Consider the following equilibrium, 0.1 mol of CO along with a catalyst is present in a 2 flask maintained at 500 K. Hydrogen is introduced into the flask until the pressure is 5 bar and 0.04 mol of is formed. The is ______ (nearest integer).

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Solution

Q.
Consider the following equilibrium,

$CO(g) + 2 H_{2} (g) \to {CH}_{3} OH(g)$

0.1 mol of CO along with a catalyst is present in a 2 ${d m}^{3}$ flask maintained at 500 K. Hydrogen is introduced into the flask until the pressure is 5 bar and 0.04 mol of $C H_{3} O H$ is formed. The $K_{p}^{0}$ is ______ $\times 10^{- 3}$ (nearest integer).

Given : R = 0.08 ${d m}^{3}$ bar $K^{- 1}$ ${mol}^{- 1}$

Assume only methanol is formed as the product and the system follows ideal gas behaviour. [2025]

Ans.
(74)

$\begin{matrix} CO(g) & + & 2 H_{2} (g) & ⇌ & {CH}_{3} OH(g) \\ t=0 (moles) & 0.1 & x & 0 \\ t=∞ (moles) & 0.1 - y & x - 2 y & y \end{matrix}$

As moles of $C H_{3} O H = y = 0.04$ mol

$\begin{matrix} CO(g) & + & 2 H_{2} (g) & ⇌ & {CH}_{3} OH(g) \\ t =∞ (moles) & 0.1 - 0.04 & x - 0.08 & 0.04 \\ = 0.06 \end{matrix}$

Total moles = $0.06 + x - 0.08 + 0.04 = (0.02 + x)$ mol

By ideal gas equation at equilibrium:

$P V = n R T$

$5 \times 2 = (0.02 + x) \times 0.08 \times 500$

$x = 0.23 mol$

Thus:

$\begin{matrix} CO(g) & + & 2 H_{2} (g) & ⇌ & {CH}_{3} OH(g) \\ t =∞ (moles) & 0.06 & 0.23 - 0.08 = 0.15 & 0.04 \\ t =∞ (mole fraction) & \frac{6}{25} & \frac{15}{25} & \frac{4}{25} \\ t =∞ (partial pressure) & \frac{6}{25} \times 5 = \frac{6}{5} & \frac{15}{25} \times 5 = 3 & \frac{4}{25} \times 5 = \frac{4}{5} \end{matrix} \begin{matrix} Total=0.25 \end{matrix}$

$K_{p} = \frac{P_{{CH}_{3} OH}}{P_{CO} \times {(P_{H_{2}})}^{2}}$ $= \frac{4 / 5}{(\frac{6}{5}) \times {(3)}^{2}} = 74 \times 10^{- 3}$

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