Question

A reaction occurs spontaneously if :

• A. ${ T \Delta S < \Delta H } $ and both $\Delta H$ and $\Delta S$ are +ve
• B. ${ T \Delta S > \Delta H } $ and both $\Delta H$ and $\Delta S$ are +ve
• C. ${ T \Delta S = \Delta H } $ and both $\Delta H$ and $\Delta S$ are +ve
• D. ${ T \Delta S > \Delta H } $ and $\Delta H$ is + ve and $\Delta S$ is -ve

Answer 1

The Correct Option is B

To determine when a reaction occurs spontaneously, we need to consider the Gibbs Free Energy change, denoted as $\Delta G$. The reaction is spontaneous if the change in Gibbs Free Energy is negative, i.e., $\Delta G < 0$.

The Gibbs Free Energy is defined by the equation:

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

Where:

• $\Delta H$ is the change in enthalpy.
• $T$ is the temperature in Kelvin.
• $\Delta S$ is the change in entropy.

For a reaction to be spontaneous at a given temperature, $\Delta G$ should be less than zero, which implies:

$\Delta H - T \Delta S < 0$

Simplifying this inequality, we get:

$T \Delta S > \Delta H$

Now, let's analyze the options given:

• ${ T \Delta S < \Delta H } $ and both $\Delta H$ and $\Delta S$ are +ve: This condition suggests that the Gibbs Free Energy will be positive, making the reaction non-spontaneous.
• ${ T \Delta S > \Delta H } $ and both $\Delta H$ and $\Delta S$ are +ve: Given this condition, the Gibbs Free Energy becomes negative, hence the reaction will be spontaneous. This is the correct condition for spontaneity.
• ${ T \Delta S = \Delta H } $ and both $\Delta H$ and $\Delta S$ are +ve: In this case, the Gibbs Free Energy is zero, indicating equilibrium, not spontaneity.
• ${ T \Delta S > \Delta H } $ and $\Delta H$ is +ve and $\Delta S$ is -ve: Here, $\Delta G$ would generally be positive, since $\Delta S$ is negative, implying non-spontaneity.

Thus, the correct answer is: ${ T \Delta S > \Delta H } $ and both $\Delta H$ and $\Delta S$ are +ve.