# Does the activation energy depend upon the stoichiometric coefficient?

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}

then

(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?

• It will become obvious as soon as you look at the reaction rate formula. Aug 23, 2019 at 8:56
• The reaction 1 and 2 are the same reaction... So, the activation energy of both must be equal.
– Koba
Aug 24, 2019 at 2:10
• @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? Aug 24, 2019 at 4:25
• chemistry.stackexchange.com/q/38167/81005 Here is an answer I took as reference to my previous comment. Aug 24, 2019 at 4:27
• 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. Aug 24, 2019 at 7:33

The rate constant is an experimentally measured quantity and knowing the species involved we invent a possible mechanism, as shown by the two reactions you give. Another reaction could be $$\mathrm{N_2O_5\to N_2O_3+O_2}$$, you can invent more. The molecule will only react by one mechanism at a given temperature and we generally use our intuition to guess what happens (i.e. we write down a scheme by guestimation) but always many experiments have to be done to work out exactly what happens. In other words we finish up by writing down the reaction scheme after experiments show what happens, not start with it. Once you have a scheme you should use the rule $$\displaystyle rate =\frac{1}{a}\frac{dA}{dt}=\frac{1}{b}\frac{dB}{dt}\cdots$$ so that we know how stoicheiometic factors $$a, b$$ etc. are treated.