Solveeit Logo
Login

Question

Chemistry Question on Thermodynamics

For a certain reaction at 300 K, K=10K = 10, then ΔG∘\Delta G^\circ for the same reaction is − ×10−1 kJ mol−1.- \, \\_ \times 10^{-1} \, \text{kJ mol}^{-1}.(Given R=8.314 JK−1mol−1R = 8.314 \, \text{JK}^{-1} \text{mol}^{-1})

Answer

Step-by-step Calculation:
The standard Gibbs free energy change ΔG∘\Delta G^\circ for a reaction is related to the equilibrium constant KK by the equation:
ΔG∘=−RTln⁡K\Delta G^\circ = -RT \ln K
where:
R=8.314 J K−1mol−1R = 8.314 \, \text{J K}^{-1} \text{mol}^{-1} (Universal gas constant)
T=300 KT = 300 \, \text{K}
K=10K = 10
Substituting the values into the equation:
ΔG∘=−8.314×300×ln⁡(10)\Delta G^\circ = -8.314 \times 300 \times \ln(10)
We know that ln⁡(10)≈2.303\ln(10) \approx 2.303.
Therefore:
ΔG∘=−8.314×300×2.303\Delta G^\circ = -8.314 \times 300 \times 2.303
ΔG∘=−8.314×690.9≈−5730 J mol−1\Delta G^\circ = -8.314 \times 690.9 \approx -5730 \, \text{J mol}^{-1}
Converting to kJ mol−1\text{kJ mol}^{-1}:
ΔG∘=−5.73 kJ mol−1\Delta G^\circ = -5.73 \, \text{kJ mol}^{-1}
Expressing in the required format:
ΔG∘=−57×10−1 kJ mol−1\Delta G^\circ = -57 \times 10^{-1} \, \text{kJ mol}^{-1}
Conclusion: The value of ΔG∘\Delta G^\circ for the reaction is −57×10−1 kJ mol−1-57 \times 10^{-1} \, \text{kJ mol}^{-1}.