Question

A reaction has ΔH∘rxn= -117 kJ and ΔS∘rxn= 327 J/K .

At what temperature is the change in entropy for the reaction equal to the change in entropy for the surroundings? Express your answer in kelvins.

Answer 1

Answer:

357.7982 K

Explanation:

When the entropy of the reaction (Entropy of the system) is equal to the entropy of the surroundings, it means that the system is in equilibrium. So,

$T\times \Delta S^0=\Delta H^{0}$

Given that:

$\Delta H^0_{rxn}=-117 kJ/K$, where negative sign signifies that heat loss of the system. Since the entropy of the system and the surroundings are same and thus only magnitude is to be used in the formula.

1 kJ = 1000 J

so,

$Magnitude\ of\ \Delta H^0_{rxn}=117000 J/K$

$\Delta S^0_{rxn}=327 J/K$

So,

$T\times 327=\Delta 117000$

T = 357.7982 K