Question

One mole H\textsubscript{2O(g) and one mole CO(g) are taken in 1L flask and heated to 725K. At equilibrium, 40\% (by mass) of water reacted with CO(g) as follows:} \[ \text{H}_2\text{O(g)} + \text{CO(g)} \rightleftharpoons \text{H}_2\text{(g)} + \text{CO}_2\text{(g)} \] Its $K_c$ value is

• A. 2.22
• B. 0.444
• C. 4.44
• D. 0.222

Hint:

Use ICE (Initial, Change, Equilibrium) tables for equilibrium problems and apply the expression for \(K_c\) accordingly.

Answer 1

The Correct Option is B

Let the initial moles of H\textsubscript{2}O and CO be 1 mole each. At equilibrium, 40\% of H\textsubscript{2}O has reacted, i.e., \(0.4\) moles. So, \[ \begin{aligned} \text{H}_2\text{O(g)} + \text{CO(g)} &\rightleftharpoons \text{H}_2\text{(g)} + \text{CO}_2\text{(g)}\\ \text{Initial (mol)} &:\quad 1 \quad\quad 1 \quad\quad 0 \quad\quad 0 \\ \text{Change (mol)} &:\quad -0.4 \quad -0.4 \quad +0.4 \quad +0.4 \\ \text{Equilibrium (mol)} &:\quad 0.6 \quad 0.6 \quad 0.4 \quad 0.4 \\ \end{aligned} \] \[ K_c = \frac{[\text{H}_2][\text{CO}_2]}{[\text{H}_2\text{O}][\text{CO}]} = \frac{0.4 \times 0.4}{0.6 \times 0.6} = \frac{0.16}{0.36} = 0.444 \]