Calculate $\Delta G _{ r \times n }$ at 298 K under the conditions shown below for the following reaction.



$SO_3(g)+H_2 O(g) \rightarrow H_2 SO_4(l) \quad \Delta G^{\circ}=-90.5 kJ$

$P \left( SO _3\right)=0.20 atm, P \left( H _2 O \right)=0.88 atm$


A+87.8 kJ
B-86.2 kJ
C-87.8 kJ
D+93.2 kJ
E-93.2 kJ

Verified Answer :

  • Calculate the reaction quotient \(Q = \frac{1}{P_{\text{SO}_3} \cdot P_{\text{H}_2\text{O}}} \approx 5.6818\).
  • Determine the adjustment term \(RT \ln(Q) \approx 4.3042\text{ kJ}\).
  • Compute \(\Delta G_{\text{rxn}} = \Delta G^{\circ} + RT \ln(Q) \approx -90.5 + 4.3042 = -86.1958\text{ kJ}\).
  • Identify the correct option as \(\boxed{-86.2\text{ kJ}}\).

Explanation

  1. Problem Analysis and Thermodynamic Formula Selection

To find the Gibbs free energy change \(\Delta G_{\text{rxn}}\) for the reaction \(\text{SO}_3(g) + \text{H}_2\text{O}(g) \rightarrow \text{H}_2\text{SO}_4(l)\), we use the thermodynamic relation:
\[\Delta G_{\text{rxn}} = \Delta G^{\circ} + RT \ln(Q)\]
Given values are:

  • \(\Delta G^{\circ} = -90.5\text{ kJ}\)
  • \(T = 298\text{ K}\)
  • \(R = 8.314 \times 10^{-3}\text{ kJ/(mol K)}\)
  • \(P(\text{SO}_3) = 0.20\text{ atm}\)
  • \(P(\text{H}_2\text{O}) = 0.88\text{ atm}\)

The reaction quotient \(Q\) is defined based on the partial pressures of the gaseous reactants, while the activity of the pure liquid product \(\text{H}_2\text{SO}_4(l)\) is taken as 1:
\[Q = \frac{1}{P(\text{SO}_3) \cdot P(\text{H}_2\text{O})}\]

  1. Calculation of Reaction Quotient and Gibbs Energy Term

First, we calculate the reaction quotient \(Q\):
\[Q = \frac{1}{0.20 \cdot 0.88} = \frac{1}{0.176} \approx 5.6818\]

Next, we calculate the term \(RT \ln(Q)\):
\[RT \ln(Q) = (8.314 \times 10^{-3}\text{ kJ/(mol K)}) \times (298\text{ K}) \times \ln(5.6818)\]
\[RT \ln(Q) \approx 2.47757 \times 1.7373 \approx 4.3042\text{ kJ}\]

  1. Final Calculation and Conclusion The final result is \(\boxed{-86.2\text{ kJ}}\).

Finally, we determine \(\Delta G_{\text{rxn}}\) by substituting the standard Gibbs free energy change and the calculated term:
\[\Delta G_{\text{rxn}} = -90.5\text{ kJ} + 4.3042\text{ kJ} = -86.1958\text{ kJ}\]
Rounding to one decimal place, we get \(\Delta G_{\text{rxn}} \approx -86.2\text{ kJ}\). Comparing this result with the given options, the correct value is \(-86.2\text{ kJ}\).

Examples

The calculation of Gibbs free energy change under non-standard conditions is fundamental for chemical engineers designing industrial processes, such as the production of sulfuric acid. By understanding how changes in reactant pressure influence the thermodynamic spontaneity of a reaction, engineers can optimize operating conditions to ensure efficiency and control reaction yields in real-world chemical plants.