Consider the reaction CO(g) + 2H₂(g) $\leftrightharpoons$ CH₃OH(l) at 25^$\circ$C.
$\Delta$ G^$\circ$_f (CO) = -137.3 kJ/mol
$\Delta$ G^$\circ$_f (CH₃OH) = -166.3 kJ/mol
$\Delta$ H^$\circ$_f (CO) = -110.5 kJ/mol
$\Delta$ H^$\circ$_f (CH₃OH) = -238.7 kJ/mol
S^$\circ$(CO) = 197.9 J/K·mol
S^$\circ$(CH₃OH) = 126.8 J/K·mol
Calculate S^$\circ$(H₂(g) ) _.

Question

Consider the reaction CO(g) + 2H₂(g)

\leftrightharpoons

CH₃OH(l) at 25^$\circ$C.

\Delta

G^$\circ$_f (CO) = -137.3 kJ/mol

\Delta

G^$\circ$_f (CH₃OH) = -166.3 kJ/mol

\Delta

H^$\circ$_f (CO) = -110.5 kJ/mol

\Delta

H^$\circ$_f (CH₃OH) = -238.7 kJ/mol
S^$\circ$(CO) = 197.9 J/K·mol
S^$\circ$(CH₃OH) = 126.8 J/K·mol
Calculate S^$\circ$(H₂(g) ) _.

Quizplus · Accepted Answer

We can use the formula ΔG° = ΔH° - TΔS° to solve for the standard entropy of H₂(g) at 25°C. First, we need to find ΔG° of the reaction: ΔG° = ΔH° - TΔS° ΔG° = [-166.3 kJ/mol - (-137.3 kJ/mol) - 2(-110.5 kJ/mol)] - (298 K)([126.8 J/K·mol] + 2[197.9 J/K·mol]) ΔG° = -98.1 kJ/mol Now we can rearrange the formula to solve for ΔS°: ΔS° = (ΔH° - ΔG°) / T ΔS° = [(-166.3 kJ/mol - (-137.3 kJ/mol) - 2(-110.5 kJ/mol)) - (-98.1 kJ/mol)] / (298 K) ΔS° = 131 J/K·mol Therefore, the standard entropy of H₂(g) is 131 J/K·mol, which corresponds to choice B.

Consider the Reaction CO(g) + 2H2(g) ⇋\leftrightharpoons⇋ CH3OH(l) at 25 ∘\circ∘ C Δ\DeltaΔ

Consider the Reaction CO(g) + 2H₂(g) $\leftrightharpoons$ CH₃OH(l) at 25^$\circ$C $\Delta$