A reaction is spontaneous at low temperature but non- spontaneous at high temperature. Which of the following is true for the reaction?

Options

A. `triangleH gt 0, triangleS gt 0`
B. `triangleH lt 0, triangleS gt 0`
C. `triangleH gt 0, triangleS=0`
D. `triangleH lt 0, triangleS lt 0`

Answer

To determine which statement is true for a reaction that is spontaneous at low temperature but non-spontaneous at high temperature, we can analyze the Gibbs free energy equation:

\[ \Delta G = \Delta H - T \Delta S \]

Where:
- \(\Delta G\) = Gibbs free energy change
- \(\Delta H\) = Enthalpy change
- \(T\) = Temperature in Kelvin
- \(\Delta S\) = Entropy change

### Step-by-Step Solution:

1. **Understanding Spontaneity**:
- A reaction is spontaneous when \(\Delta G < 0\) (negative).
- A reaction is non-spontaneous when \(\Delta G > 0\) (positive).

2. **Analyzing the Conditions**:
- The problem states that the reaction is spontaneous at low temperatures and non-spontaneous at high temperatures. This implies that at low temperatures, \(\Delta G < 0\) and at high temperatures, \(\Delta G > 0\).

3. **Setting Up the Equation**:
- At low temperature, we want \(\Delta G < 0\):
\[ \Delta H - T \Delta S < 0 \implies \Delta H < T \Delta S \]
- At high temperature, we want \(\Delta G > 0\):
\[ \Delta H - T \Delta S > 0 \implies \Delta H > T \Delta S \]

4. **Inferring the Signs of \(\Delta H\) and \(\Delta S\)**:
- For the reaction to be spontaneous at low temperatures and non-spontaneous at high temperatures, we can deduce:
- \(\Delta H\) must be negative (exothermic reaction).
- \(\Delta S\) must also be negative (decrease in entropy).

5. **Conclusion**:
- The only condition that satisfies both scenarios (spontaneous at low temperature and non-spontaneous at high temperature) is when \(\Delta H < 0\) and \(\Delta S < 0\).
- Therefore, the correct option is that \(\Delta H < 0\) and \(\Delta S < 0\).

### Final Answer:
The correct statement for the reaction is that \(\Delta H < 0\) and \(\Delta S < 0\).