✔ 最佳答案
5.
The answer is : a) ΔrH° < 0 and loweringthe temperature favours products
At T1 = 25°C = 298 K, K1 = 7.1
At T2 = 70°C = 343 K, K2 = 5.5
A form of Arrhenius equation:
ln(K1/K2) = -(ΔrH°/R)(1/T1- 1/T2)
ln(7.1/5.5) = -(ΔrH°/R)(1/298 - 1/343)
ΔrH° = -580R (J mol⁻¹ K⁻¹)
Hence, ΔrH° < 0
The forward reaction is exothermic, and the backward reaction is endothermic.
When the temperature is lowered, by Le Chatelier's principle, it favours the exothermicreaction and thus favours products (forward reaction).
..... Choose a)
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6.
The answer is : d) 3.7
Refer to Q.5.
At 25°C, PN2O4/(PNO2)^2= 7.1
(1/2)N2O4(g) ⇌ NO2(g)
Kp at 25°C
= PNO2/(PN2O4)^1/2
= √{1/[PN2O4/(PNO2)^2]}
= √(1/7.1)
= 0.38 ...... Choose d), the best answer.
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7.
The answer is : c) +63 kJ mol⁻¹
6H2(g) + 6C(s) → C6H12(l) .. ΔH° = -123 kJ mol⁻¹
C6H12(l) → C6H6(l) + 3H2 .. ΔH° = -(-206 kJ mol⁻¹)
Add the above two thermochemical equations together and cancel C6H12(l)and 3H2(g) on both sides.
3H2(g) + 6C(s) → C6H6(l) .. ΔHf°298
ΔHf°298 = -123 - (-206) = +83 kJ mol⁻¹ ...... Choose c)
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9.
The answer is : a) 3.0
2A(g) ⇌ 3B(g)
Initial concentrations:
[A]o = 2.0/10 = 0.2 mol/L
[B]o = 3.0/10 = 0.3 mol/L
At eqm.
[A] = 0.14 mol/L
Change in [A] = 0.14 - 0.20 = -0.06 mol/L
Change in [B] = +0.06 ´ (3/2) = +0.09 mol/L
[B] = 0.30 + 0.09 = 0.39 mol/L
Kc = [B]³/[A]² = 0.39³/0.14² = 3.0 ...... Choose a)
