For `N_(2)(g) + 3H_(2)(g) rarr 2NH_(3)(g) + 22kcal`, `E_(a)` for the reaction is `70 kcal`. Hence, the activation energy for `2NH_(3)(g) rarr N_(2)(g) + 3H_(2) (g)` is :

Options

A. `92 kcal`
B. `70 kcal`
C. `48 kcal`
D. `22 kcal`

Explanation

To find the activation energy for the backward reaction \(2NH_3(g) \rightarrow N_2(g) + 3H_2(g)\), we can use the relationship between the activation energies of the forward and backward reactions and the enthalpy change (\(\Delta H\)) of the reaction.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- The forward reaction is:
\[ N_2(g) + 3H_2(g) \rightarrow 2NH_3(g) + 22 \text{ kcal} \]
- The activation energy (\(E_a\)) for the forward reaction is given as \(70 \text{ kcal}\).
- The enthalpy change (\(\Delta H\)) for the reaction is \(-22 \text{ kcal}\) (since energy is released, it is negative).

2. **Understand the Relationship:**
- The relationship between the activation energies of the forward and backward reactions is given by:
\[ E_a(\text{backward}) = E_a(\text{forward}) + |\Delta H| \]
- Here, \(|\Delta H|\) is the absolute value of the enthalpy change.

3. **Calculate the Activation Energy for the Backward Reaction:**
- Substitute the known values into the equation:
\[ E_a(\text{backward}) = 70 \text{ kcal} + 22 \text{ kcal} \]
- Perform the addition:
\[ E_a(\text{backward}) = 92 \text{ kcal} \]

4. **Conclusion:**
- The activation energy for the backward reaction \(2NH_3(g) \rightarrow N_2(g) + 3H_2(g)\) is \(92 \text{ kcal}\).

### Final Answer:
The activation energy for \(2NH_3(g) \rightarrow N_2(g) + 3H_2(g)\) is \(92 \text{ kcal}\).

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