For a chemical reaction,`DeltaH` is negative and `DeltaS` is negative and `DeltaS` is negative . This reaction is

Options

A. spontaneous al all temperatures
B. nonspontaneous at all temperature
C. spontaneous only at high temperature
D. spontaneous only at low temperature

Answer

To solve the question regarding the spontaneity of a chemical reaction with given thermodynamic parameters, we follow these steps:

### Step-by-Step Solution:

1. **Understanding Given Values**:
- We are given that ΔH (enthalpy change) is negative.
- We are also given that ΔS (entropy change) is negative.

2. **Using Gibbs Free Energy Equation**:
- The Gibbs free energy (ΔG) is calculated using the formula:
\[ \Delta G = \Delta H - T \Delta S \]
- Here, T is the temperature in Kelvin.

3. **Analyzing the Sign of ΔG**:
- For a reaction to be spontaneous, ΔG must be negative.
- Since ΔH is negative and ΔS is negative, we need to analyze how these values affect ΔG.

4. **Substituting Values into the Gibbs Equation**:
- Substituting the values into the Gibbs equation:
\[ \Delta G = \text{(negative)} - T \times \text{(negative)} \]
- This simplifies to:
\[ \Delta G = \text{(negative)} + T \times \text{(positive)} \]
- As temperature (T) increases, the term \(T \Delta S\) becomes positive, which can potentially make ΔG positive.

5. **Determining Conditions for Spontaneity**:
- For ΔG to remain negative:
\[ \Delta H - T \Delta S < 0 \]
- Rearranging gives:
\[ \Delta H < T \Delta S \]
- Since ΔS is negative, \(T \Delta S\) becomes negative, and thus, the reaction will be spontaneous only at low temperatures where the negative contribution of ΔH outweighs the positive contribution of \(T \Delta S\).

6. **Conclusion**:
- Therefore, the reaction is spontaneous only at low temperatures.

### Final Answer:
The reaction is spontaneous only at low temperatures.