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 ΔH° 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 following equation:

\[ \Delta H° = \sum \Delta H_f^\circ \text{(products)} - \sum \Delta H_f^\circ \text{(reactants)} \]

### Step 1: Identify the standard enthalpy of formation for each substance.

- For the products:
- CCl₄: ΔH_f° = +z
- SO₂: ΔH_f° = -r (since there are 2 moles of SO₂, we will consider this as 2 * (-r) = -2r)
- N₂: ΔH_f° = 0 (standard enthalpy of formation for elements in their standard state is zero)

- For the reactants:
- CS₂: ΔH_f° = -X
- NOCl: ΔH_f° = -y (since there are 4 moles of NOCl, we will consider this as 4 * (-y) = -4y)

### Step 2: Substitute the values into the equation.

Now, we can substitute these values into the ΔH° equation:

\[ \Delta H° = \left( z + (-2r) + 0 \right) - \left( (-X) + (-4y) \right) \]

This simplifies to:

\[ \Delta H° = z - 2r + X + 4y \]

### Step 3: Rearrange the equation.

Rearranging gives us:

\[ \Delta H° = z - 2r + X + 4y \]

### Step 4: Final expression.

Thus, the final expression for ΔH° is:

\[ \Delta H° = z - 2r + X + 4y \]