Question

The air pollutant NO is produced in automobile engines from the high temperature reaction $N_2(g) + O_2(g) \rightleftharpoons 2NO(g); K_c=16$ at 2300 K. If the initial concentrations of $N_2$ and $O_2$ at 2300 K are both 1.5 M, what are the concentrations of $N_2$ (in M) when the reaction mixture reaches equilibrium?

• A. 0.5 M
• B. 1.0 M
• C. 1.5 M
• D. 0.75 M

Answer 1

Answer: (A)

0.5 M

Answer 2

For the reaction $N_2(g) + O_2(g) \rightleftharpoons 2NO(g)$, with $K_c=16$ and initial concentrations $[N_2]_0 = [O_2]_0 = 1.5$ M, we set up an ICE table. Let $x$ be the decrease in $[N_2]$. The equilibrium concentrations are $[N_2]_{eq} = 1.5-x$, $[O_2]_{eq} = 1.5-x$, and $[NO]_{eq} = 2x$. The equilibrium constant expression is $K_c = \frac{[NO]^2}{[N_2][O_2]}$. Substituting the equilibrium concentrations: $16 = \frac{(2x)^2}{(1.5-x)^2}$ Taking the square root of both sides gives $4 = \frac{2x}{1.5-x}$. Solving for $x$: $4(1.5-x) = 2x \implies 6 - 4x = 2x \implies 6x = 6 \implies x = 1$. The equilibrium concentration of $N_2$ is $[N_2]_{eq} = 1.5 - x = 1.5 - 1 = 0.5$ M.