Hydrogen sulfide decomposes according to the following reaction
2H₂S(g) $\to$ 2H₂(g) + S₂(g)
For this reaction at 298K $\Delta$ S°= 78.1 J/K, $\Delta$ H°= 169.4 kJ, and $\Delta$ G° = 146.1 kJ. What is the value of $\Delta$ G° at 900 K?

Question

Hydrogen sulfide decomposes according to the following reaction
2H₂S(g)

\to

2H₂(g) + S₂(g)
For this reaction at 298K

\Delta

S°= 78.1 J/K,

\Delta

H°= 169.4 kJ, and

\Delta

G° = 146.1 kJ. What is the value of

\Delta

G° at 900 K?

Quizplus · Accepted Answer

We can use the equation: ΔG°(T2) = ΔG°(T1) + ΔH°(T1 to T2) - TΔS°(T1 to T2)

where T1 = 298 K, T2 = 900 K, ΔG°(T1) = 146.1 kJ, ΔH°(T1 to T2) = 169.4 kJ/mol, and ΔS°(T1 to T2) = 78.1 J/K/mol.

Plugging in these values, we get:

ΔG°(900 K) = 146.1 kJ + (169.4 kJ/mol)(900 K/298 K) - (900 K)(0.0781 kJ/K/mol)
ΔG°(900 K) = -449.5 kJ/mol

Converting to kJ, we get -449.5 kJ/mol = -449500 J/mol = -449.5 kJ

Therefore, the answer is (C) 99.1 kJ (rounded to one decimal place).

Hydrogen Sulfide Decomposes According to the Following Reaction 2H2S(g) →\to→