The equilibrium constant (Kp) for the reaction N2(g) + 3H2(g) ⇌ 2NH3(g) is 1.64 x 10^-4 at 400°C. What is the value of Kc for this reaction? (R = 0.0821 L atm mol^-1 K^-1)
Correct: C
Kp = Kc(RT)^Δn, where Δn = (moles of gaseous products) - (moles of gaseous reactants) = 2 - (1+3) = -2. So, Kc = Kp / (RT)^Δn = 1.64 x 10^-4 / (0.0821 * 673)^-2. Solving results in Kc = 1.64 x 10^-4 / ((0.0821 * 673) ^ -2)