Question

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 K_p is ____ × 10^-3 (nearest integer).

Given: R = 0.08 dm³ bar K^-1mol^-1

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

Hint:

When calculating equilibrium constants, use the ideal gas law to determine the total moles and pressure. For equilibrium calculations, ensure that you account for the changes in the concentration of reactants and products.

Answer 1

Correct Answer: 74

This problem requires the calculation of the equilibrium constant, \( K_p^0 \), for the synthesis of methanol from carbon monoxide and hydrogen. The provided information includes the initial conditions for CO and the final equilibrium conditions for the gas mixture.

Concept Used:

The solution is based on the following principles:

1. Ideal Gas Law: This law relates the macroscopic properties of a gas mixture, expressed as \( PV = nRT \).
2. Dalton's Law of Partial Pressures: In a gas mixture, the partial pressure of a component is its mole fraction multiplied by the total pressure: \( P_i = X_i P_{\text{total}} = \left( \frac{n_i}{n_{\text{total}}} \right) P_{\text{total}} \).
3. Equilibrium Constant (\( K_p \)): For the reaction \( \text{CO(g)} + 2\text{H}_2\text{(g)} \rightleftharpoons \text{CH}_3\text{OH(g)} \), the equilibrium constant in terms of partial pressures is \( K_p = \frac{P_{\text{CH}_3\text{OH}}}{P_{\text{CO}} \times (P_{\text{H}_2})^2} \). The dimensionless thermodynamic equilibrium constant \( K_p^0 \) is obtained by dividing each partial pressure by the standard pressure \( P^0 = 1 \) bar. Numerically, \( K_p^0 \) equals \( K_p \) when pressures are expressed in bars.

Step-by-Step Solution:

Step 1: Calculate the total moles of gas at equilibrium using the Ideal Gas Law.

The given equilibrium conditions are:

• Total pressure, \( P = 5 \text{ bar} \)
• Volume, \( V = 2 \text{ dm}^3 \)
• Temperature, \( T = 500 \text{ K} \)
• Gas constant, \( R = 0.08 \, \text{dm}^3 \, \text{bar} \, \text{K}^{-1} \, \text{mol}^{-1} \)

Using \( PV = n_{\text{total}}RT \), the total number of moles is:

\[n_{\text{total}} = \frac{PV}{RT} = \frac{(5 \text{ bar}) \times (2 \text{ dm}^3)}{(0.08 \text{ dm}^3 \text{ bar K}^{-1} \text{ mol}^{-1}) \times (500 \text{ K})} = \frac{10}{40} = 0.25 \text{ mol}\]

Step 2: Determine the equilibrium moles of each species.

The reaction is: \( \text{CO(g)} + 2\text{H}_2\text{(g)} \rightleftharpoons \text{CH}_3\text{OH(g)} \).

Equilibrium moles are determined as follows:

• Initial moles of CO = 0.1 mol.
• Equilibrium moles of CH\(_3\)OH = 0.04 mol.

Based on stoichiometry, 0.04 mol of CO reacted to produce 0.04 mol of CH\(_3\)OH.

\[n_{\text{CO, equilibrium}} = n_{\text{CO, initial}} - n_{\text{CO, reacted}} = 0.1 - 0.04 = 0.06 \text{ mol}\]

The moles of H\(_2\) are calculated using the total moles at equilibrium:

\[n_{\text{total}} = n_{\text{CO}} + n_{\text{H}_2} + n_{\text{CH}_3\text{OH}}\]\[0.25 = 0.06 + n_{\text{H}_2} + 0.04\]\[0.25 = 0.10 + n_{\text{H}_2}\]\[n_{\text{H}_2, \text{equilibrium}} = 0.15 \text{ mol}\]

Step 3: Calculate the partial pressure of each gas at equilibrium.

Using Dalton's Law (\( P_i = (n_i / n_{\text{total}}) \times P_{\text{total}} \)), with \( P_{\text{total}} = 5 \text{ bar} \) and \( n_{\text{total}} = 0.25 \text{ mol} \):

Partial pressure of CO:

\[P_{\text{CO}} = \left( \frac{0.06}{0.25} \right) \times 5 = 1.2 \text{ bar}\]

Partial pressure of H\(_2\):

\[P_{\text{H}_2} = \left( \frac{0.15}{0.25} \right) \times 5 = 3.0 \text{ bar}\]

Partial pressure of CH\(_3\)OH:

\[P_{\text{CH}_3\text{OH}} = \left( \frac{0.04}{0.25} \right) \times 5 = 0.8 \text{ bar}\]

Step 4: Calculate the equilibrium constant \( K_p^0 \).

The expression for \( K_p^0 \) is:

\[K_p^0 = \frac{(P_{\text{CH}_3\text{OH}}/P^0)}{(P_{\text{CO}}/P^0) \times (P_{\text{H}_2}/P^0)^2}\]

With \( P^0 = 1 \) bar, the numerical calculation is:

\[K_p^0 = \frac{0.8}{(1.2) \times (3.0)^2} = \frac{0.8}{1.2 \times 9} = \frac{0.8}{10.8}\]

Final Computation & Result

The value of \( K_p^0 \) is computed as:

\[K_p^0 = \frac{0.8}{10.8} = \frac{8}{108} = \frac{2}{27} \approx 0.074074\]

The result is requested in the format \( \_\_\_ \times 10^{-3} \).

\[0.074074 = 74.074 \times 10^{-3}\]

Rounding 74.074 to the nearest integer gives 74.

Therefore, \( K_p^0 = \textbf{74} \times 10^{-3} \).