Question

Choose the incorrect combination.

$\begin{array}{ccccc}& \Delta H& \Delta S& \text{Temperature}& \text{Spontaneity}\\ \text{I.}& \oplus ve& \ominus ve& \text{Any temperature}& \text{Non-spontaneous}\\ \text{II.}& \ominus ve& \ominus ve& \text{Low temperature}& \text{Non-spontaneous}\\ \text{III.}& \oplus ve& \oplus ve& \text{Low temperature}& \text{Spontaneous}\\ \text{IV.}& \ominus ve& \oplus ve& \text{Any temperature}& \text{Spontaneous}\end{array}$

• A. $I,II$
• B. $II,III$
• C. $III,IV$
• D. $I,IV$

Answer 1

The correct option is B $II,III$

We know, $\Delta G=\Delta H–T\Delta S$
For a spontaneous process $\Delta G$ should be negative.
I) If $\Delta H$ is positive and $\Delta S$ is negative, then $\Delta G$ is positive at all temperatures, hence it is a non-spontaneous process.

II) If $\Delta H$ and $\Delta S$ are both negative, then $\Delta G$ will be negative at low temperature. So it will be a spontaneous process at low temperature.

III) If $\Delta H$ and $\Delta S$ both are positive then at high temperatures, the reaction will be spontaneous.

IV) If $\Delta H$ is negative and $\Delta S$ is positive, then $\Delta G$ will be negative for both high and low temperatures. So it will be spontaneous in both cases.