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Should the molecularity be equal or less than the reaction order? Consider the reaction:

$$\ce{H2(g) + I2(g) → 2 HI(g)}$$

If we consider the mechanism of this multi-step reaction, the molecularity is 3 (termolecular reaction):

$$ \begin{align} \ce{I2(g) &<=> 2I(g)} &\quad&\text{(equilibrium reaction)} \\ \ce{2I(g) + H2(g) &-> 2HI(g)} &\quad&\text{(rate-determining step)} \end{align} $$

The rate equation for the reaction is

$$\mathrm{rate} = k[\ce{H2(g)}][\ce{I2(g)}].$$

The order of the reaction is 2 here. The molecularity is greater than the reaction order. Is it possible? Or is the actual mechanism for the above reaction is different?

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    $\begingroup$ For eventual writing and formatting of chemical/mathematical formulas or equations, see MathJax usage guide It is not mandatory, but highly recommended and it gives huge advantage for writers and readers. $\endgroup$
    – Poutnik
    Jan 24, 2021 at 9:38
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    $\begingroup$ It is not $2 \ce{ I}$ that reacts with $\ce{H2}$, but plain $\ce{I2}$. The collision $\ce{H2 + I2}$ produces an intermediate excited complex $\ce{H_2I_2}$ which can get decomposed into $2 \ce{HI}$ or back to $\ce{H_2 + I_2}$ Here the order an the molecularity are both 2. $\endgroup$
    – Maurice
    Jan 24, 2021 at 9:59
  • $\begingroup$ @Maurice Are you sure ? As synthesis of HBr has more complicated mechanism with non integer reaction order. $\endgroup$
    – Poutnik
    Jan 24, 2021 at 10:23
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    $\begingroup$ @Poutnik. I know that the law governing the synthesis of $\ce{HBr}$ is rather complicated, and has no order. I also know that the synthesis of $\ce{HCl}$ is a chain mechanism. So all halogens behave differently when reacting with $\ce{H2}$. But the existence of a transition state$\ce{H2I2}$ has been developed and proved by Eyring, which has shown that the rate constant of $\ce{HI}$ synthesis is $\ce{2.0·10^9 √T e^{- 42500/RT}}$, and that the transition state $\ce{H2I2}^*$ is decomposed with a vibration constant equal to $\ce{210 cm^{-1}}$ $\endgroup$
    – Maurice
    Jan 24, 2021 at 10:47

1 Answer 1

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Molecularity applies to single elementary reaction steps and is seldom higher than 2.

The reaction order is the "effective molecularity" of the whole set of linked reactions, if it had been a single step reaction. It need not to be integer and has just indirect relation to molecularities of elementary steps.

The reaction order is either determined experimentally, either derived from the reaction schema via determined conditions for steady states of intermediate products.

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  • $\begingroup$ Does an elementary reaction $\ce{A->B}$ remains elementary for $\ce{2A->2B}$ ? If no, then the order (which is equal to molecularity) will not change… $\endgroup$
    – Apurvium
    Jun 11, 2023 at 5:33
  • $\begingroup$ @Apurvium If there is mechanism of collision 2 A, forming 2 B, then it is elementary. OTOH, A->B would not be. $\endgroup$
    – Poutnik
    Jun 11, 2023 at 6:10
  • $\begingroup$ Sorry, I did not get it. It is given that A forming B is elementary. On that basis I was asking… $\endgroup$
    – Apurvium
    Jun 11, 2023 at 8:14
  • $\begingroup$ @Apurvium Why then 2A -> 2B? Just a formal multiple for matching a bigger reaction schema, or for state function change evaluation? If so, then obviously the elementary reaction, molecularity and order remain, as the multiple is just formal. $\endgroup$
    – Poutnik
    Jun 11, 2023 at 9:06
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    $\begingroup$ @Apurvium They do while the above do not. $\endgroup$
    – Poutnik
    Jun 11, 2023 at 9:22

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