in which first step remains essentially in equilibrium. If $\Delta \mathrm{H}$ is the enthalpy change for the first reaction and $\mathrm{E}_{0}$ is the activation energy for the second reaction, the activation energy of the overall reaction will be given by
(a)$\mathrm{E}_{0}$
(b)$\mathrm{E}_{0}-\Delta \mathrm{H}$
(c)$\mathrm{E}_{0}+\Delta \mathrm{H}$
(d)$\mathrm{E}_{0}+2 \Delta \mathrm{H}$
Answer
SELF-PRACTICE — the source book printed no answer.
Nothing is invented here, so this question has no answer on record.