Step 1: Understanding the Concept:
For an exothermic reaction, \(E_a(\text{backward}) = E_a(\text{forward}) - \Delta H\).
Step 2: Detailed Explanation:
Given: \(\Delta H = +22\) kcal (endothermic as written, but reaction is exothermic? Check sign: N\(_2\) + 3H\(_2\) → 2NH\(_3\) is exothermic, so \(\Delta H = -22\) kcal).
Assuming \(\Delta H = -22\) kcal (exothermic),
\(E_a(\text{backward}) = E_a(\text{forward}) - \Delta H = 50 - (-22) = 50 + 22 = 72\) kcal.
Step 3: Final Answer:
+72 kcal