Question

The rate of the reaction between A and B increases by a factor of 100 when the concentration of A is increased 10 folds. The order of reaction with respect to A is:
A ) 10
B ) 2
C ) 1
D ) 20

Answer 1

Rate of reaction is defined as the speed of reaction in which reactants are converted into products.
For a nth order reaction, the rate law expression is given as ${\text{R = k}}{\left[ {\text{A}} \right]^n}$. Here, R is the rate of the reaction and n is the order of the reaction. K is the rate constant of the reaction.

Complete step by step answer:
Consider the reaction${\text{A}} \to {\text{B}}$. Here, A is the reactant and B is the product.
Assume that the reaction is of nth order. Write the rate law expression
${\text{R = k}}{\left[ {\text{A}} \right]^n}$
Suppose that the reaction is carried out at two different concentrations ${\left[ {\text{A}} \right]_1}{\text{ and }}{\left[ {\text{A}} \right]_2}$ and has rates ${{\text{R}}_1}{\text{ and }}{{\text{R}}_2}$ at these concentrations respectively. Write the rate law expression for these two concentrations:

{{\text{R}}_2}{\text{ = k}}\left[ {\text{A}} \right]_2^n{\text{ }}...{\text{ }}...\left( 2 \right) \\\ $$ Divide equation (2) with equation (1) $$\dfrac{{{{\text{R}}_2}}}{{{{\text{R}}_1}}} = \dfrac{{{\text{k}}\left[ {\text{A}} \right]_2^n}}{{{\text{k}}\left[ {\text{A}} \right]_1^n}} \\\ \dfrac{{{{\text{R}}_2}}}{{{{\text{R}}_1}}} = \dfrac{{\left[ {\text{A}} \right]_2^n}}{{\left[ {\text{A}} \right]_1^n}} \\\ \dfrac{{{{\text{R}}_2}}}{{{{\text{R}}_1}}} = {\left( {\dfrac{{{{\left[ {\text{A}} \right]}_2}}}{{{{\left[ {\text{A}} \right]}_1}}}} \right)^n}......\left( 3 \right) \\\\$$ The rate of the reaction increases by a factor of 100 $$\dfrac{{{{\text{R}}_2}}}{{{{\text{R}}_1}}} = 100......\left( 4 \right)$$ The concentration of A has increased 10 folds. $$\dfrac{{{{\left[ {\text{A}} \right]}_2}}}{{{{\left[ {\text{A}} \right]}_1}}} = 10......\left( 5 \right)$$ Substitute equations (4) and (5) in equation (3) $$\dfrac{{{{\text{R}}_2}}}{{{{\text{R}}_1}}} = {\left( {\dfrac{{{{\left[ {\text{A}} \right]}_2}}}{{{{\left[ {\text{A}} \right]}_1}}}} \right)^n} \\\ 100 = {\left( {10} \right)^n}......\left( 6 \right) \\\\$$ But $$100 = {\left( {10} \right)^2}......\left( 7 \right)$$ Substitute equation (7) in equation (6) $${\left( {10} \right)^2} = {\left( {10} \right)^n} \\\ n = 2 \\\\$$ Hence, the reaction is of second order. _**Hence, the correct option will be the option B ).** _ **Note:** For second order reaction, when the reactant concentration is increased to a factor of x, then the rate of the reaction is increased to a factor of $${x^2}$$. Thus, if the reactant concentration is doubled, the rate of the reaction becomes four times. If the reactant concentration is trebled, the rate of the reaction becomes nine times.