Question

The enthalpy change for a given reaction at $298K$ is $-xcal/mol$. If the reaction occurs spontaneously at $298K$, the entropy change at that temperature:

• A. Can be negative but numerically larger than $x/298cal{K}^{-1}mo{l}^{-1}$
• B. Can be negative but numerically smaller than $x/298cal{K}^{-1}mo{l}^{-1}$
• C. Cannot be negative
• D. Cannot be positive

Answer 1

The correct option is A Can be negative but numerically larger than $x/298cal{K}^{-1}mo{l}^{-1}$

Solution:- (B) Can be negative but numerically smaller than $\frac{x}{298}cal/K-mol$

As we know that,

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

For spontaneity,

$\Delta G<0$

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

Given:- $\Delta H=-xcal/mol\Rightarrow -veT=298K$

Therefore,

For $\Delta G<0\Delta S$ can be negative but the numerical value must be less than $\frac{x}{298}cal/K-mol$.