JEE Main · 2025 · 02 Apr 2025 · Shift 1 · Equilibrium

Consider the following equilibrium,

CO (g) + 2 H_{2} (g) ⇌ CH_{3} OH (g)

0.1 mol of CO 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 of

CH_{3} OH

is formed. The

K_{p}^{0}

is
______

\times 1 0^{- 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.

Answer: 74

Solution

At equilibrium

0.04

mol methanol is present, so

0.04

mol CO has reacted to leave

0.06

mol and

0.08

mol hydrogen has been used. The total gas at

5

bar follows from

n = P V / R T = (5 \times 2) / (0.08 \times 500) = 0.25

mol, which places the hydrogen at

0.25 - 0.06 - 0.04 = 0.15

mol. The mole fractions are

0.16

for methanol,

0.24

for CO and

0.60

for hydrogen, so with the total pressure

5

bar the standard constant is

K_{p}^{\circ} = \frac{x _{CH X_{3} OH}}{x _{CO} x _{H X_{2}}^{2}} \cdot \frac{1}{( P / P ^{\circ} ) ^{2}} = \frac{0.16}{0.24 \times 0.36} \times \frac{1}{25} = 0.074

, that is

74 \times 1 0^{- 3}

JEE Main · 2025 · 02 Apr 2025 · Shift 1 · Integer answer