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-1mol-1
Assume only methanol is formed as the product and the system follows ideal gas behaviour.
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-1mol-1
Assume only methanol is formed as the product and the system follows ideal gas behaviour.
Show Hint
Correct Answer: 74
Solution and Explanation
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:
- Ideal Gas Law: To relate the macroscopic properties of the gas mixture. The equation is: \[ PV = nRT \]
- 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. \[ P_i = X_i P_{\text{total}} = \left( \frac{n_i}{n_{\text{total}}} \right) P_{\text{total}} \]
- 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 \( \times 10^{-3} \).
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