Question

The signs of $\Delta$H and $\Delta$s for a reaction to be spontaneous at all temperatures respectively are

• A. Positive , positive
• B. Positive , negative
• C. Negative , negative
• D. Negative , positive

Hint:

Think of it this way: Nature favors lower energy ($-\Delta H$) and higher disorder ($+\Delta S$). When both conditions are met, the reaction has no choice but to be spontaneous!

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The spontaneity of a reaction is determined by the Gibbs free energy change (\(\Delta G\)).
A reaction is spontaneous only when \(\Delta G<0\).
Step 2: Key Formula or Approach:
The relationship is given by the Gibbs-Helmholtz equation:
\[ \Delta G = \Delta H - T\Delta S \]
Where \(T\) is absolute temperature in Kelvin (which is always positive).
Step 3: Detailed Explanation:
To ensure \(\Delta G\) is always negative regardless of the value of \(T\):
1. The enthalpy term \(\Delta H\) should be negative (exothermic).
2. The term \(-T\Delta S\) should be negative. Since \(T\) is positive, \(\Delta S\) must be positive.
Let's check:
\[ \Delta G = (\text{Negative Value}) - (\text{Positive Value}) \times (\text{Positive Value}) \]
\[ \Delta G = \text{Negative} - \text{Positive} = \text{Always Negative} \]
This means the reaction will be spontaneous at every temperature.
Step 4: Final Answer:
The signs required are \(\Delta H\) = negative and \(\Delta S\) = positive.