Question

Determine the order of reaction on the basis of following data for the reaction $A + B \to C$

Experiment	[A]	[B]	Rate of reaction
1	$0.1$	$0.1$	$2 \times {10^{ - 3}}mol{L^{ - 1}}{\sec ^{ - 1}}$
2	$0.4$	$0.1$	$0.4 \times {10^{ - 2}}mol{L^{ - 1}}{\sec ^{ - 1}}$
3	$0.1$	$0.2$	$1.4 \times {10^{ - 2}}mol{L^{ - 1}}{\sec ^{ - 1}}$

A) $\dfrac{1}{2}$
B) $2.83$
C) $\dfrac{{10}}{3}$
D) $3.33$

Answer 1

Hint: As we all know that order of a reaction is the sum of the power of concentration and pressure terms raised in the rate law expression. It may be $0,1,2,3..$and even fractional or negative and rate of reaction is the rate of change of concentration of any one of reactants or product per unit time.

Complete step by step solution: We are very well aware with the order of a reaction which is the sum of the power of concentration and pressure terms raised in the rate law expression. It may be $0,1,2,3..$and even fractional or negative and we also know that rate of reaction is the rate of change of concentration of any one of reactants or products per unit time.
We know that, $Rate = K{[A]^a}{[B]^b}$ where, $K$ is the rate constant and \[\left[ A \right]\] and \[\left[ B \right]\] are the concentrations of the reactants and $a$and $b$are the power raised to concentrations which we call here the order of reaction.
Comparing the first and second experiment and after putting the values we get:
$\dfrac{{Rat{e_1} = K{{[A]}^a}{{[B]}^b}}}{{Rat{e_2} = K{{[A]}^a}{{[B]}^b}}}$
$
  \Rightarrow \dfrac{{2 \times {{10}^{ - 3}} = K{{[0.1]}^a}{{[0.1]}^b}}}{{0.4 \times {{10}^{ - 2}} = K{{[0.4]}^a}{{[0.1]}^b}}} \\
\Rightarrow \dfrac{1}{2} = \dfrac{{{{[1]}^a}}}{{{{[4]}^a}}} \\
 $
$\Rightarrow {(\dfrac{1}{4})^{\dfrac{1}{2}}} = {(\dfrac{1}{4})^a}$
Therefore, $a = \dfrac{1}{2}$.
Similarly we can compare the first and third experiments and we will get:
$
  \dfrac{{Rat{e_1} = K{{[A]}^a}{{[B]}^b}}}{{Rat{e_2} = K{{[A]}^a}{{[B]}^b}}} \\
  \Rightarrow \dfrac{{2 \times {{10}^{ - 3}} = K{{[0.1]}^a}{{[0.1]}^b}}}{{1.4 \times {{10}^{ - 2}} = K{{[0.1]}^a}{{[0.2]}^b}}} \\
\Rightarrow \dfrac{1}{7} = \dfrac{{{{[1]}^b}}}{{{{[2]}^b}}} \\
\Rightarrow {(\dfrac{1}{7})^{2.80}} = {(\dfrac{1}{2})^b} \\
 \Rightarrow b = 2.80 \\
 $
Now, as we know that order of reaction is the sum of powers of concentrations raised. So we can calculate the order as the sum of $a$ and $b$ thus we get the order of reaction as:
Order of reaction = a + b
$\Rightarrow$ Order of reaction = 0.5 + 2.80
$\Rightarrow$ Order of Reaction = 3.3

Therefore the correct answer is (D).

Note: Always remember that it is not necessary that the concentration of every reactant changes with passage of time, therefore only that reactant whose concentration has changed with passage of time will be involved in the rate expression. Hence, if any reactant is present in excess then order of reaction with respect to that reactant is zero order.