1
$\begingroup$

In introductory chemical kinetics, an elementary reaction is taken for instance $$\ce{A -> B}$$ Then, we go on to derive integrated rate kinetics equations for them by assuming $$\text{rate} = k [\ce{A}]^n$$

Could it possible in a more generic case that : $\text{rate}=k[\ce{A}]^n [\ce{B}]^m$ and the total order of the reaction be $m+n$? In other words, can the rate of reaction depend on both the molar concentrations of reactants and products independently?

Improve this question
$\endgroup$
2
  • 1
    $\begingroup$ For complicated mechanism, it is actually quite common to have a rate law which is dependent on the concentration of the products. However, their orders will be negative, meaning that they slow down the rate of the reaction which makes sense. $\endgroup$ – Nanoputian Mar 9 '16 at 20:45
  • 1
    $\begingroup$ Not necessarily negative; autocatalytic reactions are known as well. $\endgroup$ – Ivan Neretin Mar 9 '16 at 21:15
2
$\begingroup$

There exist autocatalytic reactions than can show such a behaviour. E.g. a reaction $\ce{A + B \rightarrow 2\ B}$ might show the rate $k[\ce{A}] [\ce{B}]$.

The reaction of permanganate with oxalic acid is a well known autocatalytic reaction. $$ \mathrm{ 5\ (COOH)_2 + 2\ MnO_4^- + 6\ H^+ \rightarrow 10\ CO_2 + 2\ Mn^{2+} + 8\ H_2O} $$ Here the generated manganese(II) ions catalyze the reaction. The reaction starts slowly if no initial manganese(II) ions are present, and then speeds up.

Improve this answer
$\endgroup$
1
  • $\begingroup$ However, logically speaking if I have a fixed concentration of reactant A and I am varying concentration of B with reference to the general reaction A --> B. Then shouldn't the rate also depend on concentration of B? $\endgroup$ – Ramesh Agarwal Dec 12 '19 at 4:39
0
$\begingroup$

No, because the reaction order is defined by how the rate of reaction changes when the concentrations of one reactant is changed (though rate of product formation can be considered as well), and can be thought of as a number the sum of the possible successful collisions that the atoms can make. For example, in the reaction $$\ce{A +B->products}$$ There order of the reaction can be:

Reaction Order

[source]

If $\ce{B}$ is not a reactant, then it has no place in a rate equation with $\ce{A}$ because the two aren't colliding in order to react.

Improve this answer
$\endgroup$
3
  • $\begingroup$ What about a reaction that is autocatalytic? (I just saw there is a new answer discussing this, not sure how I missed that, but the question still stands). $\endgroup$ – SendersReagent Mar 9 '16 at 23:16
  • $\begingroup$ The rate equation would change if B is a reactant. In the question OP asked the equation was not $\ce{A +B ->2B}$ $\endgroup$ – ringo Mar 9 '16 at 23:44
  • $\begingroup$ I was thinking more of a reaction that can be catalyzed by acid that makes acid as a product. For instance, one of the reactions I teach is the reaction of aniline and acetic anhydride to make acetanilide and acetic acid. I realize this reaction is quite fast without an acid catalyst and a wake acid like AcOH probably isn't a very strong catalyst, but it technically should catalyze the reaction to some extent. $\endgroup$ – SendersReagent Mar 10 '16 at 0:02
0
$\begingroup$

You have given a reaction for an irreversible reaction: $$\ce{A -> B}$$ so the products would not really come into consideration. For a reversible reaction: $$\ce{A <=> B}$$ the relative rates of reaction of the forward and backward reactions would yield the equilibrium constant, $\ce{k_{eq}}$, which is given by: $$\ce{k_{eq} = \dfrac{[B]}{[A]}}$$. The forward reaction would be $$\ce{ r_{A} = k_{A} [A]^{n_{A}}}$$ and the reverse reaction or B would be $$\ce{ r_{B} = k_{B} [B]^{n_{B}}}$$ The quirk here is that $\ce{n_{A}}$ and $\ce{n_{B}}$ do not need to be integers for the rate equation, but in the equilibrium equation the exponents for the concentrations of A and B would be the same as the integers in the stoichiometry of the chemical equation.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Equilibrium ($\ce{ <=> }$), not resonance ($\ce{<->}$). $\endgroup$ – SendersReagent Mar 9 '16 at 23:14
  • $\begingroup$ ok, fixed formatting to use double arrow... $\endgroup$ – MaxW Mar 10 '16 at 0:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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