Oxidising power of chlorine in aqueous solution can be determined by the parameters indicated below
`(1)/(2)CL_(2)(g)overset((1)/(2)Delta_(diss)H^(Theta))(rarr)Cl(g)overset(DeltaH_(Eg)^(Theta))(rarr)`
`Cl^(-)(g)overset(Delta_(hyd)H^(Theta))(rarr)Cl^(-)(aq)`
The energy involved in the conversion of `(1)/(2)Cl_(2)(g)` to
`Cl^(-)(aq)`
(Using the data `Delta_(diss)H_(Cl_(2))^(Theta)=240KJ mol^(-1)`)
`Delta_(Eg)H_(Cl)^(Theta)=-349KJmol^(-1)` ,
`Delta_(Eg)H_(Cl)^(Theta)=-381KJmol^(-1)`) will be

Options

A. `+152 kJ mol^(-1)`
B. `-610 kJ mol ^(-1)`
C. `850 kJ mol ^(-1)`
D. `+ 120 kJ mol ^(-1)`

Answer

To determine the energy involved in the conversion of \( \frac{1}{2}Cl_2(g) \) to \( Cl^-(aq) \), we will use the provided data and apply Hess's law. Here are the steps to solve the problem:

### Step 1: Write the Reaction Steps
We need to break down the conversion of \( \frac{1}{2}Cl_2(g) \) to \( Cl^-(aq) \) into three steps:
1. Dissociation of \( Cl_2 \) into chlorine atoms.
2. Gaining an electron by chlorine atom to form chloride ion.
3. Hydration of chloride ion to form aqueous chloride ion.

### Step 2: Calculate the Enthalpy for Each Step
1. **Dissociation of \( \frac{1}{2}Cl_2(g) \)**:
\[ \frac{1}{2}Cl_2(g) \rightarrow Cl(g) \]
The enthalpy change for this step (\( \Delta H_1 \)) is given by:
\[ \Delta H_1 = \frac{1}{2} \times \Delta H_{diss}(Cl_2) = \frac{1}{2} \times 240 \, \text{kJ/mol} = 120 \, \text{kJ/mol} \]

2. **Electron Gain Enthalpy**:
\[ Cl(g) + e^- \rightarrow Cl^-(g) \]
The enthalpy change for this step (\( \Delta H_2 \)) is given by:
\[ \Delta H_2 = \Delta H_{Eg}(Cl) = -349 \, \text{kJ/mol} \]

3. **Hydration of Chloride Ion**:
\[ Cl^-(g) \rightarrow Cl^-(aq) \]
The enthalpy change for this step (\( \Delta H_3 \)) is given by:
\[ \Delta H_3 = \Delta H_{hyd}(Cl^-) = -381 \, \text{kJ/mol} \]

### Step 3: Calculate Overall Enthalpy Change
Using Hess's law, the overall enthalpy change (\( \Delta H \)) for the conversion from \( \frac{1}{2}Cl_2(g) \) to \( Cl^-(aq) \) is:
\[ \Delta H = \Delta H_1 + \Delta H_2 + \Delta H_3 \]
Substituting the values:
\[ \Delta H = 120 \, \text{kJ/mol} - 349 \, \text{kJ/mol} - 381 \, \text{kJ/mol} \]
Calculating this gives:
\[ \Delta H = 120 - 349 - 381 = -610 \, \text{kJ/mol} \]

### Final Answer
The energy involved in the conversion of \( \frac{1}{2}Cl_2(g) \) to \( Cl^-(aq) \) is \( -610 \, \text{kJ/mol} \).