Question

11) a) Calculate Kp for the reaction C(s) + H2O(g) = CO(g) + H2(g) at 990 K if the equilibrium
concentrations are as follows: [H2O) = 1.10M, [CO] = [Hz] = 0.2M, R = 0.08206 L atm k^-1 mol^-1​

Answer 1

hence value of Kp for the given reaction is 3atm

first find K_c from the given data,

chemical reaction is .... C(s) + H2O(g) = CO(g) + H2(g)

K_c = [H2][CO]/[H2O]

= 0.2M × 0.2M/1.1M

= 0.04/1.1 M

= 4/110 M

as ∆n_g = number of moles of gaseous products - number of moles of gaseous reactants

= 1 + 1 - 1 = 1

now using formula, K_P = K_c(RT)^∆n_g

= 4/110 M × (RT)¹

= 4/110 mol/L × 0.08206L.atm/K/mol × 990 K

= 2.95214 atm ≈ 3atm

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