# Rate law of the reaction of propane with oxygen

In class, I have a demonstration where I light a bubble filled with propane. As I add oxygen, the bubbles combust faster and more quickly. I have already balanced out the chemical reaction like so:

$$\ce{C3H8 + 5 O2 → 3 CO2 + 4 H2O + heat + light}$$

I have reasoned that the reaction must go faster because the limiting reagent is oxygen, and since the air is only about 20% oxygen adding more oxygen to the bubble will speed the reaction process. I have figured out the kinetics, the oxidation, the enthalpy diagram, etc.

However, the only thing I don't know how to do in this demonstration is how to find the rate law. How would I find the rate law and equation of a combustion reaction? Wouldn't it just be second order since it's a combustion reaction? I have no idea where to even begin or why it is so.

• The reaction has a multistep mechanism and the kinetics will not be trivial. – Poutnik May 28 '19 at 5:17

## 2 Answers

The oxidation of hydrocarbons ($$\ce{RH}$$) such as this one proceed via a chain mechanism. The initiation step is

$$\ce{RH + O2 -> HO2^. + R^.}$$

and the propagation

$$\ce{ R^.+O2 -> RO2^. \\ RO2^. +RH -> ROOH +R^. }$$

But it does not stop there as the hydroperoxide decomposes $$\ce{ ROOH -> RO^. + OH^.}$$ and there are other reactions such as

\begin{align} \ce{ R^. + O2 -> RO^. + O^. \\ RH + OH^. -> H2O + R^. \\ RCH2OO^. + O2 -> RCH2O + O3 \\ RH + O3 -> RO^. +HO2^.}\end{align}

and chain branching $$\ce{ \quad O^. + RH -> OH^. + R^.}$$.

The exact mechanism depends on temperature and pressure, and in a plot of $$T$$ vs $$p$$ there are regions of slow reactions, explosion and hot and cold flames. (An original paper is by Newitt & Thorne, J. Chem. Soc 1937, p1656). Thus you can appreciate that sorting out the mechanism is no simple task and will need much specialised equipment.

In order to determine the rate law, you need to be able to measure either the rate at which the reactants are consumed or the products are formed.

A very common error is to base the rate law on the balanced chemical equation. In fact, many high school level teachers will teach kinetics this way. Unfortunately this is a serious error.

It is possible for the exponents in the rate law to match the coefficients of the balanced chemical equation, but only when the reaction occurs in a single step. We can't know how many steps are in a reaction until we investigate the kinetics.

In terms of determining the rate law, you would need to do the following:

1. Vary the amount of propane systematically
2. Vary the amount of oxygen systematically
3. Measure the amount of time it takes for the reaction to finish with each variation

Since there are only two reactants, you should be able to do this by running three separate trials; one which we will consider as a standard (or reference), one where the amount of propane is increased, and one where the amount of oxygen is increased.

On a practical level, your instructor may be working off the assumption that the reaction is first order in propane ($$\ce{rate = k[propane]^1[oxygen]^x}$$). If this is the case, then you only need to vary the amount of oxygen. This may be simply be for practical reasons.