Question

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

Answer 1

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

Answer 2

The reaction would not be spontaneous above 6762 K.

A spontaneous reaction is one in which the change in free energy of the reaction is less than zero. Recall that; ΔG = ΔH - TΔS

• ΔG = change in free energy
• ΔH = change in enthalpy
• T = temperature
• ΔS = change in entropy

When; ΔG <0, ΔH = -1036 kJ and ΔS = -153.2 J/K

So;

(-1036 × 10^3 J) - (-153.2 J/K) × T < 0

T < 1036 × 10^3 J/153.2

T < 6762 K

Hence, the reaction would not be spontaneous above 6762 K.

Learn more about spontaneous reaction: https://brainly.com/question/1217654