In which case, process will be spontaneous at all temperature?
In which case, process will be spontaneous at all temperature?
A
`triangleHlt0,triangleSgt0`
B
`triangleHgt0,triangleSgt0`
C
`triangleHlt0,triangleSlt0`
D
`triangleHgt0,triangleSlt0`
Text Solution
AI Generated Solution
The correct Answer is:
To determine in which case a process will be spontaneous at all temperatures, we can use the Gibbs free energy equation:
\[
\Delta G = \Delta H - T \Delta S
\]
Where:
- \(\Delta G\) = change in Gibbs free energy
- \(\Delta H\) = change in enthalpy
- \(T\) = temperature (in Kelvin)
- \(\Delta S\) = change in entropy
### Step-by-Step Solution:
1. **Understand the Conditions for Spontaneity**:
For a process to be spontaneous, the change in Gibbs free energy (\(\Delta G\)) must be negative:
\[
\Delta G < 0
\]
2. **Rearranging the Gibbs Free Energy Equation**:
From the equation \(\Delta G = \Delta H - T \Delta S\), we can rearrange it to understand the conditions better:
\[
\Delta G < 0 \implies \Delta H - T \Delta S < 0
\]
3. **Analyzing the Terms**:
This inequality can be rearranged to:
\[
\Delta H < T \Delta S
\]
This means that for \(\Delta G\) to be negative, the term \(T \Delta S\) must be greater than \(\Delta H\).
4. **Considering the Signs of \(\Delta H\) and \(\Delta S\)**:
- If \(\Delta S > 0\) (entropy increases), then as temperature \(T\) increases, \(T \Delta S\) will also increase.
- If \(\Delta H < 0\) (exothermic process), this will help ensure that \(\Delta G\) remains negative.
5. **Conclusion**:
For the process to be spontaneous at all temperatures, we need:
- \(\Delta S > 0\) (positive change in entropy)
- \(\Delta H < 0\) (negative change in enthalpy)
Thus, the correct option is that the process will be spontaneous at all temperatures when \(\Delta S > 0\) and \(\Delta H < 0\).
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