Consider the following reaction at temperature T :
`CH_(2)=CH_(2)(g) +Cl_(2)(g)rarrClCH_(2)CH_(2)Cl(g)`
`Delta_(r ) H^(@)=-217.5kJ//"mol, " Delta_(r )S^(@)=-233.9J//K-"mol"`
Reaction is supported by :

Options

A. entropy enthalpy
B. enthalpy
C. both (a) & (b)
D. neither

Explanation

To determine whether the given reaction is supported by enthalpy, entropy, both, or neither, we can analyze the provided thermodynamic data:

1. **Given Data:**
- Reaction: \( \text{CH}_2=\text{CH}_2(g) + \text{Cl}_2(g) \rightarrow \text{ClCH}_2\text{CH}_2\text{Cl}(g) \)
- \( \Delta_r H^\circ = -217.5 \, \text{kJ/mol} \)
- \( \Delta_r S^\circ = -233.9 \, \text{J/(K·mol)} \)

2. **Understanding Spontaneity:**
- A reaction is spontaneous if the Gibbs free energy change (\( \Delta G \)) is negative.
- The relationship between Gibbs free energy, enthalpy, and entropy is given by:
\[ \Delta G = \Delta H - T \Delta S \]
- For spontaneity, we need \( \Delta G < 0 \).

3. **Analyzing Enthalpy (\( \Delta H \)):**
- The value of \( \Delta H \) is negative (\( -217.5 \, \text{kJ/mol} \)), indicating that the reaction is exothermic. This contributes favorably to spontaneity.

4. **Analyzing Entropy (\( \Delta S \)):**
- The value of \( \Delta S \) is negative (\( -233.9 \, \text{J/(K·mol)} \)). A negative entropy change suggests that the disorder of the system decreases, which does not favor spontaneity.

5. **Determining Gibbs Free Energy (\( \Delta G \)):**
- To assess whether \( \Delta G \) is negative, we can substitute the values into the Gibbs free energy equation:
\[ \Delta G = -217.5 \times 10^3 \, \text{J/mol} - T \times (-233.9) \, \text{J/(K·mol)} \]
- Rearranging gives:
\[ \Delta G = -217500 + 233.9T \]
- The sign of \( \Delta G \) will depend on the temperature \( T \). At sufficiently low temperatures, the negative term \( -217500 \) will dominate, making \( \Delta G < 0 \).

6. **Conclusion:**
- Since \( \Delta H \) is negative (favorable) and \( \Delta S \) is negative (unfavorable), the reaction is supported by enthalpy but not by entropy. Therefore, the correct answer is that the reaction is supported by enthalpy.

### Final Answer:
The reaction is supported by enthalpy.