The reaction X $$\rightarrow$$ Y is an exothermic reaction. Activation energy of the reaction for X into Y is 150 kJ mol$$^{-1}$$. Enthalpy of reaction is 135 kJ mol$$^{-1}$$. The activation energy for the reverse reaction, Y $$\rightarrow$$ X will be :

The reaction given is X → Y, which is exothermic. This means the enthalpy change, ΔH, is negative. The problem states the enthalpy of reaction is 135 kJ mol⁻¹, so we write ΔH = -135 kJ mol⁻¹.

The activation energy for the forward reaction (X → Y) is given as Ea(forward) = 150 kJ mol⁻¹.

We need to find the activation energy for the reverse reaction (Y → X), denoted as Ea(reverse).

Recall the relationship between activation energies and enthalpy change for a reaction: ΔH = Ea(forward) - Ea(reverse).

Substitute the known values into this equation:

ΔH = Ea(forward) - Ea(reverse)

-135 = 150 - Ea(reverse)

Now, solve for Ea(reverse). First, isolate the term with Ea(reverse) by moving 150 to the left side:

-135 - 150 = - Ea(reverse)

-285 = - Ea(reverse)

Multiply both sides by -1 to solve for Ea(reverse):

285 = Ea(reverse)

So, Ea(reverse) = 285 kJ mol⁻¹.

Comparing with the options:

A. 280 kJ mol⁻¹

B. 285 kJ mol⁻¹

C. 270 kJ mol⁻¹

D. 15 kJ mol⁻¹

Hence, the correct answer is Option B.