When `Delta H and TDeltaS` both are negative , then for spontaneous process which option is true?

Options

A. `Delta H gt Tdelta S`
B. `Delta H lt Tdelta S`
C. `Delta H = Tdelta S`
D. `Delta H = Delta U`

Answer

To determine the conditions for spontaneity when both ΔH (change in enthalpy) and TΔS (temperature times change in entropy) are negative, we can analyze the Gibbs free energy equation:

### Step-by-Step Solution:

1. **Understand the Gibbs Free Energy Equation**:
The spontaneity of a process is determined by the Gibbs free energy change (ΔG), which is given by the equation:
\[ \Delta G = \Delta H - T\Delta S \]

2. **Identify the Given Conditions**:
We are given that both ΔH and TΔS are negative:
- ΔH < 0 (exothermic reaction)
- TΔS < 0 (entropy change is negative)

3. **Analyze the Sign of ΔG**:
Since both ΔH and TΔS are negative, we can substitute these values into the Gibbs free energy equation:
\[ \Delta G = \text{(negative)} - \text{(negative)} \]
This can be simplified to:
\[ \Delta G = \text{negative} + \text{positive} \]
Here, we need to determine if ΔG will be negative.

4. **Determine the Condition for Spontaneity**:
For the process to be spontaneous, ΔG must be negative. This means:
\[ \Delta H - T\Delta S < 0 \]
Rearranging this gives:
\[ \Delta H < T\Delta S \]
However, since both ΔH and TΔS are negative, we can also express this as:
\[ |\Delta H| > |T\Delta S| \]
which implies that the magnitude of ΔH must be greater than the magnitude of TΔS.

5. **Conclusion**:
Therefore, for the process to be spontaneous when both ΔH and TΔS are negative, the correct condition is:
\[ \Delta H > T\Delta S \]

### Final Answer:
The correct option is that ΔH must be greater than TΔS for the process to be spontaneous.

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