For the reaction 2 SO₂(g) + O₂(g) $\rarr$ 2 SO₃(g), if initially P(SO₂) = 1.2 atm, P(O₂) = 1.8 atm, and P(SO₃) = 2.1 atm, calculate $\Delta$G for this reaction at 25^$\circ$C. The following data is valid at 25^$\circ$C: $\begin{array}{|l|l|} \hline & \Delta \mathrm{G}_{\mathrm{f}}{ }^{\circ}(\mathrm{kJ} / \mathrm{mol}) \ \hline \mathrm{SO}_{2} & -300.4 \ \hline \mathrm{SO}_{3} & -370.4 \ \hline \end{array}$ A) -140.0 kJ/mol B) -137.6 kJ/mol C) -138.7 kJ/mol D) 1,174.7 kJ/mol E) -141.3 kJ/mol

Question

Quizplus · Accepted Answer

To calculate the standard Gibbs free energy change, we use the formula: ΔG° = -RTlnK where R is the gas constant (8.314 J/mol·K), T is the temperature in Kelvin (25°C = 298 K), and K is the equilibrium constant. To find K, we need to use the partial pressures and stoichiometric coefficients of the reaction: K = (P(SO₃))^2 / (P(SO₂))^2(P(O₂)) Plugging in the values, we get: K = (2.1 atm)^2 / (1.2 atm)^2(1.8 atm) = 2.15 Now we can calculate ΔG°: ΔG° = -RTlnK = -(8.314 J/mol·K)(298 K)ln(2.15) = -138.7 kJ/mol Therefore, the best choice is C, with a ΔG° of -138.7 kJ/mol.

For the Reaction 2 SO2(g) + O2(g) →\rarr→ 2 SO3(g), If Initially P(SO2) = 1

For the Reaction 2 SO₂(g) + O₂(g) $\rarr$ 2 SO₃(g), If Initially P(SO₂) = 1