Write the equilibrium constant expression (K_c) for the reaction.

Question

Write the equilibrium constant expression (K_c) for the reaction.

Asked on March 28, 2026Chemistry

This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.

Accepted Answer · 2026-03-28T18:25:01.689Z

Step 1: Write the equilibrium constant expression ($K_c$) for the reaction. The given reaction is: $$\text{N}_2(\text{g}) + 3\text{H}_2(\text{g}) \rightleftharpoons 2\text{NH}_3(\text{g})$$ The equilibrium constant expression for this reaction is: $$K_c = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3}$$ Step 2: Substitute the given equilibrium concentrations into the expression. The given equilibrium concentrations are: $[\text{N}_2] = 0.5 \text{ mol/litre}$ $[\text{H}_2] = 0.5 \text{ mol/litre}$ $[\text{NH}_3] = 0.15 \text{ mol/litre}$ Substitute these values into the $K_c$ expression: $$K_c = \frac{(0.15 \text{ mol/litre})^2}{(0.5 \text{ mol/litre})(0.5 \text{ mol/litre})^3}$$ Step 3: Calculate the value of $K_c$. $$K_c = \frac{0.0225 \text{ (mol/litre)}^2}{(0.5 \text{ mol/litre})(0.125 \text{ (mol/litre)}^3)}$$ $$K_c = \frac{0.0225 \text{ (mol/litre)}^2}{0.0625 \text{ (mol/litre)}^4}$$ $$K_c = 0.36 \text{ (mol/litre)}^{-2}$$ $$K_c = 0.36 \text{ L}^2/\text{mol}^2$$ The value of $K_c$ is $\boxed{\text{0.36 L}^2/\text{mol}^2}$.

Accepted Answer · 2026-03-28T18:25:01.689Z

Step 1: Write the equilibrium constant expression ($K_c$) for the reaction. The given reaction is: $$\text{N}_2(\text{g}) + 3\text{H}_2(\text{g}) \rightleftharpoons 2\text{NH}_3(\text{g})$$ The equilibrium constant expression for this reaction is: $$K_c = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3}$$ Step 2: Substitute the given equilibrium concentrations into the expression. The given equilibrium concentrations are: $[\text{N}_2] = 0.5 \text{ mol/litre}$ $[\text{H}_2] = 0.5 \text{ mol/litre}$ $[\text{NH}_3] = 0.15 \text{ mol/litre}$ Substitute these values into the $K_c$ expression: $$K_c = \frac{(0.15 \text{ mol/litre})^2}{(0.5 \text{ mol/litre})(0.5 \text{ mol/litre})^3}$$ Step 3: Calculate the value of $K_c$. $$K_c = \frac{0.0225 \text{ (mol/litre)}^2}{(0.5 \text{ mol/litre})(0.125 \text{ (mol/litre)}^3)}$$ $$K_c = \frac{0.0225 \text{ (mol/litre)}^2}{0.0625 \text{ (mol/litre)}^4}$$ $$K_c = 0.36 \text{ (mol/litre)}^{-2}$$ $$K_c = 0.36 \text{ L}^2/\text{mol}^2$$ The value of $K_c$ is $\boxed{\text{0.36 L}^2/\text{mol}^2}$.