A mixture of 1 mole of $\mathrm{H}_{2} \mathrm{O}$ and 1 mole of CO is taken in a 10 litre container and heated to 725 K . At equilibrium $40 \%$ of water by mass reacts with carbon monoxide according to the equation : $\mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{g})$. The equilibrium constant $\mathrm{K}_{\mathrm{C}} \times 10^{2}$ for the reaction is ____. (Nearest integer)

JEE Main · 2023 · 11 Apr 2023 · Shift 1 · Equilibrium

A mixture of 1 mole of

H_{2} O

and 1 mole of CO is taken in a 10 litre container and heated to 725 K . At equilibrium

40%

of water by mass reacts with carbon monoxide according to the equation :

CO (g) + H_{2} O (g) ⇌ CO_{2} (g) + H_{2} (g)

.
The equilibrium constant

K_{C} \times 1 0^{2}

for the reaction is ____. (Nearest integer)

Answer: 44

Initial moles: CO = 1 mol,

H_{2} O

= 1 mol. At equilibrium, 40% of water reacts, so x = 0.4 mol.

At equilibrium:

[CO] = [H_{2} O] = 1 - 0.4 = 0.6

mol

[CO_{2}] = [H_{2}] = 0.4

mol

K_{C} = \frac{[ CO _{2} ] [ H _{2} ]}{[ CO ] [ H _{2} O ]} = \frac{0.4 \times 0.4}{0.6 \times 0.6} = \frac{0.16}{0.36} = 0.444

K_{C} \times 1 0^{2} = 0.444 \times 100 = 44.4 \approx 44

JEE Main · 2023 · 11 Apr 2023 · Shift 1 · Integer answer