Question

One mole of $H_{2}O$ and one mole of CO are taken in 10 L vessel and heated to 725 K. At equilibrium 40% of water reacts with CO according to the equation,
$H_{2}O\left(g\right)+CO\left(g\right)\rightleftharpoons H_2\left(g\right)+C{O}_2\left(g\right)$. What is the value of equilibrium constant $\left(K_c\right)$ for the reaction?

• A. 44
• B. 4.4
• C. 0.44
• D. 2.22

Answer 1

The correct option is C 0.44

So,
$\begin{array}{cccccccc}& H_{2}O\left(g\right)& +& CO\left(g\right)& \rightleftharpoons & H_2\left(g\right)& +& C{O}_2\left(g\right)\\ t=0& 1mol& & 1mol& & 0& & 0\\ & -0.4mol& & -0.4mol& & 0.4mol& & 0.4mol\end{array}$
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$\begin{array}{ccccccccccccccccccc}t=t_{eq}& 0.6& & & & & 0.6& & & & & 0.4& & & & & & & 0.4\end{array}$

So, concentrations at equilibrium are,
$\left[H_2\right]=\frac{0.4}{10}=0.04M$
$\left[C{O}_2\right]=\frac{0.4}{10}=0.04M$
$\left[H_{2}O\right]=\frac{0.6}{10}=0.06M$
$\left[CO\right]=\frac{0.6}{10}=0.06M$

putting the values,
$K_c=\frac{0.04\times 0.04}{0.06\times 0.06}=\frac{16}{36}=\frac{4}{9}=0.44$