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 under which a process is spontaneous 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. **Write the Gibbs Free Energy Equation**:
\[ \Delta G = \Delta H - T\Delta S \]
Here, ΔG is the change in Gibbs free energy, ΔH is the change in enthalpy, T is the temperature in Kelvin, and ΔS is the change in entropy.

2. **Substitute the Given Values**:
Since both ΔH and TΔS are negative, we can denote:
\[ \Delta H < 0 \quad \text{and} \quad T\Delta S < 0 \]
This means we can express TΔS as a negative value:
\[ T\Delta S = -|T\Delta S| \quad \text{(where } |T\Delta S| \text{ is a positive value)} \]

3. **Rearranging the Gibbs Free Energy Equation**:
Substitute the negative values into the Gibbs free energy equation:
\[ \Delta G = \Delta H - (-|T\Delta S|) = \Delta H + |T\Delta S| \]

4. **Analyzing the Sign of ΔG**:
For the process to be spontaneous, ΔG must be negative:
\[ \Delta G < 0 \implies \Delta H + |T\Delta S| < 0 \]
This can be rearranged to:
\[ \Delta H < -|T\Delta S| \]
This indicates that the magnitude of ΔH must be greater than the magnitude of TΔS.

5. **Conclusion**:
Therefore, the condition for spontaneity when both ΔH and TΔS are negative is:
\[ |\Delta H| > |T\Delta S| \]
This means that the absolute value of ΔH must be greater than the absolute value of TΔS for the process to be spontaneous.