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The half-life period of catalytic decomposition of $A{B_3}$ at $50$ mm is found to be $4$ hour and at $100$ mm it is $2.0$ hour. What is the order of the reaction?
A.$3$
B.$1$
C.$2$
D.$0$

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Last updated date: 13th Jun 2024
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Answer
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Hint:Half-life and initial pressure of a reaction have an inverse relationship. Hence we can find out the order of reaction by substitution on the corresponding equation while taking all other terms as constant.

Complete step by step answer:Given in the first case,
Initial pressure, ${\left( {{P_0}} \right)_1} = 50$ mm
Half-life, \[{\left( {{t_{0.5}}} \right)_1} = 4\] hours
In the second case,
Initial pressure, ${\left( {{P_0}} \right)_2} = 100$ mm
Half-life, \[{\left( {{t_{0.5}}} \right)_2} = 2\] hours
For a reaction having order n, the relation between half-life and initial pressure is given by,
${t_{0.5}} = \dfrac{{{2^{n - 1}} - 1}}{{(n - 1)k{P_0}^{n - 1}}}$
Where k is rate constant, n is order of reaction and ${P_0}$is the initial pressure.
For a particular reaction, rate constant and rate of reaction are constant. Hence we can write,
${t_{0.5}} \propto \dfrac{1}{{{P_0}^{n - 1}}}$.
i.e. half-life and initial concentration are inversely proportional to each other.
Hence in this condition, we can write,
$\dfrac{{{{\left( {{t_{0.5}}} \right)}_1}}}{{{{\left( {{t_{0.5}}} \right)}_2}}} = \dfrac{{{{\left( {{P_0}^{n - 1}} \right)}_2}}}{{{{\left( {{P_0}^{n - 1}} \right)}_1}}} = {\left( {\dfrac{{{{\left( {{P_0}} \right)}_2}}}{{{{\left( {{P_0}} \right)}_1}}}} \right)^{n - 1}}$
The only unknown in this equation is order of reaction, n. Hence by substituting the given values we can find out the order of reaction.
Let us substitute the values.
\[\dfrac{4}{2} = {\left( {\dfrac{{100}}{{50}}} \right)^{n - 1}}\]
\[2 = {2^{n - 1}}\]
Which means that n $ = 2$.
Hence the order of the reaction is $2$.
Therefore, the correct option is C.

Additional information:
Order of a reaction is the sum of powers of concentration terms of reactants in the rate equation. Since the order for catalytic decomposition of $A{B_3}$ is calculated as two, rate law for the corresponding reaction will be,
$r = k{\left[ {A{B_3}} \right]^2}$
Where r is the rate of catalytic decomposition of $A{B_3}$.

Note:
In this problem, we are taking the ratio of values. The units get cancelled each other. Hence we do not need to concentrate on the units of values. Otherwise we should be aware of the units and should convert it into required values if necessary.