Question

In the given graph, \( E_a \) for the reverse reaction will be 

• A. 215 KJ
• B. 90 KJ
• C. 305 KJ
• D. 125 KJ

Hint:

To calculate the activation energy for the reverse reaction, use the relationship \( \Delta H = E_{a (forward)} - E_{a (reverse)} \).

Answer 1

The Correct Option is D

From the graph, the activation energy for the forward reaction is \( E_a = 215 \) KJ. The change in enthalpy \( \Delta H \) is given as \( 90 \) KJ. The activation energy for the reverse reaction is: \[ \Delta H = E_{a (forward)} - E_{a (reverse)} \] Since \( \Delta H = -90 \) KJ (because the products have lower energy than the reactants), we calculate: \[ -90 = 215 - E_{a (reverse)} \] \[ E_{a (reverse)} = 215 - 90 = 125 \text{ KJ} \]