If the enthalpy of formation and enthalpy of solution of HCl (g) are-92.3kj /mol and -75.14kJ/mol respectively then find the enthalpy of `Cl^(-)`(aq):

Options

A. `-17.16 kJ//mol`
B. `-167.44kJ//mol`
C. 17.16kj/mol
D. none of these

Answer

To find the enthalpy of formation of the chloride ion, \( \text{Cl}^- \) (aq), we can use the given enthalpy values of formation and solution for HCl (g). Here’s the step-by-step solution:

### Step 1: Write the Reaction for the Formation of HCl
The formation of HCl from its elements can be represented as:
\[ \text{H}_2(g) + \frac{1}{2} \text{Cl}_2(g) \rightarrow \text{HCl}(g) \]
The enthalpy change for this reaction is given as:
\[ \Delta H_f^\circ (\text{HCl}) = -92.3 \text{ kJ/mol} \]

### Step 2: Write the Dissolution Reaction of HCl
When HCl gas is dissolved in water, it dissociates into ions:
\[ \text{HCl}(g) \rightarrow \text{H}^+(aq) + \text{Cl}^-(aq) \]
The enthalpy change for this dissolution is given as:
\[ \Delta H_{sol}^\circ (\text{HCl}) = -75.14 \text{ kJ/mol} \]

### Step 3: Combine the Reactions
To find the enthalpy of formation of the chloride ion, we can combine the enthalpy of formation of HCl and the enthalpy of solution of HCl. We can express this as:
\[ \Delta H_f^\circ (\text{Cl}^-(aq)) = \Delta H_f^\circ (\text{HCl}) + \Delta H_{sol}^\circ (\text{HCl}) \]

### Step 4: Substitute the Values
Now, substituting the values we have:
\[ \Delta H_f^\circ (\text{Cl}^-(aq)) = -92.3 \text{ kJ/mol} + (-75.14 \text{ kJ/mol}) \]
\[ \Delta H_f^\circ (\text{Cl}^-(aq)) = -92.3 - 75.14 = -167.44 \text{ kJ/mol} \]

### Final Answer
Thus, the enthalpy of formation of \( \text{Cl}^-(aq) \) is:
\[ \Delta H_f^\circ (\text{Cl}^-(aq)) = -167.44 \text{ kJ/mol} \]

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