Question

For a reaction, $△H=+29kJmo{l}^{-1};△S=-35J{K}^{-1}mo{l}^{-1}$ at what temperature, the reaction will be spontaneous?

• A. ${828.7}^{o}C$
• B. $828.7K$
• C. Spontaneous at all temperature
• D. Non-Spontaneous at all temperature

Answer 1

The correct option is D Non-Spontaneous at all temperature

Given-$△H=+29kJmo{l}^{-1};△S=-35J{K}^{-1}mo{l}^{-1}$
The expression for the Gibbs free energy change is given as : $△G=△H-T△S$
For the reaction to be sponteneous $△G$ should be negative
But here,
$△H=+ve,△S=-ve$
So, $△G$ = $△H-T△S$ will always be positive.
Both the enthalpy and entropy factors are not favourable for the reaction to proceed.
So the reaction will always be non-spontaneous at all T.
so, defining value of T at which it is spontaneous is not possible here.


Theory:

For spontaneous process, $\Delta S_{univ}$ is greater than 0 and minimum value of $T$ is 0, $T$ will be positive.
So for spontaneity $\Delta G_{sys}$ must be negative.

For Spontaneous reaction $G_{sys}$ is less than 0
For non spontaneous reaction $G_{sys}$ is greater than 0
For equilibrium condition $G_{sys}=0$

Factors Affecting Spontaneity of a Reaction :

$\Delta G=\Delta H-T\Delta S$
1. If the reaction is exothermic, that is $\Delta H$ is negative and in the process entropy is increasing, that is $\Delta S$ is positive or greater than 0, So the sign of $\Delta G$ will be negative that means the process is spontaneous.

2.If the reaction is endothermic, that is $\Delta H$ is positive and in the process entropy is decreasing, that is $\Delta S$ is negative, So the sign of $\Delta G$ will be positive that means the process is non spontaneous.

3. If the reaction is exothermic, that is $\Delta H$ is negative and in the process entropy is decreasing, that is $\Delta S$ is negative, So the sign of $\Delta G$ may be positive or negative.

5. If the reaction is endothermic, that is $\Delta H$ is positive and in the process entropy is increasing, that is $\Delta S$ is positive, So the sign of $\Delta G$ may be positive or negative.