Question

At 1000 K, the equilibrium constant for the reaction
\(\text{CO}_2(\text{g}) + \text{H}_2(\text{g}) \rightleftharpoons \text{CO}(\text{g}) + \text{H}_2\text{O}(\text{g})\) is 0.53. In a one litre vessel, at equilibrium the mixture contains 0.25 mole of CO, 0.5 mole of \(\text{CO}_2\), 0.6 mole of \(\text{H}_2\) and \(x\) moles of \(\text{H}_2\text{O}\). The value of \(x\) is

• A. 0.563
• B. 0.363
• C. 0.636
• D. 0.736

Hint:

For reactions where \(\Delta n_g = 0\) (equal moles of gas on both sides), the equilibrium constant expression depends only on the moles, and the volume term cancels out.

Answer 1

The Correct Option is C

Step 1: Write Equilibrium Constant Expression For the reaction: \(\text{CO}_2 + \text{H}_2 \rightleftharpoons \text{CO} + \text{H}_2\text{O}\) \[ K_c = \frac{[\text{CO}][\text{H}_2\text{O}]}{[\text{CO}_2][\text{H}_2]} \] Since volume is 1 Litre, Molar concentration = Number of moles.
Step 2: Substitute values Given: \(K_c = 0.53\) \([\text{CO}] = 0.25\) \([\text{CO}_2] = 0.5\) \([\text{H}_2] = 0.6\) \([\text{H}_2\text{O}] = x\) \[ 0.53 = \frac{0.25 \times x}{0.5 \times 0.6} \]
Step 3: Solve for \(x\) \[ 0.53 = \frac{0.25x}{0.3} \] \[ 0.25x = 0.53 \times 0.3 \] \[ 0.25x = 0.159 \] \[ x = \frac{0.159}{0.25} \] \[ x = 0.159 \times 4 \] \[ x = 0.636 \] Final Answer:
0.636.