邀請中港台的高考化學精英解題

(a)
NH4HS(s) = NH3(g) + H2S(g)
This is an heterogeneous equilibrium. The concentration/pressure of NH4HS solid should NOT be considered in the equilibrium expression.

KP' = PNH3•PH2S

For the decomposition of NH4­HS(s), equal amount of NH­3 and H2S is formed.
Hence, PNH3 = PH2S.

At 10oC:
T1 = (273 + 10) K = 283 K .. and .. K1 = (0.22 atm)2 = 0.0484 atm2

At 25oC:
T2 = (273 + 25) K = 298 K .. and .. K2 = (0.64 atm)2 = 0.4096 atm2

ln(K1/K2) = -(ΔH/R)(1/T1 - 1/T2)
ln(0.0484/0.4096) = -(ΔH/8.314)(1/283 - 1/298)

ΔH = +99830 J mol-1 = +99.83 kJ mol-1


(b)
(i)
H2(g) + CO2 = CO(g) + H2O(g)

(ii)
KP = PCO•PH2O/PH2•PCO2

For Experiment 1:
KP = (70)(46)/(23)(140) = 1

For Experiment 2:
KP = (80)(48)/(32)(120) = 1

For Experiment 3:
KP = (50)(80)/(40)(100) = 1

Average KP = (1 + 1 + 1)/3 = 1
KC = KP = 1

2009-07-19 19:02:06 補充：
第一題下半部更正如下：

At 10°C: T1= 283 K
P(NH3) + P(H2S) = 0.22 atm
Hence, P(NH3) = P(H2S) = 0.11 atm
K1 = P(NH3)•P(H2S) = (0.11)^2 atm^2 = 0.0121 atm^2

At 25°C: T2= 298 K
P(NH3) + P(H2S) = 0.64 atm
Hence, P(NH3) = P(H2S) = 0.32 atm
K2 = P(NH3)•P(H2S) = (0.32)^2 atm^2 = 0.1024 atm^2

2009-07-19 19:02:25 補充：
ln(K1/K2) = -(ΔH/R)(1/T1 - 1/T2)
ln(0.0121/0.1024) = -(ΔH/8.314)(1/283 - 1/298)

ΔH = +99830 J mol^1 = +99.83 kJ mol^1

老爺子解釋的很詳盡

板大,注意一下K的定義,不難明白為什麼是平方

在NH4HS(s) = NH3(g) + H2S(g)
的氣體分子mole 是 1 : 1: 1
為何K1 和 K2要用2次方?
K1 = (0.22 atm)2 K2 = (0.64 atm)2