For the reaction
(i) `H_(2)(g)+Cl_(2)(g)=2HCl(g)+xkJ`
(ii). `H_(2)(g)+Cl_(2)(g)=2HCl(l)+ykJ`
which one of the following statements is correct?

Options

A. `x gt y`
B. `x lt y`
C. `x-y=0`
D. `x=y`

Answer

To solve the problem, we need to analyze the two reactions given and understand the relationship between the energies released (x and y) during the formation of gaseous and liquid HCl.

### Step-by-Step Solution:

1. **Identify the Reactions**:
- The first reaction is:
\[ H_2(g) + Cl_2(g) \rightarrow 2HCl(g) + x \text{ kJ} \]
- The second reaction is:
\[ H_2(g) + Cl_2(g) \rightarrow 2HCl(l) + y \text{ kJ} \]

2. **Understand the Energy Terms**:
- The term \( x \) represents the energy released when 2 moles of HCl gas are formed.
- The term \( y \) represents the energy released when 2 moles of HCl liquid are formed.

3. **Relate the Two Reactions**:
- To convert liquid HCl to gaseous HCl, we need to consider the enthalpy of vaporization (\( \Delta H_{vap} \)). This is the energy required to convert a liquid into a gas.
- The relationship can be expressed as:
\[ \Delta H_{vap} = y - x \]
- This means that the energy released when forming gaseous HCl from liquid HCl is equal to the energy released in the second reaction plus the energy required for vaporization.

4. **Rearranging the Equation**:
- Rearranging the equation gives us:
\[ x = y - \Delta H_{vap} \]

5. **Analyzing the Relationship**:
- Since \( \Delta H_{vap} \) is a positive quantity (energy is required to vaporize the liquid), it follows that:
\[ x < y \]
- This indicates that the energy released when forming gaseous HCl (x) is less than the energy released when forming liquid HCl (y).

6. **Conclusion**:
- Therefore, the correct statement is that \( y \) is greater than \( x \).

### Final Answer:
The correct statement is \( y > x \).

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