# In a reaction of combustion of H_2, when H_2 concentration is kept constant, the amount of H_2 reacting increases. Explain the order of the reaction with respect to H_2 and O_2?

Jul 20, 2016

Well, what this sounds like to me is that the order of the reaction depends only on ${\text{H}}_{2}$, and that the reaction is zero order with respect to ${\text{O}}_{2}$.

But without knowing whether the reaction is elementary or complex, that's all we can say.

If it were to be elementary, then the order would be second-order with respect to ${\text{H}}_{2}$.

The combustion reaction discussed is:

$2 \text{H"_2 + "O"_2 -> 2"H"_2"O}$

REACTION'S RESPONSE FOR H2

Based on the information given, if we keep the ${\text{H}}_{2}$ concentration constant (which I'm assuming is for the same trial), ${\text{H}}_{2}$ does keep reacting, but a steady concentration of $\setminus m a t h b f \left({\text{H}}_{2}\right)$ is maintained by slowly adding more ${\text{H}}_{2}$ as it is consumed.

That is basically like "increasing" the concentration of the reactants, which pushes the equilibrium forward as with Le Chatelier's principle.

This is reflected in more $\setminus m a t h b f \left({\text{H}}_{2}\right)$ reacting, as the question conditions state, and so, the reaction order (at least partially) depends on ${H}_{2}$.

REACTION'S RESPONSE FOR O2

As for ${\text{O}}_{2}$, if the reaction order did depend on its concentration, we would expect that having more ${\text{H}}_{2}$ around would allow more ${\text{O}}_{2}$ to react.

However, since the reactivity of ${\text{O}}_{2}$ is not stated to increase while this occurs, it means it is unaffected by the (steady) "increase" in reactant concentration.

That says that the reaction rate does not depend on the concentration of ${\text{O}}_{2}$. For that to happen, it has to not be in the rate law, basically. Its concentration decreases linearly with time.

In other words, we have a rate law of:

\mathbf(r(t)) = k_"obs"["H"_2]^m["O"_2]^0 = \mathbf(k_"obs"["H"_2]^m)

CAVEAT: NOT KNOWING WHETHER IT'S AN ELEMENTARY OR COMPLEX REACTION

Now, without knowing the true mechanism by which this takes place, we can't really give a specific order with respect to ${\text{H}}_{2}$, as we haven't been told whether this is an elementary reaction or not.

Basically, we can't say what $m$ is equal to unless:

• We know whether the reaction given is an elementary or complex reaction.
• We know the exact mechanism by which the overall reaction occurs (in which case we can use the steady-state approximation on intermediate(s)).

OR...

• We have multiple trials that give enough data to determine reaction orders.

If the overall reaction were to be elementary (i.e. it literally occurs as-written), then the order would depend on the number of molecules reacting:

$\textcolor{red}{2} \text{H"_2 + "O"_2 -> 2"H"_2"O}$

...in which case the reaction would be second order with respect to ${\text{H}}_{2}$ because for every one ${\text{O}}_{2}$, two ${\text{H}}_{2}$ stoichiometrically balance it.

But since we aren't told whether it is definitively an elementary reaction, all we can say is:

This reaction's order is the order with respect to ${H}_{2}$, and the reaction is zero order with respect to ${O}_{2}$.