What is the activation energy for the reverse of this reaction?
`N_(2)O_(4)(g)rarr2NO_(2)(g)`
Data for the given reaction is `: DeltaH=+54 kJ and E_(a)=+57.2kJ:`

Options

A. `-54 kJ`
B. `+3.2 kJ`
C. `+60.2 kJ`
D. `+111.2 kJ`

Answer

To find the activation energy for the reverse reaction of the given reaction \( N_2O_4(g) \rightarrow 2NO_2(g) \), we can use the relationship between the activation energies of the forward and reverse reactions and the enthalpy change (\( \Delta H \)) of the reaction.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Enthalpy change (\( \Delta H \)) for the reaction: \( +54 \, \text{kJ} \)
- Activation energy for the forward reaction (\( E_a \) forward): \( +57.2 \, \text{kJ} \)

2. **Understand the Relationship:**
The relationship between the activation energies and the enthalpy change is given by the formula:
\[ \Delta H = E_a \text{ (forward)} - E_a \text{ (reverse)} \]
Where:
- \( E_a \text{ (reverse)} \) is the activation energy for the reverse reaction.

3. **Rearranging the Formula:**
To find the activation energy for the reverse reaction, we can rearrange the formula:
\[ E_a \text{ (reverse)} = E_a \text{ (forward)} - \Delta H \]

4. **Substituting the Values:**
Now we can substitute the known values into the rearranged formula:
\[ E_a \text{ (reverse)} = 57.2 \, \text{kJ} - 54 \, \text{kJ} \]

5. **Calculating the Activation Energy for the Reverse Reaction:**
\[ E_a \text{ (reverse)} = 3.2 \, \text{kJ} \]

6. **Conclusion:**
The activation energy for the reverse reaction \( N_2O_4(g) \leftarrow 2NO_2(g) \) is \( +3.2 \, \text{kJ} \).

### Final Answer:
The activation energy for the reverse reaction is \( +3.2 \, \text{kJ} \).

---