At a certain temperature, the equilibrium constant $(K_c)$ is $16$ for the reaction $SO_{2(g)} + NO_

Question

At a certain temperature, the equilibrium constant $(K_c)$ is $16$ for the reaction $SO_{2(g)} + NO_{2(g)} ightleftharpoons SO_{3(g)} + NO_{(g)}$. If $1.0 	ext{ mole}$ of each of the four gases is taken in a $1.0 	ext{ L}$ container, what would be the equilibrium concentration of $NO_{(g)}$?

Professor PC Thomas Classes · Answer

The initial concentration of each gas is $1.0 	ext{ M}$. The reaction quotient $Q_c = \frac{(1.0)(1.0)}{(1.0)(1.0)} = 1$. Since $Q_c < K_c = 16$, the reaction proceeds in the forward direction. Let $x$ moles of reactants convert to products. At equilibrium, $[SO_2] = 1-x$, $[NO_2] = 1-x$, $[SO_3] = 1+x$, $[NO] = 1+x$. $K_c = \frac{(1+x)^2}{(1-x)^2} = 16$. Taking the square root of both sides gives $\frac{1+x}{1-x} = 4 \implies 1+x = 4-4x \implies 5x = 3 \implies x = 0.6$. The equilibrium concentration of $NO_{(g)}$ is $1 + 0.6 = 1.6 	ext{ M}$.