Question

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

• 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.
a) 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.
b) 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.
c) If $\Delta H$ and $\Delta S$ both are positive then at high temperatures, the reaction will be spontaneous.
d) 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.