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Why does the following reaction

$$\ce{2NO + O_2 -> 2NO_2}$$

have negative reaction coefficient, i.e., why does the rate of the reaction decrease with an increase in temperature ?

I have been taught that this is the only reaction that has this property.

Mr.HiggsBoson
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  • I find it extremely unlikely that this is not only reaction in the universe with this property. Looking at the free energy of the reaction, and using $G = H - TS$, contemplate the effect of $T$ on the reaction. – Jon Custer May 21 '19 at 16:52
  • @JonCuster How does the thermodynamic equation of G equate to kinetics – Mr.HiggsBoson May 21 '19 at 18:09
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    When this occurs it is because the reaction is more complex than the stoichiometric equation suggests. If there is a ratio of rate constants each of the Arrhenius form rate coefficient = $A_1e^{-E_1/RT}/A_2e^{-E_2/RT} \sim e^{-(E_1-E_2)/RT}$ then the measured value can increase of decrease with temperature depending on the relative size of the two energies $E_{1,2}$ – porphyrin May 21 '19 at 18:58
  • @Harsh Wasnik, the question refers to the rate constant not the equilibrium constant or reaction rate. – porphyrin May 22 '19 at 08:30
  • @porphyrin exactly – Mr.HiggsBoson May 22 '19 at 11:59
  • I believe my answer is very relevant to you: https://chemistry.stackexchange.com/q/95343/44877 – Tan Yong Boon May 27 '19 at 04:08

1 Answers1

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It's strange, but there are mechanisms whereby this type of thing can happen. Basically you have a rate determining step that involves an intermediate species, and that rate determining step must have the usual positive temperature dependence. But the preceding equilibrium that produces the intermediate species could have a negative temperature dependence that dominates.

Let's translate this into concrete terms for the reaction given. A quite plausible mechanism would look like this:

$\ce{NO + O2 <=> ONOO}$ Fast, reversible

$\ce{ONOO + NO -> 2NO2}$ Rate determining step

The equilibrium in the first step is given by

$K_\mathrm{p1} =\dfrac{p^\mathrm{eq}(\ce{ONOO})}{p^\mathrm{eq}(\ce{NO})p^\mathrm{eq}(\ce{O2})}$

Assuming $\ce{ONOO}$ is a minor species, it's formation hardly affects the equilibrium partial pressures of the other species at any time, so we may just use $p(\ce{NO})$ as the nominal nitric oxide partial pressure and similarly for the molecular oxygen. Then we can solve for the partial pressure of the $\ce{ONOO}$ intermediate:

${p^\mathrm{eq}(\ce{ONOO})}=K_\mathrm{p1}(p(\ce{NO}))(p(\ce{O2})$

Now put this result into the rate determining step with rate

$r_2=k_2{p^\mathrm{eq}(\ce{ONOO})p^\mathrm{eq}(\ce{NO})}=k_2K_\mathrm{p1}(p(\ce{NO})^2)(p(\ce{O2})$

Now for the eye-opener: put in the temperature dependence of the equilibrium and rate constants

$K_\mathrm{p1}=K_\mathrm{p1}^0\exp(-\Delta H_1/RT)$

$k_2=k_2^0\exp(-E_{a2}/RT)$

So then

$r_2=k_2^0K_\mathrm{p1}^0\exp(-\dfrac{\color{blue}{E_{a2}+\Delta H_1}}{RT})p(\ce{NO})^2p(\ce{O2})$

The activation energy then contains two terms, one from the inherent rate in Step 2 and the other from the equilibrium in Step 1. The latter is negative if Step 1 is exothermic, as it might well be if a bond is being (temporarily) formed between the reactants. Then if this enthalpy term is the absolutely larger part of the sum ... we see a reversed temperature dependence!

Oscar Lanzi
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