5

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?

laksheya
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    It will become obvious as soon as you look at the reaction rate formula. – Ivan Neretin Aug 23 '19 at 08:56
  • The reaction 1 and 2 are the same reaction... So, the activation energy of both must be equal. – Koba Aug 24 '19 at 02:10
  • @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? – laksheya Aug 24 '19 at 04:25
  • https://chemistry.stackexchange.com/q/38167/81005 Here is an answer I took as reference to my previous comment. – laksheya Aug 24 '19 at 04:27
  • 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. – Ivan Neretin Aug 24 '19 at 07:33
  • @IvanNeretin I'm sorry, I don't understand what you're trying to say. If we keep the temperature constant, and the stoichiometric coefficients change, then the arrhenius constant is the same, the rate constant changes and so does the activation energy. I'm sure I'm going wrong somewhere, but I can't figure it out. – laksheya Aug 24 '19 at 10:39
  • You are totally right. Now what if we don't keep the temperature constant? – Ivan Neretin Aug 24 '19 at 11:30
  • @IvanNeretin If we don't keep the temperature constant, the activation energy is constant but the rate constants are different. But, the question I was trying to solve only changes the stoichiometric coefficients of the reactions, so why are we considering the temperature to not be constant? – laksheya Aug 24 '19 at 14:36
  • By considering the temperature to not be constant, we may deduce the character of the rate's dependence (or independence) on the stoichiometric coefficients. – Ivan Neretin Aug 24 '19 at 18:30
  • Activation energy does not depend on number of moles of reactant(s) as it is defined for per unit mole of reactant. Thus the answer should be (4) – Suraj Sen May 01 '22 at 18:58

2 Answers2

1

No, the activation energy does not depend of the stoicheiometric coefficients. The activation energy is determined by the energy of the electrons and nuclei as the reaction proceeds to its products.

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.

porphyrin
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Rate measurements, while measuring the overall reaction, are concerned only with the rate determining step. This is usually an elementary reaction in the overall mechanism so the overall stoichiometry has no obvious effect and until one looks real closely the orders work out to be small whole numbers. That is until low pressure gas reactions, enzyme reactions, polymerization reactions, photochemical reactions are studied. Then elaborate rate laws show up.

jimchmst
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