Question

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 little with temperature, at what temperature does the reaction change from nonspontaneous to spontaneous?

Answer 1

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

Answer 2

A particular reaction, with ΔH° = 55.4 kJ and ΔS° = 94.1 J/K, will change from nonspontaneous to spontaneous at 589 K.

Relation between the standard free Gibbs energy and the spontaneity of a reaction

• When ΔG° < 0, the reaction is spontaneous.
• When ΔG° = 0, the reaction is not spontaneous nor nonspontaneous.
• When ΔG° > 0, the reaction is nonspontaneous.

We want to know at what temperature a reaction changes from nonspontaneous to spontaneous, that is, at what temperature ΔG° = 0.

We know that ΔH° = 55.4 kJ and ΔS° = 94.1 J/K, and that they change very little with the temperature. We can find at what temperature ΔG° = 0 using the following expression.

ΔG° = ΔH° - T × ΔS°

0 = 55.4 kJ - T × (94.1 J/K)

T = 589 K

A particular reaction, with ΔH° = 55.4 kJ and ΔS° = 94.1 J/K, will change from nonspontaneous to spontaneous at 589 K.

Learn more about spontaneity here: brainly.com/question/9552459