Methods of Determining Reaction Order - Practice Questions & MCQ

It is used when the rate law involves only one concentration term.

$\mathrm{t}_{1 / 2} \propto(\mathrm{a})^{1-\mathrm{n}}$ 
or $\mathrm{t}_{1 / 2} \propto 1 / \mathrm{a}^{\mathrm{n}-1}$

For two different concentrations, we have:

$\frac{\left(\mathrm{t}^{1 / 2}\right)_1}{\left(\mathrm{t}^{1 / 2}\right)_2}=\left(\frac{\mathrm{a}_2}{\mathrm{a}_1}\right)^{\mathrm{n}-1}$

On taking logarithms on both sides, we get:

$\log _{10} \frac{\left(\mathrm{t}_{1 / 2}\right)_1}{\left(\mathrm{t}_{1 / 2}\right)_2}=(\mathrm{n}-1) \log _{10}\left(\mathrm{a}_2 / \mathrm{a}_1\right)$

Hence,

$\mathrm{n}=1+\frac{\log \left(\mathrm{t}^{1 / 2}\right)_1-\log \left(\mathrm{t}^{1 / 2}\right)_2}{\log \mathrm{a}_2-\log \mathrm{a}_1}$
Here,  n is the order of the reaction.

Here graphs are plotted between rate and concentration to find the order of the reaction.

$\left[\right.$ Rate $\left.=\mathrm{k}(\text { concentration })^{\mathrm{n}}\right]$

Plots of Rate vs Concentration

If the data for time(t) and [A] is given then this method is applicable. Thus follows the steps given below to find the order of reaction by using the integrated rate law method.

• Check for First Order: 
  1. Use the formula given below to find out the two values of k as k₁ and k₂.$\mathrm{k}=\frac{2.303}{\mathrm{t}} \log _{10}\left[\frac{\mathrm{~A}_{\mathrm{o}}}{\mathrm{A}}\right]$
  2. If these two values k₁ and k₂ are same, then this given reaction is of first-order. But if k₁≠ k₂, then check for zero-order.
• Check for Zero-Order:
  1. Use the formula given below to find out the two values of k as k₁ and k₂.$\mathrm{k}=\frac{\mathrm{A}_{\mathrm{o}}-\mathrm{A}}{\mathrm{t}}$
  2. Again, if these two values k₁ and k₂ are same, then this given reaction is of zero-order. But if k₁≠ k₂, then check for second-order.
• Check for Third-Order:
  1. Use the formula given below to find out the two values of k as k₁ and k₂.$\mathrm{k}=\frac{1}{\mathrm{t}}\left[\frac{1}{\mathrm{~A}}-\frac{1}{\mathrm{~A}_{\mathrm{o}}}\right]$
  2. Further, if these two values k₁ and k₂ are same, then this given reaction is of second-order. But if k₁≠ k₂, then check for third-order and so on.