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 gaseous water (H₂O(g)) at normal pressure and 274 K is non-spontaneous, we can analyze the given entropy changes for the system and surroundings.

### Step-by-Step Solution:

1. **Identify Given Data:**
- Change in entropy of the system (ΔS_system) = -22.13 J/K
- Change in entropy of the surroundings (ΔS_surr) = +22.05 J/K
- Temperature (T) = 274 K

2. **Calculate the Change in Entropy of the Universe (ΔS_universe):**
\[ ΔS_{universe} = ΔS_{system} + ΔS_{surr} \]
Substituting the values:
\[ ΔS_{universe} = -22.13 \, \text{J/K} + 22.05 \, \text{J/K} = -0.08 \, \text{J/K} \]

3. **Determine Spontaneity:**
- For a process to be spontaneous, the change in entropy of the universe (ΔS_universe) must be greater than zero (ΔS_universe > 0).
- In this case, we found that ΔS_universe = -0.08 J/K, which is less than zero.

4. **Conclusion:**
- Since ΔS_universe is negative, the process of converting liquid water to gaseous water at 274 K is non-spontaneous.

### Final Answer:
The process is non-spontaneous because the change in entropy of the universe (ΔS_universe) is negative.

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