Question

# One mole of ${{H}_{2}}O$ and one mole of CO were heated in a 10L closed vessel at 1260K. At equilibrium, 40% of water was found to react in the equation,${{H}_{2}}O\left( g \right)+CO\left( g \right)\rightleftharpoons {{H}_{2}}\left( g \right)+C{{O}_{2}}\left( g \right)$Calculate the equilibrium constant of the reaction.

Hint: We will first find out the concentration of the reactants at t=o and then will find the concentration at ${{t}_{equillibrium}}$ . And the expression for equilibrium constant ${{k}_{C}}$ is given by the equation:
${{k}_{C}}=\frac{concentration\text{ }of\text{ }product}{concentration\text{ }of\text{ }reactant}$

- 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.

Note:
- 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.