question_answer

In a reversible reaction, the enthalpy change and the activation energy in the forward direction are  respectively\[-x\text{ }kJmo{{l}^{-1}}\] and y \[kJmo{{l}^{-1}}\]. Therefore, the energy of activation in the backward direction, in\[kJ\,mo{{l}^{-1}}\], is

A) \[(y-x)\]                              

B)  \[(x+y)\]

C) \[(x-y)\]                              

D) \[-(x+y)\]