Answer:
![T > (\Delta H^o)/(\Delta S^o)](https://img.qammunity.org/2020/formulas/chemistry/college/m3zpk7y9azqowkpxncn354wcj8w2yjdo43.png)
Step-by-step explanation:
The spontaneity of a reaction is described by the Gibbs free energy change. Several statements should be emphasized in the context of this problem:
- the change in Gibbs free energy relates enthalpy, entropy and temperature by the following equation:
; - if the change in the Gibbs free energy is negative, the reaction is spontaneous, that is:
or
; - if the change in the Gibbs free energy is positive, the reaction is non-spontaneous, that is:
or
.
In this problem, we wish to find the temperature at which the given reaction becomes spontaneous, that is, the minimum T value for which we obtain a negative value of the Gibbs free energy change:
![\Delta H^o - T\Delta S^o < 0](https://img.qammunity.org/2020/formulas/chemistry/college/erbehshh9avzgznwimyfakyp2p5k49hyfi.png)
Rearrange the inequality:
![\Delta H^o < T\Delta S^o](https://img.qammunity.org/2020/formulas/chemistry/college/u0aktne9dfp9zqyhwjqp7yyh2sckvjueyx.png)
Divide both sides by the change in entropy:
![T > (\Delta H^o)/(\Delta S^o)](https://img.qammunity.org/2020/formulas/chemistry/college/m3zpk7y9azqowkpxncn354wcj8w2yjdo43.png)
Let's take an example. Let's say that we have a change in enthalpy as 50 kJ/mol and a change in entropy of 100 J/K, then:
![T > (50\cdot 10^3 J/mol)/(100 J/K) = 500 K](https://img.qammunity.org/2020/formulas/chemistry/college/ctu42uztctyk96qub89ejqps7mo7nrjcd3.png)
Thus, for these conditions, our reaction would become spontaneous at 500 K and above.