Question

For the elementary reaction $M \to N$, the rate of disappearance of M increases by a factor of 8 upon doubling the concentration of M. The order of the reaction with respect to M is:
A.4
B.3
C.2
D.1

Answer 1

Order of reaction is the sum of power of concentration or pressure terms raised in the rate law expression. It can be 0,1,2,3 etc. It can be fractional or negative. It is determined experimentally. Rate of reaction means the speed of reaction with which it takes place.

Complete step by step answer:
The elementary reaction M → N for this reaction rate can be written as :-
$Rate{\left[ M \right]^n}$
$Rate{\text{ }} = K{\left[ M \right]^1} - {\text{ }} - {\text{ }} - {\text{ Eqn}}1$
As given in question when concentration of [M] becomes double, the rate of reaction becomes 8 times, then the reaction rate becomes:- $8{\text{ }} \times {\text{ }}Rate{\text{ }} = k{\left[ {2M} \right]^n} - {\text{ }} - {\text{ }} - {\text{ }} - Eqn2$
Now if we divide equation 1 by equation 2
$\dfrac{{Rate}}{{8Rate}} = \dfrac{{K\mathop {[M]}\nolimits^1 }}{{K\mathop {[2M]}\nolimits^n }}$
Now rate by rate get cancelled and K by k cancel from Numerator and Denominator
After simplify the equation we get
$\dfrac{1}{8} = \dfrac{1}{{\mathop 2\nolimits^n }}$
$\mathop 2\nolimits^n = 8$
$\Rightarrow n = 3$
Our required answer is B that is “3”.

Note:
The effect of concentration on the rate of reaction was studied by using the "Law of Mass Action'. According to the Law of Mass Action rate of a chemical reaction is directly proportional to the product of concentration time reacting species with each concentration term raised to the power equal to the stoichiometric coefficient.