Consider the reaction,
`4NO_(2)(g)+O_(2)(g)rarr2N_(2)O_(5)(g),Delta_(r )H=-111 kJ.` If `N_(2)O_(5)(s)` is formed instead of `N_(2)O_(5)(g)` in the above reaction, the `Delta_(r )H` value will be
(Given, `DeltaH` of sublimation for `N_(2)O_(5)` is `54 kJ mol^(-1))`

Options

A. `-165 KJ`
B. `+54 KJ`
C. `+219 KJ`
D. `-219 KJ`

Explanation

To solve the problem, we need to determine the change in enthalpy (ΔH) for the reaction when N2O5 is formed as a solid instead of a gas. We will use Hess's law, which states that the total enthalpy change for a reaction is the sum of the enthalpy changes for the individual steps.

### Step-by-Step Solution:

1. **Identify the given reaction and its ΔH**:
The reaction is:
\[ 4NO_{2}(g) + O_{2}(g) \rightarrow 2N_{2}O_{5}(g) \quad \Delta_{r}H = -111 \text{ kJ} \]

2. **Understand the sublimation process**:
The sublimation of N2O5 from solid to gas is given as:
\[ N_{2}O_{5}(s) \rightarrow N_{2}O_{5}(g) \quad \Delta H_{sublimation} = +54 \text{ kJ/mol} \]
Since we are forming 2 moles of N2O5, the total enthalpy change for sublimation will be:
\[ \Delta H_{sublimation, total} = 2 \times 54 \text{ kJ} = 108 \text{ kJ} \]

3. **Apply Hess's Law**:
According to Hess's law, the overall enthalpy change for the reaction when forming solid N2O5 (ΔH') can be calculated as:
\[ \Delta H' = \Delta H_{reaction} - \Delta H_{sublimation, total} \]
Substituting the values we have:
\[ \Delta H' = (-111 \text{ kJ}) - (108 \text{ kJ}) = -219 \text{ kJ} \]

4. **Final Answer**:
The ΔH value when N2O5 is formed as a solid instead of a gas is:
\[ \Delta H' = -219 \text{ kJ} \]

### Summary:
The enthalpy change for the reaction when N2O5 is formed as a solid instead of a gas is -219 kJ.