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

We need to calculate the equilibrium constant, \( K_p^0 \), for the synthesis of methanol from carbon monoxide and hydrogen. The problem provides initial conditions for CO and final equilibrium conditions for the mixture.

Concept Used:

The solution involves the following principles:

1. Ideal Gas Law: To relate the macroscopic properties of the gas mixture. The equation is: \[ PV = nRT \]
2. Dalton's Law of Partial Pressures: The partial pressure of a component gas in a mixture is the product of its mole fraction and 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 given 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 thermodynamic equilibrium constant \( K_p^0 \) is dimensionless, calculated by dividing each partial pressure in bars by the standard pressure \( P^0 = 1 \) bar. Numerically, it is the same as \( K_p \) when pressures are in bars.

Step-by-Step Solution:

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

Given equilibrium conditions:

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

From \( PV = n_{\text{total}}RT \):

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

Step 2: Determine the number of moles of each species at equilibrium.

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

Let's analyze the moles at equilibrium:

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

From the stoichiometry, to form 0.04 mol of CH\(_3\)OH, 0.04 mol of CO must have reacted.

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

Now, we use the total moles at equilibrium to find the moles of H\(_2\).

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

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:

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

Since \( P^0 = 1 \) bar, the numerical value 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

Calculating the value of \( K_p^0 \):

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

The question asks for the answer in the format \( \_\_\_ \times 10^{-3} \).

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

The nearest integer to 74.074 is 74.

The \( K_p^0 \) is 74 \( \times 10^{-3} \).