- Calculate the standard Gibbs free energy change of the reaction as \(\Delta G^\circ = \Delta G^\circ_f(\text{CH}_3\text{OH}) - [\Delta G^\circ_f(\text{CO}) + 2\Delta G^\circ_f(\text{H}_2)] = -24.8 \text{ kJ/mol}\).
- Determine the reaction quotient using the given partial pressures: \(Q = \frac{0.805}{0.125 \cdot 0.200^2} = 161\).
- Apply the Gibbs free energy formula for non-standard conditions: \(\Delta G = \Delta G^\circ + RT \ln(Q)\).
- Compute the final result as \(\Delta G = -24.8 + (8.314 \times 10^{-3}) \times 298.15 \times \ln(161) \approx -12.2 \text{ kJ}\).
- \(\boxed{-12.2 \text{ kJ}}\)
Explanation
- Calculation of Standard Gibbs Free Energy Change
The balanced equation for the synthesis of methanol is:
\[\text{CO}(g) + 2\text{H}_2(g) \rightleftharpoons \text{CH}_3\text{OH}(g)\]
The standard Gibbs free energy change of the reaction, \(\Delta G^\circ\), is calculated from the standard Gibbs free energies of formation (\(\Delta G^\circ_f\)) of the reactants and products:
\[\Delta G^\circ = \Delta G^\circ_f(\text{CH}_3\text{OH}) - [\Delta G^\circ_f(\text{CO}) + 2\Delta G^\circ_f(\text{H}_2)]\]
Using the standard values:
\(\Delta G^\circ_f(\text{CH}_3\text{OH}) = -162.0 \text{ kJ/mol}\)
\(\Delta G^\circ_f(\text{CO}) = -137.2 \text{ kJ/mol}\)
\(\Delta G^\circ_f(\text{H}_2) = 0 \text{ kJ/mol}\)
The standard reaction Gibbs free energy is:
\[\Delta G^\circ = -162.0 - (-137.2 + 2 \times 0) = -162.0 + 137.2 = -24.8 \text{ kJ/mol}\]
- Determination of the Reaction Quotient Q
The reaction quotient \(Q\) is defined by the partial pressures of the gases involved in the reaction:
\[Q = \frac{P_{\text{CH}_3\text{OH}}}{P_{\text{CO}} \cdot P_{\text{H}_2}^2}\]
Given the partial pressures:
\(P_{\text{CH}_3\text{OH}} = 0.805 \text{ atm}\)
\(P_{\text{CO}} = 0.125 \text{ atm}\)
\(P_{\text{H}_2} = 0.200 \text{ atm}\)
Substituting these values:
\[Q = \frac{0.805}{0.125 \cdot (0.200)^2} = \frac{0.805}{0.125 \cdot 0.04} = \frac{0.805}{0.005} = 161\]
- Calculation of Non-standard Gibbs Free Energy Change ΔG
The non-standard Gibbs free energy change, \(\Delta G\), is related to the standard Gibbs free energy change, \(\Delta G^\circ\), by the equation:
\[\Delta G = \Delta G^\circ + RT \ln(Q)\]
Where:
\(R = 8.314 \times 10^{-3} \text{ kJ/(mol}\cdot\text{K)}\)
\(T = 298.15 \text{ K}\)
\(\ln(Q) = \ln(161) \approx 5.0814\)
Calculating the term \(RT \ln(Q)\):
\[RT \ln(Q) = (8.314 \times 10^{-3} \text{ kJ/(mol}\cdot\text{K)}) \times 298.15 \text{ K} \times 5.0814 \approx 12.596 \text{ kJ/mol}\]
Finally, adding the values:
\[\Delta G = -24.8 \text{ kJ/mol} + 12.596 \text{ kJ/mol} = -12.204 \text{ kJ/mol}\]
Rounding to three significant figures, we get \(\Delta G = -12.2 \text{ kJ/mol}\).
- Final Answer
The non-standard Gibbs free energy change \(\Delta G\) for the given reaction at 25°C is \(\boxed{-12.2 \text{ kJ}}\).
Examples
In industrial chemistry, this calculation is vital for determining the feasibility and yield of a reaction under specific conditions. By understanding how changes in pressure affect the Gibbs free energy, engineers can optimize processes like methanol production, ensuring that conditions are adjusted to favor the desired product and improve overall efficiency. This application is fundamental to chemical manufacturing and reactor design.