Question

A reaction will be spontaneous only at low temperatures if both ΔH and ΔS are negative. For a reaction in which ΔH = −310.1 kJ/mol and ΔS = −89.00 J/K · mol, determine the temperature (in °C) below which the reaction is spontaneous.

Answer 1

Answer:

3211.12 K

Explanation:

The expression for the standard change in free energy is:

$\Delta G=\Delta H-T\times \Delta S$

Where,  

$\Delta G$ is the change in the Gibbs free energy.

T is the absolute temperature. (T in kelvins)

$\Delta H$ is the enthalpy change of the reaction.

$\Delta S$ is the change in entropy.

For reaction to be spontaneous, $\Delta G$

Given, $\Delta H=-310.1\ kJ/mol=-310100\ J/mol$

$\Delta S=-89.00\ J/K.mol$

So,

$\Delta H-T\times \Delta S$

Thus, applying values as:-

$-310100-T\times (-89.00)$

So, T = 3484.27 K

The conversion of T( °C) to T(K) is shown below:

T( °C)  = T(K) - 273.15  

So,  

T = (3484.27 - 273.15) K = 3211.12 K

The temperature below which the reaction is spontaneous is:- 3211.12 K