Consider the following equilibrium, CO(g) + 2H2(g) = CH3OH(g) 0.1 mol of CO along with a catalyst is present in a 2 dm3 flask maintained at 500 K. Hydrogen is introduced into the flask until the pressure is 5 bar and 0.04 mol of CH3OH is formed. The Kp is _________ ×10-3 (nearest integer). Given : R = 0.08 dm3 bar K-1 mol-1 Assume only methanol is formed as the product and the system follows ideal gas behaviour.

The correct answer is Option 74. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Chemistry Q73 – JEE-MAIN 2025

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Home Chemistry EquilibriumQuestion #441

Consider the following equilibrium,

C O (g) + 2 H_{2} (g) ⇌ C H_{3} O H (g)

0.1 mol

C O

along with a catalyst is present in a

2 {dm}^{3}

flask maintained at

500 K

. Hydrogen is introduced into the flask until the pressure is

5 bar

and

0.04 mol

C H_{3} O H

is formed. The

K_{p}^{θ}

is _________

\times 10^{- 3}

(nearest integer). Given :

R = 0.08 {dm}^{3} {bar K}^{- 1} {mol}^{- 1}

Assume only methanol is formed as the product and the system follows ideal gas behaviour.

Enter Calculated Numerical Value:

Clue 1 of 3: Equilibrium Setup

Start by writing down the balanced chemical equation and setting up an ICE table (Initial, Change, Equilibrium) for the moles of each species.

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Step 1: Determine equilibrium moles of all species✦ Active

The reaction is $C O (g) + 2 H_{2} (g) ⇌ C H_{3} O H (g)$ . Initial moles: $n_{C O, i} = 0.1 mol$ , $n_{H_{2}, i} = x mol$ (unknown), $n_{C H_{3} O H, i} = 0 mol$ . At equilibrium, $n_{C H_{3} O H, e q} = 0.04 mol$ . Change in moles: $Δ n_{C O} = - 0.04 mol$ , $Δ n_{H_{2}} = - 2 \times 0.04 = - 0.08 mol$ , $Δ n_{C H_{3} O H} = + 0.04 mol$ . Equilibrium moles: $n_{C O, e q} = 0.1 - 0.04 = 0.06 mol$ . $n_{H_{2}, e q} = x - 0.08 mol$ . Total moles at equilibrium, $n_{t o t a l, e q} = n_{C O, e q} + n_{H_{2}, e q} + n_{C H_{3} O H, e q} = 0.06 + (x - 0.08) + 0.04 = x + 0.02 mol$ . Using the ideal gas law $P_{t o t a l} V = n_{t o t a l, e q} R T$ : $5 bar \times 2 {dm}^{3} = (x + 0.02) mol \times 0.08 {dm}^{3} {bar K}^{- 1} {mol}^{- 1} \times 500 K$ $10 = (x + 0.02) \times 40 ⟹ x + 0.02 = 0.25 ⟹ x = 0.23 mol$ . Therefore, $n_{H_{2}, e q} = 0.23 - 0.08 = 0.15 mol$ . The equilibrium moles are: $n_{C O} = 0.06 mol$ , $n_{H_{2}} = 0.15 mol$ , $n_{C H_{3} O H} = 0.04 mol$ . Total moles at equilibrium, $n_{t o t a l} = 0.06 + 0.15 + 0.04 = 0.25 mol$ .

💡 Teacher's Secret Hint

Ensure consistent units for $R$ , $P$ , $V$ , and $T$ .

Step 2: Calculate partial pressures of each gas○ Expand

Partial pressure $P_{i} = \frac{n_{i}}{n_{t o t a l}} P_{t o t a l}$ . $P_{C O} = \frac{0.06}{0.25} \times 5 bar = 1.2 bar$ . $P_{H_{2}} = \frac{0.15}{0.25} \times 5 bar = 3.0 bar$ . $P_{C H_{3} O H} = \frac{0.04}{0.25} \times 5 bar = 0.8 bar$ .

💡 Teacher's Secret Hint

Verify that the sum of partial pressures equals the total pressure.

Step 3: Calculate the equilibrium constant

K_{p}

○ Expand

The expression for $K_{p}$ is $K_{p} = \frac{P_{C H_{3} O H}}{P_{C O} \times (P_{H_{2}})^{2}}$ . Substituting the partial pressures:

K_{p} = \frac{0.8}{1.2 \times (3.0)^{2}} = \frac{0.8}{1.2 \times 9} = \frac{0.8}{10.8} = \frac{8}{108} = \frac{2}{27}

$K_{p} \approx 0.074074$ . To express this as $\times 10^{- 3}$ : $0.074074 = 74.074 \times 10^{- 3}$ . Rounding to the nearest integer, the value is $74$ .

💡 Teacher's Secret Hint

Pay attention to the stoichiometric coefficients when writing the $K_{p}$ expression.

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Chemistry Question 73 – JEE-MAIN 2025

Clue 1 of 3: Equilibrium Setup

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