Question

At 773k the equilibrium constant Kc for the reaction is N2+3H2_2NH3 6.02×10^-2L^2/mol^2.Calculate the value of Kp at the same temperature

Answer 1

Kp at the same temperature will be - 1.5*10^-5 atm-2

Kp and Kc are related as follows -

Kp = Kc*(R*T)^∆n

Here, Kp represents equilibrium constant in terms of pressure,

kc represents equilibrium constant in terms of molarity, R represents gas constant, t is for temperature and ∆n denotes change in the number of moles.

Given values in question are -

Kc = 6.02×10^-2 L^2/mol^2

∆n = 2NH3 - N2+3H2

∆n = 2-4 = -2.

T = 773 K

R = 0.082 L atm mol-1 K-1

Keeping the values in equation -

Kp = 6.02×10^-2*(0.082*773)^-2

Kp = 1.5*10^-5 atm-2

Thus, value of equilibrium constant in terms of pressure is 1.5*10^-5 atm-2.