# How do I find the order for each reactant?

May 24, 2017

It must be done for two trials where one of the reactants' concentrations are held constant. (Otherwise there are too many unknowns.)

A general rate law is:

$r \left(t\right) = k {\left[A\right]}^{m} {\left[B\right]}^{n}$

where $\left[\text{ }\right]$ is the concentration in $\text{mol/L}$ or $\text{M}$, $r \left(t\right)$ is the initial rate of reaction in $\text{M/s}$, and $m$ and $n$ are reaction orders (the weights of each reactant on the overall reaction order).

For any reaction, we need its kinetics data. Bare minimum we need:

• $\left[A\right]$ for three trials
• $\left[B\right]$ for three trials
• $r \left(t\right)$ for three trials

If we consider ${\left[\text{ }\right]}_{i}$ and ${r}_{i} \left(t\right)$ to be the concentrations and initial rates of reaction for trial $i$, then we can write for all three trials:

${r}_{1} \left(t\right) = k {\left[A\right]}_{1}^{m} {\left[B\right]}_{1}^{n}$

${r}_{2} \left(t\right) = k {\left[A\right]}_{2}^{m} {\left[B\right]}_{2}^{n}$

${r}_{3} \left(t\right) = k {\left[A\right]}_{3}^{m} {\left[B\right]}_{3}^{n}$

where the rate constant $k$ is the same for the same reaction at the same temperature.

To find the orders, it tends to be easiest to choose trials in which $\boldsymbol{\setminus \frac{{\left[A\right]}_{i}}{{\left[A\right]}_{j}} = 1}$ or $\boldsymbol{\setminus \frac{{\left[B\right]}_{i}}{{\left[B\right]}_{j}} = 1}$, so that one term cancels out. If the data are sufficiently nice (that is, if the ratios are close to integers), no logarithms are necessary.

In the case that no trial data are sufficiently nice, then for trials $i$ and $j$, where $i \ne j$:

$\textcolor{b l u e}{\frac{{r}_{i} \left(t\right)}{{r}_{j} \left(t\right)}} = \frac{{\left[A\right]}_{i}^{m} {\left[B\right]}_{i}^{n}}{{\left[A\right]}_{j}^{m} {\left[B\right]}_{j}^{m}}$

$= \textcolor{b l u e}{{\left(\frac{{\left[A\right]}_{i}}{{\left[A\right]}_{j}}\right)}^{m} {\left(\frac{{\left[B\right]}_{i}}{{\left[B\right]}_{j}}\right)}^{n}}$

For example, if we had trials such that $\frac{{\left[A\right]}_{i}}{{\left[A\right]}_{j}} = 1$, then ${1}^{m} = 1$ and:

$\frac{{r}_{i} \left(t\right)}{{r}_{j} \left(t\right)} = {\left(\frac{{\left[B\right]}_{i}}{{\left[B\right]}_{j}}\right)}^{n}$

Assuming the data are not sufficiently nice, take the $\ln$ of both sides:

$\ln \left(\frac{{r}_{i} \left(t\right)}{{r}_{j} \left(t\right)}\right) = \ln {\left(\frac{{\left[B\right]}_{i}}{{\left[B\right]}_{j}}\right)}^{n}$

Using the property that $\ln {a}^{b} = b \ln a$, we get

$\implies n \ln \left(\frac{{\left[B\right]}_{i}}{{\left[B\right]}_{j}}\right)$.

Therefore, as long as trials were chosen that held $\boldsymbol{\left[A\right]}$ constant:

$\textcolor{b l u e}{n = \ln \frac{\frac{{r}_{i} \left(t\right)}{{r}_{j} \left(t\right)}}{\ln \left(\frac{{\left[B\right]}_{i}}{{\left[B\right]}_{j}}\right)}}$

Or, as long as trials were chosen that held $\boldsymbol{\left[B\right]}$ constant:

$\textcolor{b l u e}{m = \ln \frac{\frac{{r}_{i} \left(t\right)}{{r}_{j} \left(t\right)}}{\ln \left(\frac{{\left[A\right]}_{i}}{{\left[A\right]}_{j}}\right)}}$