QJEEM23Cheabb: JEE Main 2023 Chemistry - Class 11 JEEM23Cheabb Solution: A mixture of 1 mole of and 1 mole of is taken in a 10 litre…

The given reaction is :

$\mathrm{CO(g)} + \mathrm{H_{2}O(g)} \rightleftharpoons \mathrm{CO_{2}(g)} + \mathrm{H_{2}(g)}$

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

The molecular weight of water (H₂O) is approximately $18 \, \mathrm{g/mol}$ , 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 $7.2 \, \mathrm{g} / 18 \, \mathrm{g/mol} = 0.4 \, \mathrm{mole}$ .

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 :

$K_{c} = \frac{[\mathrm{CO_{2}}][\mathrm{H_{2}}]}{[\mathrm{CO}][\mathrm{H_{2}O}]}$

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 $K_{c} \times 10^{2}$ :

$K_{c} \times 10^{2} = 0.44 \times 10^{2} = 44 \, (\text{rounded to the nearestinteger})$

Therefore, $K_{c} \times 10^{2} = 44$ .