KCET · Chemistry · Chemical Kinetics

The activation energy for the reaction \(X \rightarrow Y\) is \(150\text{ kJ mol}^{-1}\). The change in enthalpy for the above reaction is \(-135\text{ kJ mol}^{-1}\). Then the activation energy for \(Y \rightarrow X\) is

A \(280\text{ kJ mol}^{-1}\)
B \(285\text{ kJ mol}^{-1}\)
C \(270\text{ kJ mol}^{-1}\)
D \(15\text{ kJ mol}^{-1}\)

Verified Solution

Answer & Solution

Correct Answer

(B) \(285\text{ kJ mol}^{-1}\)

Step-by-step Solution

Detailed explanation

The change in enthalpy for a reaction is related to the activation energies of the forward and backward reactions by the equation:

\(\Delta H = E_{a(f)} - E_{a(b)}\)

Given:
\(E_{a(f)} = 150\text{ kJ mol}^{-1}\)
\(\Delta H = -135\text{ kJ mol}^{-1}\)

Substituting the values into the equation:

\(-135 = 150 - E_{a(b)}\)

\(E_{a(b)} = 150 + 135\)

\(E_{a(b)} = 285\text{ kJ mol}^{-1}\)

The activation energy for the reverse reaction \(Y \rightarrow X\) is \(285\text{ kJ mol}^{-1}\).

Answer: \(285\text{ kJ mol}^{-1}\)

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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