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

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 as a gas. We will use Hess's Law to do this.

### Step-by-Step Solution:

1. **Understand the Given Reaction:**
The reaction provided is:
\[ 4NO_2(g) + O_2(g) \rightarrow 2N_2O_5(g), \quad \Delta_rH = -111 \, \text{kJ} \]
This indicates that when 4 moles of NO₂ and 1 mole of O₂ react to form 2 moles of gaseous N₂O₅, 111 kJ of energy is released.

2. **Identify the Change in State:**
We need to find the ΔH when N₂O₅ is formed as a solid (N₂O₅(s)) instead of as a gas (N₂O₅(g)).

3. **Use the Given Sublimation Energy:**
The sublimation energy of N₂O₅ is given as:
\[ \Delta H_{sublimation} = 54 \, \text{kJ/mol} \]
This means that to convert 1 mole of N₂O₅ from solid to gas, 54 kJ of energy is required.

4. **Apply Hess's Law:**
According to Hess's Law, the total enthalpy change for a reaction is the sum of the enthalpy changes for the individual steps. We can express the new reaction as:
\[ 4NO_2(g) + O_2(g) \rightarrow 2N_2O_5(s) \]
The ΔH for this reaction can be calculated as:
\[ \Delta H_{reaction} = \Delta H_{reaction \, (g)} - \Delta H_{sublimation} \]
Where:
- \(\Delta H_{reaction \, (g)} = -111 \, \text{kJ}\)
- \(\Delta H_{sublimation} = 54 \, \text{kJ}\)

5. **Calculate the New ΔH:**
Substitute the values into the equation:
\[ \Delta H_{reaction} = -111 \, \text{kJ} - 54 \, \text{kJ} = -165 \, \text{kJ} \]

6. **Conclusion:**
The value of ΔH for the reaction when N₂O₅ is formed as a solid is:
\[ \Delta H = -165 \, \text{kJ} \]