Question

At T(K), the equilibrium constant for the reaction \[ {N}_2(g) + 3{H}_2(g) \rightleftharpoons 2{NH}_3(g) \] is \(6 \times 10^{-2}\). At equilibrium, if the molar concentrations of \({H}_2\) and \({NH}_3\) are 0.25 M and 0.06 M respectively, the equilibrium concentration of dinitrogen (in mol L\(^{-1}\)) is:

• A. 2.84
• B. 3.84
• C. 8.34
• D. 4.82

Hint:

Use equilibrium expression and known concentrations to find unknown equilibrium concentrations.

Answer 1

The Correct Option is B

Equilibrium constant expression: \[ K = \frac{[{NH}_3]^2}{[{N}_2][{H}_2]^3} \] Rearranged to find \([{N}_2]\): \[ [{N}_2] = \frac{[{NH}_3]^2}{K [{H}_2]^3} \] Substitute values: \[ [{N}_2] = \frac{(0.06)^2}{6 \times 10^{-2} \times (0.25)^3} = \frac{0.0036}{6 \times 10^{-2} \times 0.015625} = \frac{0.0036}{0.0009375} = 3.84 \, {mol L}^{-1} \]