Question

At 700 K,$C{{O}_{2}}$ and ${{H}_{2}}$react to form $CO$ and ${{H}_{2}}O$. For this process Kc is 0.11. If a mixture of 0.45 mole of $C{{O}_{2}}$ and 0.45 mole of ${{H}_{2}}$is heated at 700 K.
(i) Find out the amount of each gas at equilibrium state.
(ii) After equilibrium is reached another 0.34 mole of $C{{O}_{2}}$and 0.34 mole of ${{H}_{2}}$ are added to the reaction mixture. Find the composition of the mixture at the new equilibrium state.

Answer 1

In a reaction at equilibrium the concentration of the both reactants and products are the same. If we will change the concentration of reactants, the concentration of the product also changes.

Complete step by step answer:
In the question it was mentioned that carbon dioxide ($C{{O}_{2}}$) and hydrogen gas (${{H}_{2}}$) reacts and forms carbon monoxide ($CO$) and water (${{H}_{2}}O$).
$\text{ }C{{O}_{2}}+{{H}_{2}}\underset{{}}{\leftrightarrows}CO+{{H}_{2}}O$
i) Find out the amount of each gas at equilibrium state.
At the initial stage 0.45 moles of carbon dioxide reacts with 0.45 moles of Hydrogen.
The chemical equation is as follows

& at\text{ }time\text{ }0\text{ }\underset{0.45}{\mathop{C{{O}_{2}}}}\,+\underset{0.45\text{ }}{\mathop{{{H}_{2}}}}\,\overset{{}}{leftrightarrows}\underset{-}{\mathop{CO}}\,+\underset{-}{\mathop{{{H}_{2}}O}}\, \\\ & at\text{ }equilibrium\text{ }\underset{0.45-x}{\mathop{C{{O}_{2}}}}\,+\underset{0.45-x\text{ }}{\mathop{{{H}_{2}}}}\,\overset{{}}{leftrightarrows}\underset{x}{\mathop{CO}}\,+\underset{x}{\mathop{{{H}_{2}}O}}\, \\\ \end{aligned}$$ At time zero the concentration of the products are also zero. At equilibrium the concentration of the products is equal to the concentration of the reactants. Therefore equilibrium constant is equal to $$\begin{aligned} & Kc\text{ }=\text{ }\dfrac{[CO]\text{ }[{{H}_{2}}O]}{[C{{O}_{2}}]\text{ }[{{H}_{2}}]} \\\ & \text{ }=\text{ }\dfrac{{{x}^{2}}}{{{(0.45-x)}^{2}}} \\\ & \text{ }x\text{ }=\text{ }0.112\text{ }mole \\\ \end{aligned}$$ In the reaction number of moles of $$C{{O}_{2}}$$ = number of moles of $${{H}_{2}}$$ Therefore substitute value of x in 0.45-x = 0.45-0.112 = 0.338 mole Therefore, the number of moles of $$CO$$ = moles of $${{H}_{2}}O$$ = 0.112 mole at equilibrium. The amount of reactants at equilibrium is $$C{{O}_{2}}$$= 0.338 mole, $${{H}_{2}}$$= 0.338 mole. ii) After equilibrium is reached another 0.34 mole of $$C{{O}_{2}}$$and 0.34 mole of $${{H}_{2}}$$ are added to the reaction mixture. So, the concentration of the reactants and products are going to change. $$\begin{aligned} & \text{at time =0 }\underset{\text{0}\text{.79}}{\mathop{C{{O}_{2}}}}\,+\underset{\text{0}\text{.79 }}{\mathop{{{H}_{2}}}}\,\underset{{}}{\leftrightarrows}\underset{\text{-}}{\mathop{CO}}\,+\underset{\text{-}}{\mathop{{{H}_{2}}O}}\, \\\ & \text{at equilibrium }\underset{\text{0}\text{.79-x}}{\mathop{C{{O}_{2}}}}\,+\underset{\text{0}\text{.79-x }}{\mathop{{{H}_{2}}}}\,\underset{{}}{\leftrightarrows}\underset{\text{x}}{\mathop{CO}}\,+\underset{\text{x}}{\mathop{{{H}_{2}}O}}\, \\\ & \text{ } \\\ \end{aligned}$$ Therefore equilibrium constant is equal to $$\begin{aligned} & Kc\text{ }=\text{ }\dfrac{[CO]\text{ }[{{H}_{2}}O]}{[C{{O}_{2}}]\text{ }[{{H}_{2}}]} \\\ & \text{ }=\text{ }\dfrac{{{x}^{2}}}{{{(0.79-x)}^{2}}} \\\ & \text{ }x=\text{ }0.197\text{ }moles \\\ \end{aligned}$$ We know that number moles of $$C{{O}_{2}}$$ = number of moles of $${{H}_{2}}$$ Substitute the value of x in 0.79-x to get the concentration of reactants at equilibrium = 0.79-0.197 = 0.593 moles Therefore the composition of the products and reactants at the new equilibrium state is, $$C{{O}_{2}}$$= 0.593 moles, $${{H}_{2}}$$= 0.593 moles, $$CO$$= 0.197 mole, and $${{H}_{2}}O$$= 0.197 mole. **Note:** Don’t be confused with the word equilibrium. Equilibrium: “In a chemical reaction, chemical equilibrium is the state in which both reactants and products are present in concentrations which have no further tendency to change with time”.