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Naddik [55]
4 years ago
9

For a particular reaction, Δ H ∘ = 55.4 kJ ΔH∘=55.4 kJ and Δ S ∘ = 94.1 J/K. ΔS∘=94.1 J/K. Assuming these values change very lit

tle with temperature, at what temperature does the reaction change from nonspontaneous to spontaneous?
Chemistry
1 answer:
nadezda [96]4 years ago
7 0

Answer:

The temperature at which the reaction changes from non-spontaneous to spontaneous is 588.735 K

Explanation:

The spontaneity of a reaction is determined by the change in Gibbs Free Energy, \Delta G^{0}.

\Delta G^{0} =\Delta H^{0} -T\Delta S^{0}

If \Delta G^{0} is greater than zero, then a reaction is feasible.

If \Delta G^{0} is less than zero, then a reaction is not feasible.

To determine the temperature at which the reaction changes from non-spontaneous to spontaneous, we should equate the \Delta G^{0} to zero.

We take \Delta G^{0}=0 as the limiting condition.

T=\frac{\Delta H^{0}}{\Delta S^{0}}=\frac{55.4\times10^{3}}{94.1}=588.735K

Therefore, the temperature is: 588.735K

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