Question

The equilibrium constant ($K_c$) at a certain temperature for a reaction, N$_2$(g) + O$_2$ (g) $\rightleftharpoons$ 2NO (g) is 0.04. If we take 1.5 mole of NO and 2 mole each of N$_2$ and O$_2$ in a 5 litre container, what would be the equilibrium concentration of NO in mol/litre ?

• A. 0.1 mol/litre
• B. 0.3 mol/litre
• C. 0.4 mol/litre
• D. 0.5625 mol/litre

Answer 1

Answer: (A)

0.1 mol/litre

Answer 2

1. Calculate initial concentrations: [N$_2$] = 2 mol / 5 L = 0.4 M, [O$_2$] = 0.4 M, [NO] = 1.5 mol / 5 L = 0.3 M.
2. Calculate the reaction quotient $Q_c = \frac{[\text{NO}]^2}{[\text{N}_2][\text{O}_2]} = \frac{(0.3)^2}{(0.4)(0.4)} = 0.5625$.
3. Since $Q_c > K_c$ (0.5625 > 0.04), the reaction will shift in the reverse direction (towards reactants) to reach equilibrium.
4. Let the equilibrium concentrations be: [N$_2$] = 0.4 + x, [O$_2$] = 0.4 + x, [NO] = 0.3 - 2x.
5. Substitute these into the $K_c$ expression: $K_c = \frac{(0.3 - 2x)^2}{(0.4 + x)^2} = 0.04$.
6. Taking the square root of both sides gives $\frac{0.3 - 2x}{0.4 + x} = 0.2$.
7. Solving this equation for x yields $x = 0.1$.
8. The equilibrium concentration of NO is [NO]$_{eq} = 0.3 - 2x = 0.3 - 2(0.1) = 0.1$ M.