Consider the following data for the given reaction 2 HI _ ( g ) → H...

The initial concentrations of $HI$ as $0 . 005, 0 . 01, \text{and}0 . 02$ mol/L, respectively, and the corresponding initial rates as given:

$\begin{cases}HI\left({molL}^{- 1}\right) & 0 . 005 & 0 . 01 & 0 . 02\end{cases}$

Rate \left({molL}^{- 1}{s}^{- 1}\right)7 . 5 \times {10}^{- 4}3 . 0 \times {10}^{- 3}1 . 2 \times {10}^{- 2}$</p><p> For the reaction, rate law is<br/>$rate = k{\left[HI\right]}^{P}$<br/>$P is the order of the reaction.
From the given data
$7 . 5 \times {10}^{- 4} = k{\left[0 . 05\right]}^{P}$-------(1)
$3 \times {10}^{- 3} = k{\left[0 . 01\right]}^{P}$  --------(2)
$1 . 2 \times {10}^{- 2} = k{\left[0 . 02\right]}^{P}$ ---------(3)
On dividing Equation (3) by equation (2)
$\dfrac{12 \times {10}^{- 3}}{3 \times {10}^{- 3}} = \dfrac{k{\left[0 . 02\right]}^{P}}{k{\left[0 . 01\right]}^{P}}$
$⇒ {2}^{P} = {2}^{2}$
$⇒ P = 2$