Question

The value of \( K_c \) for the equilibrium reaction \[ 2 {NO}_2(g) \rightleftharpoons {N}_2{O}_4(g) \] is \(2 \times 10^{-40} \, {mol}^{-1} \, {dm}^3\) at 298 K. If the equilibrium concentration of \({NO}_2\) is \(2 \times 10^{-2}\) M, the concentration of \({N}_2{O}_4\) is:

• A. \(6 \times 10^{-42} \, {M}\)
• B. \(12 \times 10^{-44} \, {M}\)
• C. \(8 \times 10^{-44} \, {M}\)
• D. \(2 \times 10^{-44} \, {M}\)
• E. \(4 \times 10^{-44} \, {M}\)

Hint:

Use the equilibrium expression to solve for unknown concentrations when given \(K_c\).

Answer 1

The Correct Option is C

The equilibrium constant \( K_c \) for the reaction is given by: \[ K_c = \frac{[{N}_2{O}_4]}{[{NO}_2]^2} \] Given that the concentration of \({NO}_2\) at equilibrium is \(2 \times 10^{-2} \, {M}\) and that \( K_c = 2 \times 10^{-40} \, {mol}^{-1} \, {dm}^3 \), we can solve for the concentration of \({N}_2{O}_4\): \[ 2 \times 10^{-40} = \frac{[{N}_2{O}_4]}{(2 \times 10^{-2})^2} \] \[ [{N}_2{O}_4] = 2 \times 10^{-40} \times (4 \times 10^{-4}) = 8 \times 10^{-44} \, {M} \]