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${{H}_{2}}O\left( g \right)+CO\left( g \right)\rightleftharpoons {{H}_{2}}\left( g \right)+C{{O}_{2}}\left( g \right)$

Calculate the equilibrium constant of the reaction.

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\[{{k}_{C}}=\frac{concentration\text{ }of\text{ }product}{concentration\text{ }of\text{ }reactant}\]

- We are given with the value of volume of water as 10 L.

- We will write the reaction given as:

${{H}_{2}}O\left( g \right)+CO\left( g \right)\rightleftharpoons {{H}_{2}}\left( g \right)+C{{O}_{2}}\left( g \right)$

- As we know the volume of water, so we will find the concentration of water as:

$\begin{align}

& \left[ {{H}_{2}}O \right]=\frac{1}{10} \\

& =0.1 \\

\end{align}$

- Similarly we will find the concentration of carbon monoxide, that will be:

\[\begin{align}

& \left[ CO \right]=\frac{1}{10} \\

& =0.1 \\

\end{align}\]

- So, at t=0, the concentration of water and carbon monoxide is 0.1

- At ${{t}_{equilibrium}}$ we will consider that x mole of water will decompose . So, the concentration of water will be 0.1-x and the concentration of carbon monoxide will be also 0.1-x. And concentration of ${{H}_{2}}$will be x, and concentration of $C{{O}_{2}}$ will be x.

- So, according to the given information value of x will be 0.04, as only 40% of water is reacting, so 40% of 0.1 is 0.04. We can write it as:

\[\begin{align}

& \left( \frac{40}{100}\times 0.1 \right) \\

& =0.04 \\

\end{align}\]

- Now, we will write the expression for equilibrium constant ${{k}_{C}}$:

\[{{k}_{C}}=\frac{concentration\text{ }of\text{ }product}{concentration\text{ }of\text{ }reactant}\]

\[\begin{align}

& {{k}_{C}}=\frac{x\times x}{{{\left( 0.1x \right)}^{2}}} \\

& =\frac{0.04\times 0.04}{0.06\times 0.06} \\

& =\frac{4}{9} \\

\end{align}\]

- Hence, we can conclude that the value of equilibrium constant of the reaction is $\frac{4}{9}$or we can say 0.44.

- It is found that the equilibrium constant depends upon temperature. In case of exothermic reactions increasing the temperature will reduce ${{k}_{C}}$, and in endothermic reactions increasing the temperature will increase ${{k}_{C}}$.

- And change in concentration, catalyst, pressure etc. have no effect on equilibrium constant.