Question #174969

The initial partial pressures are PA = 0.000 torr, PB = 0.000 torr, and

PC = 150. torr, and the total pressure at equilibrium is 200.0 torr. What are

the values for Kp and Kc at 800. K temperature?


1
Expert's answer
2021-03-25T04:08:19-0400

A(g) + B(g) \leftrightarrow C(g)

If total pressure is 200 torr, then PA = x torr, PB = x torr, PC = (150 - x) torr

200 = 2x + (150 - x)

x = 50

Kp = PCPAPB\dfrac{P_C}{P_AP_B} =10050×50\dfrac{100}{50\times50} = 0.04 (torr-1)

Kc = Kp(RT)Δn\dfrac{Kp}{(RT)^{\Delta n}} =0.04(8.134×800)1\dfrac{0.04}{(8.134\times800)^{-1}} = 266.05


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