- Calculate the standard Gibbs free energy change \(\Delta G^\circ = -24.8 \text{ kJ/mol}\).
- Determine the reaction quotient \(Q = 161\) using the provided partial pressures.
- Compute the non-standard Gibbs free energy \(\Delta G = \Delta G^\circ + RT \ln Q\).
- The final Gibbs free energy change for the reaction is \(\boxed{-12.2 \text{ kJ/mol}}\).
Explanation
- Calculating Standard Gibbs Free Energy Change (\(\Delta G^\circ\))
To determine the Gibbs free energy change (\(\Delta G\)) for the reaction \(\text{CO(g)} + 2\text{H}_2\text{(g)} \rightarrow \text{CH}_3\text{OH(g)}\) at \(25^\circ\text{C}\), we start by identifying the standard Gibbs free energy change (\(\Delta G^\circ\)). The standard Gibbs free energy change of the reaction is calculated from the standard Gibbs free energies of formation (\(\Delta G_f^\circ\)) of the reactants and products. Using representative values: \(\Delta G_f^\circ(\text{CO}) = -137.2 \text{ kJ/mol}\), \(\Delta G_f^\circ(\text{CH}_3\text{OH}) = -162.0 \text{ kJ/mol}\), and \(\Delta G_f^\circ(\text{H}_2) = 0 \text{ kJ/mol}\). The standard change for the reaction is \(\Delta G^\circ = \Delta G_f^\circ(\text{CH}_3\text{OH}) - [\Delta G_f^\circ(\text{CO}) + 2 \cdot \Delta G_f^\circ(\text{H}_2)] = -162.0 - (-137.2 + 0) = -24.8 \text{ kJ/mol}\).
- Calculating the Reaction Quotient (\(Q\))
Next, we calculate the reaction quotient \(Q\) using the given partial pressures: \(P_{\text{CH}_3\text{OH}} = 0.805 \text{ atm}\), \(P_{\text{CO}} = 0.125 \text{ atm}\), and \(P_{\text{H}_2} = 0.200 \text{ atm}\). The reaction quotient is defined as \(Q = \frac{P_{\text{CH}_3\text{OH}}}{P_{\text{CO}} \cdot (P_{\text{H}_2})^2}\). Substituting the values, we find \(Q = \frac{0.805}{0.125 \cdot (0.200)^2} = \frac{0.805}{0.125 \cdot 0.04} = \frac{0.805}{0.005} = 161\).
- Calculating Non-Standard Gibbs Free Energy Change (\(\Delta G\))
Now, we use the formula relating \(\Delta G\) to \(\Delta G^\circ\): \(\Delta G = \Delta G^\circ + RT \ln Q\), where \(R = 8.314 \times 10^{-3} \text{ kJ/(mol}\cdot\text{K)}\) and \(T = 298.15 \text{ K}\) (converting \(25^\circ\text{C}\) to Kelvin). The term \(RT \ln Q\) is \( (8.314 \times 10^{-3} \text{ kJ/(mol}\cdot\text{K)}) \times 298.15 \text{ K} \times \ln(161) \approx 2.4788 \times 5.0814 \approx 12.60 \text{ kJ/mol}\). Finally, \(\Delta G = -24.8 \text{ kJ/mol} + 12.60 \text{ kJ/mol} = -12.2 \text{ kJ/mol}\).
Examples
This type of thermodynamic calculation is essential in industrial chemistry, such as in the synthesis of methanol from syngas. Engineers must account for the specific pressures and temperatures in a chemical reactor to predict whether a reaction will proceed spontaneously (if \(\Delta G < 0\)) or require external energy input, thereby optimizing the conditions for maximum efficiency and yield. Understanding these principles ensures that chemical processes operate under stable and economically viable conditions.