Question

Carbon monoxide in water gas, reacts with steam according to the reaction:
$CO\left(g\right)+H_{2}O\left(g\right)\to C{O}_2\left(g\right)+H_2\left(g\right);Keq=1.0$
If three volumes of water gas (containing 1:2 of $CO$ and $H_2$) are mixed with 3 volumes of steam, then the volume ratio of the four gases present at equilibrium is

• A. $1:5:1:5$
• B. $1:4:3:1$
• C. $1:5:1:1$
• D. $1:1:1:2$

Answer 1

Answer: (A)

$1:5:1:5$

$CO\left(g\right)+H_{2}O\left(g\right)\to C{O}_2\left(g\right)+H_2\left(g\right)$;

$\begin{array}{ccccc}1& 3& & 0& 2\\ 1-x& 3-x& & x& 2+x\end{array}$

$K_{eq}=\frac{x(2+x)}{(1-x)(3-x)}=1$

$x^2+2x=3-4x+x^2$

$x=0.5$

Hence, the ratio for the gases, $CO\left(g\right):H_{2}O\left(g\right):C{O}_2\left(g\right):H_2\left(g\right)=0.5:2.5:0.5:2.5$

Or $1:5:1:5$

Option A is correct.