Assuming $\Delta$S^$\circ$ and $\Delta$H^$\circ$ do not vary with temperature, at what temperature will the reaction shown below become spontaneous C(s) + H₂O(g)$\rarr$H₂(g) + CO(s) ($\Delta$S^$\circ$ = 133.6 J/K·mol; $\Delta$H^$\circ$ = 131.3 kJ/mol) A) 670^$\circ$C B) 690^$\circ$C C) 710^$\circ$C D) 730^$\circ$C E) None of the above

Question

Quizplus · Accepted Answer

For the reaction to be spontaneous, the Gibbs free energy change (ΔG) must be negative. ΔG can be calculated using the equation: ΔG = ΔH - TΔS, where ΔH is the enthalpy change, ΔS is the entropy change, and T is the temperature in kelvin.

Using the given values: 
ΔH = -131.3 kJ/mol 
ΔS = -133.6 J/K·mol 
ΔG = ΔH - TΔS

At equilibrium, ΔG = 0. Therefore: 
0 = -131.3 kJ/mol - T(-133.6 J/K·mol) 
0 = -131.3 kJ/mol + 0.1336 kJ/K·mol · T 
T = 983 K = 710 °C

Therefore, the answer is choice C, 710°F1F1F1S1° F1F1F10C.

Assuming Δ\DeltaΔ S ∘\circ∘ And Δ\DeltaΔ H ∘\circ∘ Do Not Vary with Temperature, at What Temperature Will

Assuming $\Delta$ S^$\circ$ And $\Delta$ H^$\circ$ Do Not Vary with Temperature, at What Temperature Will