AP Chemistry
Unit 8 Review Guide Key
1.
2.
3.
12.
S= entropy increasing +, H= exothermic –, and
G= spontaneous at all temps. –

S
H
only spontaneous at high temps
x (at all temps)
never spontaneous
x (spontaneous at low temp)
+
+
+
+
Temperature
low temp
high temp
high temp
low temp
°
°
°
∆𝑆𝑟𝑥𝑛
= ∑ ∆𝑆𝑝𝑟𝑜𝑑
− ∑ ∆𝑆𝑟𝑒𝑎𝑐𝑡
°
∆𝑆𝑟𝑥𝑛 = 42.63 − [. 5(205.14) + (33.15)]
𝐽
°
∆𝑆𝑟𝑥𝑛
= −93.09
𝐾 𝑚𝑜𝑙
∆𝐻
∆𝑆
119 𝑘𝐽
= .263
= 452.5 𝐾
𝐽/𝐾
4.
𝑇=
5.
(C) is the best choice, the entropy is decreasing. (B) is
not a choice as the complexity of the compound is
higher than the constituent elements it is formed from.
6.
a) ∆𝐺 ° = ∆𝐻 − 𝑇∆𝑆
8.
9.
13. a) ∆H = 4(BEC–H) + BEC=C + 3(BEO=O) – [4(BEC=O) + 4(BEO–H)]
-1323 = 4(413) + BEC=C + 3(495) – 4(799) – 4(463)
-1323 = -1911 + BEC=C
BEC=C = 588 kJ
b) ∆H = 2∆HfoCO2 + 2∆HfoH2O –∆HfoC2H4 – 3∆HfoO2
-1323 kJ = 2(-393.5) + 2(-241.8) – ∆HfoC2H4 – 3(0)
∆HfoC2H4 = +52.4 kJ
c) ∆S = 2∆SoCO2 + 2∆SoH2O – ∆SoC2H4 + 3∆SoO2]
∆S = 2(0.2136 kJ) + 2(0.1887 kJ) – [(0.2195 +3(0.2050)]
∆S = -0.0299 kJ/K
d) Tthreshold = ∆H/∆S = -1323 kJ/-0.0299 kJ/K = 44,000 K
All temperatures below 44,000 K
. 0929𝑘𝐽
= 30.84 𝑘𝐽 − 298𝐾 (
) = 3.16𝑘𝐽
𝐾
∆𝐻
30.84 𝑘𝐽
b) 𝑇 =
=
= 332𝐾
14. ∆𝐺 ° = ∆𝐻 − 𝑇∆𝑆
∆𝐺 ° = −638𝑘𝐽 − 298𝐾(.1569 𝐽/𝐾
∆𝐺 ° = −684.8 𝑘𝐽, 𝑠𝑝𝑜𝑛𝑡𝑎𝑛𝑒𝑜𝑢𝑠
a) ∆𝐻 = −, 𝑒𝑥𝑜𝑡ℎ𝑒𝑟𝑚𝑖𝑐. If reaction is endothermic in
one direction , it’s exothermic in the other.
b) ∆𝑆 = − , 𝑑𝑒𝑐𝑟𝑒𝑎𝑠𝑖𝑛𝑔. Randomness, disorder is
decreasing. Solid state is more ordered.
c) spontaneous at low tempertures.
15. 𝑞 = (𝐶𝑐𝑎𝑙 + 𝑚𝐶𝑠 )∆𝑇
𝐽
𝑞 = (837 + 1200𝑔 × 4.18
∆𝑆
7.
a) divide each coefficient by 2 -23/2
b) flip rxn, divide by 6
39/6
c) flip rxn, divide by 3
-18/3
∆𝐻 = −11 𝑘𝐽
.0929 𝑘𝐽/𝐾
K ∆𝑮° = ∑ ∆𝑮°𝒑𝒓𝒐𝒅 − ∑ ∆𝑮°𝒓𝒆𝒂𝒄𝒕
∆𝑮° = [−95.3𝑘𝐽 + (−80.3𝑘𝐽)]— 60.5 𝑘𝐽 + 0
= −115.1 𝑘𝐽
a) Br2(g)
b) C3H8(g)
c) MgO(s) (stronger ionic attraction)
d) KOH(aq)
10. ..
a) ∆𝑆 > 0
b) ∆𝑆 > 0
c) no apparent change
d) ∆𝑆 > 0
e) ∆𝑆 > 0
f) ∆𝑆 < 0
g) ∆𝑆 > 0
11.
a) spontaneous at all temp
b) spontaneous at low temp
c) spontaneous at high temp
d) spontaneous at all temp
e) nonspontaneous at all temp
°𝐶
𝐽
𝑔 °𝐶
) 8.2°𝐶
47,994 𝐽 𝑜𝑟 48 𝑘𝐽
1𝑚𝑜𝑙
1.00𝑔𝐶8 𝐻18 ×
= .0088 𝑚𝑜𝑙
114.26𝑔
48𝑘𝐽
∆𝐻𝑟𝑥𝑛 =
= 5454.5 𝑘𝐽
.0088𝑚𝑜𝑙