Question

By using relation \( \Delta G = \Delta H - T \Delta S \), which of the following is incorrect for a spontaneous reaction at a given temperature?

• A. \( \Delta H>0, \Delta S>0 \)
• B. \( \Delta H>0, \Delta S<0 \)
• C. \( \Delta H<0, \Delta S>0 \)
• D. \( \Delta H<0, \Delta S<0 \)

Answer 1

The Correct Option is B

The spontaneity of a reaction is determined by the Gibbs free energy change \( \Delta G \), given by: \[ \Delta G = \Delta H - T \Delta S \] For a reaction to be spontaneous, \( \Delta G \) must be negative. Let’s analyze each case: 

1. \( \Delta H>0, \Delta S>0 \): Here, the enthalpy is positive and the entropy is positive. At high temperatures, \( T \Delta S \) can outweigh \( \Delta H \), making \( \Delta G \) negative, thus spontaneous at high temperatures. This case is possible for a spontaneous reaction at high temperature. 

2. \( \Delta H>0, \Delta S<0 \): Here, both \( \Delta H \) and \( \Delta S \) are positive and negative, respectively. No matter the temperature, \( \Delta G \) will always be positive, so this cannot represent a spontaneous reaction. 

3. \( \Delta H<0, \Delta S>0 \): Here, both \( \Delta H \) and \( \Delta S \) are negative and positive, respectively. For this case, \( \Delta G \) will always be negative, making the reaction spontaneous. 

4. \( \Delta H<0, \Delta S<0 \): In this case, both \( \Delta H \) and \( \Delta S \) are negative, and the spontaneity depends on the temperature. At low temperatures, the term \( T \Delta S \) will be small enough to make \( \Delta G \) negative, thus spontaneous. Hence, the incorrect statement is Option (2) where \( \Delta H>0 \) and \( \Delta S<0 \). Thus, the correct answer is \( \boxed{2} \).