2.50 mol NOCl was placed in a 2.50 L reaction vessel at 400ºC. After equilibrium was established, it was found that 28% of the NOCl had dissociated according to the equation 2NOCl(g) $\rightleftharpoons$ 2NO(g) + Cl₂(g).Calculate the equilibrium constant, K_c, for the reaction.

Question

Quizplus · Accepted Answer

The equilibrium constant, Kc, is given by the expression:Kc = [NO]^2[Cl2]/[NOCl]^2At equilibrium, the concentration of NOCl is:[NOCl] = (1 - 0.28) * 2.50 mol / 2.50 L = 1.80 mol/LThe concentration of NO and Cl2 is:[NO] = [Cl2] = 0.28 * 2.50 mol / 2.50 L = 0.70 mol/LSubstituting these values into the expression for Kc, we get:Kc = (0.70 mol/L)^2 / (1.80 mol/L)^2 = 0.021

250 Mol NOCl Was Placed in a 2 ⇌\rightleftharpoons⇌ 2NO(g) + Cl2(g)

250 Mol NOCl Was Placed in a 2 $\rightleftharpoons$ 2NO(g) + Cl₂(g)