Find Kc and Kp at equilibrium

Question

Consider the equilibrium system described by the chemical reaction below. At equilibrium, a 5.0 L reaction vessel contained a mixture of 1.2 mol CH4, 2.8 mol H2S, 0.63 mol CS2 and 0.042 mol H2 at 93.0 celsius. What are the values of Kc and Kp for the reaction at this temperature.

CH4 (g) + 2 H2S (g) -> CS2 (s) +4 H2 (g)

Answer 1

CH4 (g) + 2 H2S (g) -> CS2 (s) +4 H2 (g)

Let's find the concentrations of the species:

[CH4] = 1.2 mol / 5 L = 0.24 M

[H2S] = 2.8 mol / 5 L = 0.56 M

[CS2] = 0.63 mol / 5 L= 0.126 M

[H2] = 0.042 mol / 5 L = 0.0084 M

Writing the Kc expression we have

Kc = [CS2][H2]⁴ / [CH4][H2S]²

Plugging in values and solving for Kc, we have

Kc = (0.126)(0.0084)⁴ / (0.24)(0.56)²

Kc = 8x10^-8

To find Kp, we can use the equation showing the relationship between Kp and Kc, which is...

Kp = Kc(RT]^∆n

R = gas constant = 0.0821 Latm/Kmol

T = temp in Kelvin = 93C + 273 = 366K

∆n = change in number of moles = 5 - 3 = 2

Kp = 8x10^-8 (0.0821*366)²

Kp = 8x10^-8 (903)

Kp = 7.2x10^-5