The value of `Delta H^(@)` in kJ for the reaction will be `CS_(2)(l) + 4NOCl(g) rightarrow CCCl_(4)(l) + 2SO_(2)(g) + 2N_(2)(g)` if
`Delta H_(t)^(@)(CS_(2)) = -X`
`DeltaH_(l)^(@)(NOCl) = -y`
`Delta H_(l)^(@)(CCCL_(4)) = + z`
`Delta H_(l)^(@)(SO_(2)) = - r`

Options

A. ` x + 4y -z - 2r`
B. `r + z+ 4y -x`
C. `2r + z +4y + x`
D. `x + 4y + z -2r`

Answer

To find the value of \(\Delta H^\circ\) for the reaction:

\[ \text{CS}_2(l) + 4\text{NOCl}(g) \rightarrow \text{CCl}_4(l) + 2\text{SO}_2(g) + 2\text{N}_2(g) \]

we will use the standard enthalpy of formation values provided:

- \(\Delta H_f^\circ(\text{CS}_2) = -X\)
- \(\Delta H_f^\circ(\text{NOCl}) = -y\)
- \(\Delta H_f^\circ(\text{CCl}_4) = +z\)
- \(\Delta H_f^\circ(\text{SO}_2) = -r\)
- \(\Delta H_f^\circ(\text{N}_2) = 0\) (since it is in its standard state)

### Step-by-Step Solution:

1. **Write the expression for \(\Delta H^\circ\) of the reaction:**
\[ \Delta H^\circ_{\text{reaction}} = \sum \Delta H_f^\circ \text{(products)} - \sum \Delta H_f^\circ \text{(reactants)} \]

2. **Identify the products and their enthalpies:**
- For \(\text{CCl}_4\), the enthalpy of formation is \(+z\).
- For \(2\text{SO}_2\), the total enthalpy of formation is \(2 \times (-r) = -2r\).
- For \(2\text{N}_2\), the enthalpy of formation is \(2 \times 0 = 0\).

Thus, the total enthalpy of formation for the products is:
\[ \Delta H_f^\circ(\text{products}) = z - 2r + 0 = z - 2r \]

3. **Identify the reactants and their enthalpies:**
- For \(\text{CS}_2\), the enthalpy of formation is \(-X\).
- For \(4\text{NOCl}\), the total enthalpy of formation is \(4 \times (-y) = -4y\).

Thus, the total enthalpy of formation for the reactants is:
\[ \Delta H_f^\circ(\text{reactants}) = -X - 4y \]

4. **Combine the expressions:**
Now, substitute the values into the expression for \(\Delta H^\circ_{\text{reaction}}\):
\[ \Delta H^\circ_{\text{reaction}} = (z - 2r) - (-X - 4y) \]
Simplifying this gives:
\[ \Delta H^\circ_{\text{reaction}} = z - 2r + X + 4y \]

5. **Final expression:**
Therefore, the final expression for \(\Delta H^\circ\) is:
\[ \Delta H^\circ_{\text{reaction}} = z - 2r + X + 4y \]

### Final Answer:
\[ \Delta H^\circ = z - 2r + X + 4y \]