Answer to Question #170714 in General Chemistry for Luke Walbourne

Question

Initially, 1.0 mol of NO(g) and 1 mol of Cl₂(g) were added to a 1 L container. As a result of the reaction the equilibrium concentration of NOCl(g) became 0.96 M. Using the ICE table methodology determine the value of the equilibrium constant KC for this reaction. . 2NO(g) + Cl₂(g) ⇋2NOCl(g)

Your answer

Answer 1

2NO(g) + Cl₂(g) $\leftrightarrow$ 2NOCl

1mol/L____1mol/L____0mol/L

-2x______-x_________+2x

[NOCl]_eq = 2x = 0.96M

x = 0.48 M

Kc = $\frac{[NOCl]^2}{[NO]^2[Cl]}$ = $\frac{[0.48]^2}{[1-2x]^2[1-x]}$

Kc = $\frac{0.2304}{[1-0.96]^2[1-0.48]}$ = $\frac{0.2304}{[0.04]^2[0.52]}$

Kc = $\frac{0.2304}{(0.0016)×0.52}$ = $\frac{0.2304}{0.000832}$ = 276.92