5
$\begingroup$

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?

$\endgroup$
10
  • 1
    $\begingroup$ It will become obvious as soon as you look at the reaction rate formula. $\endgroup$ Aug 23, 2019 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, 2019 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, 2019 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, 2019 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. $\endgroup$ Aug 24, 2019 at 7:33

2 Answers 2

1
$\begingroup$

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.

$\endgroup$
-1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.