JEE Main 2025 Apr 2 Shift 1, Chemistry Q24: Law of Chemical Equilibrium And Equilibrium Constant

ICE table (in moles):

$\text{Initial: }{n}_{CO}=0.1,n_{H_2}=a,n_{C{H}_{3}OH}=0$

$\text{Equilibrium: }{n}_{CO}=0.1-x,n_{H_2}=a-2x,n_{C{H}_{3}OH}=x=0.04$

So $n_{CO}=0.06$ mol at equilibrium.

Find $a$ from total pressure: total equilibrium moles $n_T=(0.1-x)+(a-2x)+x=0.1+a-2x=0.02+a$.

Using $PV=n_{T}RT$:

$$5\times 2=(0.02+a)(0.08)(500)\Rightarrow 10=40(0.02+a)$$

$$0.02+a=0.25\Rightarrow a=0.23\text{ mol}$$

Then $n_{H_2}=a-2x=0.23-0.08=0.15$ mol, and $n_T=0.25$ mol.

Mole fractions:

$x_{CO}=\frac{0.06}{0.25}=0.24,x_{H_2}=\frac{0.15}{0.25}=0.60,x_{C{H}_{3}OH}=\frac{0.04}{0.25}=0.16$

$K_p$ using partial pressures $P_i=x_i{P}_T$ with $P_T=5$ bar:

$$K_p=\frac{P_{C{H}_{3}OH}}{P_{CO}{P}_{H_2}^2}=\frac{x_{C{H}_{3}OH}}{x_{CO}{x}_{H_2}^2}\cdot \frac{1}{P_T^2}$$

$$K_p=\frac{0.16}{0.24\times (0.60{)}^2\times 25}=\frac{0.16}{0.24\times 0.36\times 25}$$

$$K_p=\frac{0.16}{2.16}\approx 0.0741{\text{ bar}}^{-2}$$

$\therefore$ $K_p^0\approx 74\times 10^{-3}$, so the integer answer is 74.