c) Ssystem is positive, since a gas is produced.
(d) Ssystem is negative, since a solid is formed.
19.34) (a) S = (1 mol)(94.6 J/molK) – [(1 mol)(192.5 J/molK) + (1 mol)(186.69
J/molK)] = -284.6 J/K
There are fewer moles of gas in the products.
(d) S = [(1 mol)(229.5 J/molK) + (1 mol)(130.58 J/molK)] – (2 mol)(186.3
J/molK) = -12.5 J/molK
The change is small due to the number of molecules of gas remaining the same;
thus, the formation of a simple molecule such as H2 with fewer degrees of
freedom lowers the entropy.
19.38) (a) endothermic rxn (H is +)
(b) disorder of system increases (S is +)
(c) G = H - TS = 15.4 kJ – (298 K)(0.0647 kJ/K) = -3.9 kJ
(d) Since the G is negative, the rxn is spontaneous.
19.50) (a) H = [(1 mol)(-393.5 kJ/mol) + (1 mol)(-74.8 kJ/mol)] – (-487.0 kJ/mol) =
8.7 kJ
S = [(1 mol)(213.6 J/molK) + (1 mol)(1186.3 J/molK)] – (1 mol)(159.8
J/molK) = 240.1 J/K = 0.2401 kJ/K
G = H - TS
0 > 18.7 kJ – T(0.2401 kJ/K)
T > 77.9 K
19.52) (a) H = (1 mol)(-201.2 kJ/mol) – [(1 mol)(-110.5 kJ/mol) + 0] = -90.7 kJ
S = (1 mol)(-237.6 J/molK) – [(1 mol)(197.9 J/molK) + (2 mol)(130.58
J/molK)] = -221.5 J/ K
(b) Since H is negative (favorable) and S is positive (unfavorable), as
temperature increases, G will increase (unfavorable).
(c) G = H - TS = -90.7 kJ – (298 K)(-0.22146 kJ/K) = -24.7 kJ
rxn is spontaneous
(d) G = H - TS = -90.7 kJ – (500 K)(-0.22146 kJ/K) = +20.0 kJ
rxn is nonspontaneous
19.58) (a) more H2 (reactant)  Q decreases  For G = G + RT lnQ, G decreases
which favors the forward rxn
(b) more H2 (product)  Q increases  For G = G + RT lnQ, G increases
which favors the reverse rxn
(c) more H2 (reactant)  Q decreases  For G = G + RT lnQ, G decreases
which favors the reverse rxn
19.60) (a) G = (2 mol)(-270.70 kJ/mol) – [(1 mol)(0) + (1 mol)(0)] = -541.40 kJ
(b) G = G + RT lnQ
19.66) (a) CH3NH2(aq) + H2O(l)  CH3NH3+(aq) + OH-(aq)
(b) G = -RT lnKb = -(8.31 J/molK)(298 K)ln(4.4 x 10-4) = 19,000 J/mol
(c) G = 0
(d) [H+] = 3.0 x 10-10 M, [OH-] = 1.0 x 10-14/3.0 x 10-10 = 3.33 x 10-5 M,
[CH3NH3+] = 8.0 x 10-3 M, [CH3NH2] = 0.070 M
Q = [CH3NH3+][OH-]/[CH3NH2] = (8.0 x 10-3)(3.33 x 10-5)/(0.070) = 3.806 x 10-6
G = G + RT lnQ = 19,148 J/mol + (8.31 J/molK)(298 K) ln(3.806 x 10-6) =
-1200 J/mol