For the decomposition of $\ce{N2O5(g)}$ it is given that

$$ \begin{align} \ce{2 N2O5(g) &→ 4 NO2(g) + O2(g)} &\quad\text{activation energy} &= E_\mathrm{a}\\ \ce{N2O5(g) &→ 2 NO2(g) + 1/2 O2(g)} &\quad\text{activation energy} &= E'_\mathrm{a} \end{align} $$


(1) $E_\mathrm{a} = 2E'_\mathrm{a}$
(2) $E_\mathrm{a} > E'_\mathrm{a}$
(3) $E_\mathrm{a} < E'_\mathrm{a}$
(4) $E_\mathrm{a} = E'_\mathrm{a}$

In the definition of activation energy, it doesn't say if that is the minimum energy required for 1 mole of reactant or otherwise. Can anyone help me understand this?

  • 1
    $\begingroup$ It will become obvious as soon as you look at the reaction rate formula. $\endgroup$ – Ivan Neretin Aug 23 '19 at 8:56
  • $\begingroup$ The reaction 1 and 2 are the same reaction... So, the activation energy of both must be equal. $\endgroup$ – Koba Aug 24 '19 at 2:10
  • $\begingroup$ @IvanNeretin In the reaction rate formula, the rate constant changes with different stoichiometric coefficients. So, going by the arrhenius equation, if rate constant changes, the activation energy should change too. But that's wrong according to the answer given. Can you explain further? $\endgroup$ – laksheya Aug 24 '19 at 4:25
  • $\begingroup$ chemistry.stackexchange.com/q/38167/81005 Here is an answer I took as reference to my previous comment. $\endgroup$ – laksheya Aug 24 '19 at 4:27
  • $\begingroup$ Don't look at the rate constant per se. (Of course it will change, but then again, it's not the activation energy, is it?) Look at its temperature dependence. $\endgroup$ – Ivan Neretin Aug 24 '19 at 7:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.