Consider the following data for the given reaction, The order of the reaction is

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Solution

Q.
Consider the following data for the given reaction

$2 {HI}_{(g)} \to {H_{2}}_{(g)} + {I_{2}}_{(g)}$

1 2 3

$HI (mol L^{- 1})$ 0.005 0.01 0.02

$Rate (mol L^{- 1} s^{- 1})$ $7.5 \times 10^{- 4}$ $3.0 \times 10^{- 3}$ $1.2 \times 10^{- 2}$

The order of the reaction is ______________ . [2024]

Ans.
(2)

$Let rate law expression is r = k {[HI]}^{n}$

   $r_{1} = 7.5 \times 10^{- 4} mol L^{- 1} s^{- 1} = k \times {(0.005)}^{n}$

   $r_{2} = 3 \times 10^{- 3} mol L^{- 1} s^{- 1} = k \times {(0.01)}^{n}$

   $r_{3} = 1.2 \times 10^{- 2} mol L^{- 1} s^{- 1} = k \times {(0.02)}^{n}$

Divide $r_{2}$ by $r_{1}$

$\frac{3 \times 10^{- 3}}{7.5 \times 10^{- 4}} = \frac{{(0.01)}^{n}}{{(0.005)}^{n}}$

   $\frac{30 \times 10^{- 4}}{7.5 \times 10^{- 4}} = 2^{n} \Rightarrow 4 = 2^{n} \Rightarrow n = 2$

Divide $r_{3}$ by $r_{2}$

$\frac{1.2 \times 10^{- 2}}{3 \times 10^{- 3}} = \frac{{(0.02)}^{n}}{{(0.01)}^{n}}$

$\frac{12 \times 10^{- 3}}{3 \times 10^{- 3}} = \frac{{(0.02)}^{n}}{{(0.01)}^{n}} \Rightarrow 4 = 2^{n} \Rightarrow n = 2$

  Thus rate law expression is: $r = k {[HI]}^{2}$

It is a second order reaction

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