JEE Main 2023 — Chemical Equilibrium Question with Solution

The given reaction is :

$$\mathrm{CO}(\mathrm{g})+{\mathrm{H}}_2\mathrm{O}(\mathrm{g})\rightleftharpoons \mathrm{C}{\mathrm{O}}_2(\mathrm{g})+{\mathrm{H}}_2(\mathrm{g})$$

We are given that $40\%$ of water by mass reacts with carbon monoxide.

The molecular weight of water (H₂O) is approximately $18g/mol$, so $1\mathrm{mole}$ of water weighs $18\mathrm{g}$. Therefore, the mass of the reacted water is $0.40\times 18\mathrm{g}=7.2\mathrm{g}$.

Since from the stoichiometry of the reaction we can see that $1\mathrm{mole}$ of CO reacts with $1\mathrm{mole}$ of H₂O to form $1\mathrm{mole}$ of CO₂ and $1\mathrm{mole}$ of H₂, this means that $7.2\mathrm{g}$ of H₂O is equivalent to $7.2\mathrm{g}/18g/mol=0.4\mathrm{mole}$.

We start with $1\mathrm{mole}$ of CO and $1\mathrm{mole}$ of H2O. At equilibrium, we have :

- CO: $1\mathrm{mole}-0.4\mathrm{mole}=0.6\mathrm{mole}$

- H₂O : $1\mathrm{mole}-0.4\mathrm{mole}=0.6\mathrm{mole}$

- CO₂ : $0\mathrm{mole}+0.4\mathrm{mole}=0.4\mathrm{mole}$

- H₂ : $0\mathrm{mole}+0.4\mathrm{mole}=0.4\mathrm{mole}$

The volume of the container is $10\mathrm{litres}$. Therefore, we can convert the moles to concentrations (in M = moles/L) as :

- [CO] = $0.6\mathrm{M}$

- [H₂O] = $0.6\mathrm{M}$

- [CO₂] = $0.4\mathrm{M}$

- [H₂] = $0.4\mathrm{M}$

The equilibrium constant $K_c$ for the reaction can be calculated as :

$$K_c=\frac{[\mathrm{C}{\mathrm{O}}_2][{\mathrm{H}}_2]}{[\mathrm{CO}][{\mathrm{H}}_2\mathrm{O}]}$$

So, substituting the values we get :

$$K_c=\frac{(0.4\times 0.4)}{(0.6\times 0.6)}=0.44$$

As per the question, we need to calculate the value of $K_c\times 10^2$ :

$$K_c\times 10^2=0.44\times 10^2=44(\text{rounded to the nearest integer})$$

Therefore, $K_c\times 10^2=44$.