Consider the following reaction :- (a) `H_(aq)^(+)+CI_(aq)^(-)toHCI_(l),DeltaH=-X_(1) KJmol^(-1)` (b) `H_2(g)^(+)+CI_2(g)to2HCI_(l),DeltaH=-X_(2) KJmol^(-1)` ( c) `(1)/(2)H_(2(g))+(1)/(2)CI_(2(g))toHCI_(l),DeltaH=-X_(3) KJmol^(-1)`
(d) `H_((g))+CI_((g))toHCI_(l),DeltaH=-X_(4) KJmol^(-1)`
Enthalpy of formation of `HCI_(l)` is :-

Options

A. `-X_(1)Kjmol^(-1)`
B. `-X_(2)Kjmol^(-1)`
C. `-X_(3)Kjmol^(-1)`
D. `-X_(4)Kjmol^(-1)`

Answer

To find the enthalpy of formation of HCl (l), we will analyze the given reactions step by step.

### Step 1: Understand the Definition of Enthalpy of Formation
The enthalpy of formation (ΔH_f) of a compound is defined as the change in enthalpy when one mole of the compound is formed from its elements in their standard states.

### Step 2: Identify the Correct Reaction
We need to find which of the given reactions represents the formation of HCl from its elements in their standard states. The relevant reaction for the formation of HCl is:

**Reaction (a):**
\[ H^+_{(aq)} + Cl^-_{(aq)} \rightarrow HCl_{(l)} \]
This reaction shows the formation of HCl from its ions in aqueous solution.

### Step 3: Analyze Other Reactions
- **Reaction (b):**
\[ H_2(g) + Cl_2(g) \rightarrow 2 HCl_{(l)} \]
This reaction shows the formation of 2 moles of HCl from hydrogen and chlorine gases, but we need the formation of 1 mole.

- **Reaction (c):**
\[ \frac{1}{2} H_2(g) + \frac{1}{2} Cl_2(g) \rightarrow HCl_{(l)} \]
This reaction correctly represents the formation of 1 mole of HCl from its elements in their standard states.

- **Reaction (d):**
\[ H_{(g)} + Cl_{(g)} \rightarrow HCl_{(l)} \]
This reaction also represents the formation of HCl from its gaseous elements, but hydrogen and chlorine do not exist as single atoms in nature; they exist as diatomic molecules (H2 and Cl2).

### Step 4: Conclusion
The correct reaction that represents the formation of HCl from its elements in their standard states is:
\[ \frac{1}{2} H_2(g) + \frac{1}{2} Cl_2(g) \rightarrow HCl_{(l)} \]
Thus, the enthalpy of formation of HCl (l) is given by:
\[ \Delta H_f = -X_3 \text{ KJ/mol} \]

### Final Answer
The enthalpy of formation of HCl (l) is \(-X_3 \text{ KJ/mol}\).

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