Question

For a reaction, AH = -75 kJ/mol and AS = -0.081 kJ/(K-mol). At whattemperatures is this reaction spontaneous?O A. T< 930 KB. T< 100 KOC. At all temperaturesOD. T> 930 KSUBMIT

Answer 1

B. T< 100 K

Step-by-step explanation:

1st) It is necessary to calculate the temperature with the given condition of ΔH and ΔS:

`\begin{gathered} T=(ΔH)/(ΔS) \\ T=\frac{(-75\text{ kJ/mol}){}}{(-0.081\text{ kJ/K*mol})} \\ T=926K \end{gathered}`

2nd) The reaction with negative enthalpy change and negative entropy change will be spontaneous only if the absolute value of the product between T and ΔS is less than ΔH:

`\begin{gathered} \lvert{T*\Delta S}\rvert<\Delta H \\ \lvert{75}\rvert<\Delta H \end{gathered}`

At 926K the product will be equal to

So, at temperatures less than 100K the reaction will be spontaneous.