Consider the following data for the given reaction 2 HI ( g ) → H 2 ( g ) + I 2 ( g ) HI molL - 1 0 . 005 0 …

The initial concentrations of $\mathrm{HI}$ as $0.005, 0.01, \text{and} 0.02$ mol/L, respectively, and the corresponding initial rates as given:

$\begin{array}{llll}\mathrm{HI}\left({\mathrm{molL}}^{-1}\right)& 0.005& 0.01 & 0.02\end{array}$

Rate $\left({\mathrm{molL}}^{-1} {\mathrm{s}}^{-1}\right) 7.5\times {10}^{-4} 3.0\times {10}^{-3} 1.2\times {10}^{-2}$

 For the reaction, rate law is
$\mathrm{rate}=\mathrm{k}{\left[\mathrm{HI}\right]}^{\mathrm{P}}$
$\mathrm{P}$ is the order of the reaction.
From the given data
$7.5\times {10}^{-4}=\mathrm{k}{\left[0.05\right]}^{\mathrm{P}}$-------(1)
$3\times {10}^{-3}=\mathrm{k}{\left[0.01\right]}^{\mathrm{P}}$  --------(2)
$1.2\times {10}^{-2}=\mathrm{k}{\left[0.02\right]}^{\mathrm{P}}$ ---------(3)
On dividing Equation (3) by equation (2)
$\frac{12\times {10}^{-3}}{3\times {10}^{-3}}=\frac{\mathrm{k}{\left[0.02\right]}^{\mathrm{P}}}{\mathrm{k}{\left[0.01\right]}^{\mathrm{P}}}$
$\Rightarrow 2^{\mathrm{P}}=2^2$
$$$\Rightarrow \mathrm{P}=2$