Given `S_(C_(2)H_(6))^(@) = 225 J mol^(-1) K^(-1)` ,
`S_(C_(2)H_(4)^(@) = 220 J mol^(-1) K^(-1), S_(H_(2)^(@) = 130 J mol^(-1) K^(-1)`
Then `Delta S^(@)` for the process
`C_(2)H_(4) + H_(2) rightarrow C_(2)H_(6)` is

Options

A. `+25 J`
B. `-125 J`
C. `135 J`
D. `315 J`

Answer

To find the standard entropy change (ΔS°) for the reaction:

\[ C_2H_4 + H_2 \rightarrow C_2H_6 \]

we can use the standard entropy values provided for each substance involved in the reaction.

### Step-by-step solution:

1. **Write down the standard entropy values:**
- \( S^\circ_{C_2H_6} = 225 \, \text{J mol}^{-1} \text{K}^{-1} \)
- \( S^\circ_{C_2H_4} = 220 \, \text{J mol}^{-1} \text{K}^{-1} \)
- \( S^\circ_{H_2} = 130 \, \text{J mol}^{-1} \text{K}^{-1} \)

2. **Use the formula for the standard entropy change:**
The standard entropy change for the reaction can be calculated using the formula:
\[ \Delta S^\circ = S^\circ_{\text{products}} - S^\circ_{\text{reactants}} \]
For our reaction:
\[ \Delta S^\circ = S^\circ_{C_2H_6} - (S^\circ_{C_2H_4} + S^\circ_{H_2}) \]

3. **Substitute the values into the equation:**
\[ \Delta S^\circ = 225 \, \text{J mol}^{-1} \text{K}^{-1} - (220 \, \text{J mol}^{-1} \text{K}^{-1} + 130 \, \text{J mol}^{-1} \text{K}^{-1}) \]

4. **Calculate the sum of the reactants' entropy:**
\[ S^\circ_{C_2H_4} + S^\circ_{H_2} = 220 + 130 = 350 \, \text{J mol}^{-1} \text{K}^{-1} \]

5. **Now calculate ΔS°:**
\[ \Delta S^\circ = 225 - 350 = -125 \, \text{J mol}^{-1} \text{K}^{-1} \]

### Final Answer:
\[ \Delta S^\circ = -125 \, \text{J mol}^{-1} \text{K}^{-1} \]