Entropy changesh for the process, `H_(2)O(l) to H_(2)O` at normal pressure and 274K are given below
`triangleS_("system") =-22.13, triangleS_("surr")=+22.05`, the process is non spontaneous because

Options

A. `triangleS_("system")" is "-ve`
B. `triangleS_("surr")" is "+ve`
C. `triangleS_("u")" is "-ve`
D. `triangleS_("system") ne triangleS_("surr")`

Answer

To determine why the process of converting liquid water (H₂O(l)) to water vapor (H₂O(g)) at 274 K is non-spontaneous, we need to analyze the changes in entropy and the Gibbs free energy.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Entropy change of the system (ΔS_system) = -22.13 J/K
- Entropy change of the surroundings (ΔS_surr) = +22.05 J/K

2. **Calculate the Total Entropy Change of the Universe:**
- The total entropy change of the universe (ΔS_universe) is the sum of the entropy change of the system and the surroundings:
\[ ΔS_{universe} = ΔS_{system} + ΔS_{surr} \]
- Substituting the values:
\[ ΔS_{universe} = -22.13 \, \text{J/K} + 22.05 \, \text{J/K} = -0.08 \, \text{J/K} \]

3. **Determine the Spontaneity of the Process:**
- A process is spontaneous if the total entropy change of the universe is positive (ΔS_universe > 0). In this case:
\[ ΔS_{universe} = -0.08 \, \text{J/K} < 0 \]
- Since ΔS_universe is negative, the process is non-spontaneous.

4. **Relate Entropy Change to Gibbs Free Energy:**
- The Gibbs free energy change (ΔG) is given by the equation:
\[ ΔG = ΔH - TΔS_{universe} \]
- For a process to be spontaneous, ΔG must be less than zero (ΔG < 0). Since ΔS_universe is negative, this implies that ΔG will be greater than ΔH (if ΔH is positive) or will not be negative enough to make ΔG < 0.

5. **Conclusion:**
- The process of converting liquid water to water vapor at 274 K is non-spontaneous because the total entropy change of the universe is negative (ΔS_universe < 0), indicating that the process does not favor spontaneity.