Here's the derivation for the free energy change for the given reaction. The reaction is: $$\text{A}_{(\text{g})} + \text{B}_{(\text{g})} \rightarrow \text{C}_{(\text{g})} + \text{D}_{(\text{g})}$$ Step 1: Define the change in Gibbs free energy for the reaction. For a chemical reaction, the change in Gibbs free energy ($\Delta G$) is the sum of the chemical potentials of the products minus the sum of the chemical potentials of the reactants, each multiplied by its stoichiometric coefficient. $$\Delta G = (\mu_C + \mu_D) - (\mu_A + \mu_B)$$ where $\mu_i$ is the chemical potential of component $i$. Step 2: Express the chemical potential for an ideal gas. For an ideal gas, the chemical potential ($\mu_i$) of component $i$ at a given partial pressure $P_i$ and temperature $T$ is related to its standard chemical potential ($\mu_i^{\circ}$) by: $$\mu_i = \mu_i^{\circ} + RT \ln\left(\frac{P_i}{P^{\circ}}\right)$$ Here, $R$ is the ideal gas constant, $T$ is the absolute temperature, and $P^{\circ}$ is the standard pressure (usually $1 \, \text{bar}$ or $1 \, \text{atm}$). Step 3: Substitute the chemical potential expressions into the $\Delta G$ equation. Substitute the expression for $\mu_i$ for each component into the $\Delta G$ equation: $$\Delta G = \left[\mu_C^{\circ} + RT \ln\left(\frac{P_C}{P^{\circ}}\right)\right] + \left[\mu_D^{\circ} + RT \ln\left(\frac{P_D}{P^{\circ}}\right)\right] - \left[\mu_A^{\circ} + RT \ln\left(\frac{P_A}{P^{\circ}}\right)\right] - \left[\mu_B^{\circ} + RT \ln\left(\frac{P_B}{P^{\circ}}\right)\right]$$ Step 4: Rearrange and simplify the expression. Group the standard chemical potential terms and the logarithmic terms: $$\Delta G = (\mu_C^{\circ} + \mu_D^{\circ} - \mu_A^{\circ} - \mu_B^{\circ}) + RT \left[\ln\left(\frac{P_C}{P^{\circ}}\right) + \ln\left(\frac{P_D}{P^{\circ}}\right) - \ln\left(\frac{P_A}{P^{\circ}}\right) - \ln\left(\frac{P_B}{P^{\circ}}\right)\right]$$ The term $(\mu_C^{\circ} + \mu_D^{\circ} - \mu_A^{\circ} - \mu_B^{\circ})$ represents the standard Gibbs free energy change for the reaction, $\Delta G^{\circ}$. $$\Delta G^{\circ} = \sum (\nu_p \mu_p^{\circ}) - \sum (\nu_r \mu_r^{\circ})$$ Combine the logarithmic terms using the properties of logarithms ($\ln x + \ln y = \ln(xy)$ and $\ln x - \ln y = \ln(x/y)$): $$RT \left[\ln\left(\frac{P_C}{P^{\circ}}\right) + \ln\left(\frac{P_D}{P^{\circ}}\right) - \ln\left(\frac{P_A}{P^{\circ}}\right) - \ln\left(\frac{P_B}{P^{\circ}}\right)\right] = RT \ln\left[\frac{\left(\frac{P_C}{P^{\circ}}\right)\left(\frac{P_D}{P^{\circ}}\right)}{\left(\frac{P_A}{P^{\circ}}\right)\left(\frac{P_B}{P^{\circ}}\right)}\right]$$ The term inside the logarithm is the reaction quotient ($Q_p$) for partial pressures, where the partial pressures are divided by the standard pressure $P^{\circ}$: $$Q_p = \frac{\left(\frac{P_C}{P^{\circ}}\right)\left(\frac{P_D}{P^{\circ}}\right)}{\left(\frac{P_A}{P^{\circ}}\right)\left(\frac{P_B}{P^{\circ}}\right)}$$ Since the sum of the stoichiometric coefficients for products ($1+1=2$) equals the sum for reactants ($1+1=2$), the $P^{\circ}$ terms cancel out in this specific case, simplifying $Q_p$ to: $$Q_p = \frac{P_C P_D}{P_A P_B}$$ Therefore, the expression for the free energy change for the reaction is: $$\boxed{\Delta G = \Delta G^{\circ} + RT \ln Q_p}$$ That's 2 down. 3 left today — send the next one.