Chem AL question Kp

[匿名]

2010-04-23 8:31 pm

SO2(g) and O2(g) were mixed in the mole ratio of 3:1 at 1000K in the presence of a catalyst. When equilibrium was attained at 272 kPa pressure, one half of the SO2(g) had been converted into SO3(g).

Calculate Kp for the above reaction under the given condition.

回答 (1)

Uncle Michael

2010-04-23 9:36 pm

✔ 最佳答案

Let the initial no. of moles of SO2 added = 3n mol
Then the initial no. of moles of O2 added = n mol

2SO2(g) + O2(g) = 2SO3(g)

In the system at equilibrium:
No. of moles of SO2 = 3n x [ 1 - (1/2)] = 1.5n mol
No. of moles of O2 = n - 1.5n x (1/2) = 0.25n mol
No. of moles of SO3 = 1.5n mol
Total no. of moles = 1.5n + 0.25n + 1.5n = 3.25n mol

At equilibrium, by Dalton's law of partial pressure:
Partial pressure of SO2, PSO2 = 272 x (1.5n/3.25n) = 408/3.25 kPa
Partial pressure of O2, PO2 = 272 x (0.25n/3.25n) = 68/3.25 kPa
Partial pressure of SO2, PSO3 = 272 x (1.5n/3.25n) = 408/32.5 kPa

Equilibrium constant, Kp
= (PSO3)² / (PSO2)²(PO2)
= (408/3.25)² / (408/3.25)² (68/3.25)
= 0.0478 /kPa

參考: Uncle Michael

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