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nata0808 [166]
3 years ago
10

Above what temperature does the following reaction become nonspontaneous? 2 H2S(g) + 3 O2(g) → 2 SO2(g) + 2 H2O(g) ΔH = -1036 kJ

; ΔS = -153.2 J/K
Chemistry
1 answer:
Katen [24]3 years ago
5 0

Answer:

Above 6762 K the reaction becomes nonspontaneous.

Explanation:

Let's consider the following equation.

2 H₂S(g) + 3 O₂(g) → 2 SO₂(g) + 2 H₂O(g)

This reaction will become nonspontaneous when the Gibbs free energy (ΔG) is higher than zero. ΔG is related to the enthalpy of the reaction (ΔH) and the entropy of the reaction (ΔS) through the following expression.

ΔG = ΔH - T.ΔS

If ΔG must be positive, then

ΔH - T.ΔS > 0

ΔH > T.ΔS

-1036 × 10³ J > T . (-153.2 J/K)

T > 6762 K

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\large \boxed{\text{105 kcal}}

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MM:                        30.01

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