
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.$0$
Answer
552.6k+ views
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.
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.
Recently Updated Pages
Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 Physics: Engaging Questions & Answers for Success

Master Class 12 English: Engaging Questions & Answers for Success

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Business Studies: Engaging Questions & Answers for Success

Trending doubts
Which are the Top 10 Largest Countries of the World?

What are the major means of transport Explain each class 12 social science CBSE

Draw a labelled sketch of the human eye class 12 physics CBSE

Why cannot DNA pass through cell membranes class 12 biology CBSE

Differentiate between insitu conservation and exsitu class 12 biology CBSE

Draw a neat and well labeled diagram of TS of ovary class 12 biology CBSE

