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 the driving force behind the reaction \( \text{CH}_2=\text{CH}_2(g) + \text{Cl}_2(g) \rightarrow \text{ClCH}_2\text{CH}_2\text{Cl}(g) \) with given values of \( \Delta_r H^\circ = -217.5 \, \text{kJ/mol} \) and \( \Delta_r S^\circ = -233.9 \, \text{J/K} \cdot \text{mol} \), we can analyze the spontaneity of the reaction using Gibbs free energy (\( \Delta G \)).

### Step-by-Step Solution:

1. **Identify the Gibbs Free Energy Equation**:
The Gibbs free energy change (\( \Delta G \)) for a reaction is given by the equation:
\[ \Delta G = \Delta H - T \Delta S \]

2. **Determine the Sign of \( \Delta H \)**:
From the given data, \( \Delta_r H^\circ = -217.5 \, \text{kJ/mol} \). A negative \( \Delta H \) indicates that the reaction is exothermic, which favors spontaneity.

3. **Determine the Sign of \( \Delta S \)**:
The given value for entropy change is \( \Delta_r S^\circ = -233.9 \, \text{J/K} \cdot \text{mol} \). A negative \( \Delta S \) suggests that the disorder of the system decreases, which does not favor spontaneity.

4. **Analyze the Conditions for Spontaneity**:
For the reaction to be spontaneous, \( \Delta G \) must be negative. Since \( \Delta H \) is negative (favorable) and \( \Delta S \) is negative (unfavorable), we need to consider the temperature \( T \):
- If \( T \Delta S \) is sufficiently large and positive, it can make \( \Delta G \) positive, thus making the reaction non-spontaneous at high temperatures.
- Conversely, at low temperatures, the negative \( \Delta H \) can dominate, making \( \Delta G \) negative and the reaction spontaneous.

5. **Conclusion**:
Since the reaction has a negative enthalpy change and a negative entropy change, it is supported by enthalpy. Therefore, the reaction is spontaneous at lower temperatures where the enthalpy effect outweighs the entropy effect.

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