# Relationship between molecularity and the reaction order

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?

• 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. – Poutnik Jan 24 at 9:38
• 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. – Maurice Jan 24 at 9:59
• @Maurice Are you sure ? As synthesis of HBr has more complicated mechanism with non integer reaction order. – Poutnik Jan 24 at 10:23
• @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 $\ce{210 cm^{-1}}$ – Maurice Jan 24 at 10:47