The mechanism for the reaction: \begin{align} \ce{NO_{(g)} + O_{2(g)} &<=>[k_{-1}][k_{1}] NO_{3(g)}} &\text{(fast)}\\ \ce{NO_{3(g)} + NO_{(g)} &<=>[K_{2}] 2\,NO_{2(g)}} &\text{(slow)}\\ \end{align}

The experimental rate law: $$Rate=-\dfrac{\Delta\ce{[NO]}}{\Delta t}=k\ce{[NO]}^2\ce{[O_{2}]}$$

At equilibrium, the forward and reverse rates are equal to each other. We then have $$k_{1}\ce{[NO]}\ce{[O_{2}]}=k_{-1}\ce{[NO_{3}]}$$

This can be re-arranged to $$\ce{[NO_{3}]}=\dfrac{K_{1}}{K_{-1}}\ce{[NO]}\ce{[O_{2}]}$$

Since $2\,\ce{NO}$ disappear in the overall reaction for every NO that reacts in the second step, we multiply the overall reaction rate by two and thus have:


I understand everything up to this point. What I do not understand is how we get to this equation from the previous one.



1 Answer 1


The final rate equation is generated from substituting $\ce{[NO3]}$ defined in the third to last equation:

$\ce{[NO3]} = \frac{k_1}{k_{–1}}\ce{[NO][O_2]}$

into the second to last equation:


This will give you the final rate equation defined only in terms of the starting concentrations. The $\ce{[NO_3]}$ term is replaced, $\ce{[NO]}$ squared, and $\ce{[O_2]}$ remains. None of the equilibrium constants cancel out. The intermediate $\ce{[NO_3]}$ is not included in the final equation, as it is consumed in the overall reaction.


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.