Question

In a certain gaseous reaction between $A$ and $B,$$A + 3B\xrightarrow{{}}A{B_3}$. The initial rates are reported as follows: The rate law is

[A]	[B]	Rate
$0.1\,M$	$0.1\,M$	$0.002\,M{s^{ - 1}}$
$0.2\,M$	$0.1\,M$	$0.002\,M\,{s^{ - 1}}$
$0.3\,M$	$0.2\,M$	$0.008\,M\,{s^{ - 1}}$
$0.4\,M$	$0.3\,M$	$0.018\,M\,{s^{ - 1}}$

(A)$R = K[A]{[B]^3}$
(B) $R = K{[A]^0}{[B]^2}$
(C) $R = K[A][B]$
(D) $R = K{[A]^0}{[B]^3}$

Answer 1

Hint: Rate law states that rate of reaction depends upon the concentration in which the rate of reaction actually depends, as observed experimentally. The rate law can be determined by the Ostwald isolation method.

Complete step by step solution:
In this method, concentrations of all the reactants are taken in large excess except that of one. The concentration of only one reactant will influence the rate as others are so much in excess that practically there is no change in their concentrations, i.e., the reactants taken in large excess will be considered constant.
 In the given question we have to calculate the rate law. $A + 3B\xrightarrow{{}}A{B_3}$(reaction given)
Let us the rate equation be
$Rate = K{\left[ A \right]^\alpha }{\left[ B \right]^\beta }$
From experiment (1), $0.002 = K \times {(0.1)^\alpha } \times {\left( {0.1} \right)^\beta }$ ……..(i)
From experiment (2), $0.002 = K \times {\left( {0.2} \right)^\alpha } \times {\left( {0.1} \right)^\beta }$ ………(ii)
From experiment (3), $0.008 = K \times {\left( {0.3} \right)^\alpha } \times {\left( {0.2} \right)^\beta }$ ………(iii)
From experiment (4), $0.0018 = K \times {\left( {0.4} \right)^\alpha } \times {\left( {0.3} \right)^\beta }$ ………(iv)
Dividing equation (ii) by equation (i)
$\dfrac{{0.002}}{{0.002}} = \dfrac{{K \times {{\left( {0.2} \right)}^\alpha } \times {{\left( {0.1} \right)}^\beta }}}{{K \times {{(0.1)}^\alpha } \times {{\left( {0.1} \right)}^\beta }}}$
$1 = {\left( 2 \right)^\alpha }$
$\alpha = 0$
From (ii) and (i) equations we are seeing that there is no effect on the rate when the concentration of the reactant changes. So the rate w.r.t $A$ is $rate = K{\left[ A \right]^0}$
(iii) Equation divided by (i)
$\dfrac{{0.008}}{{0.002}} = \dfrac{{K \times {{\left( {0.3} \right)}^\alpha } \times {{\left( {0.2} \right)}^\beta }}}{{K \times {{\left( {0.1} \right)}^\alpha } \times {{\left( {0.1} \right)}^\beta }}}$
$4 = {\left( 3 \right)^0} \times {\left( 2 \right)^\beta }$ because $\alpha = 0$
Hence, ${\left( 2 \right)^2} = {\left( 2 \right)^\beta }$
Hence, $\beta = 2$
Now the rate of reaction with respect to $B$
\[Rate = K {\left[ B \right]^2}\]
Now net rate of reaction$ = K{\left[ A \right]^0} {\left[ B \right]^2}$

Therefore the correct option is B. $R = K{[A]^0}{[B]^2}$

Note:
The rate law or rate equation for a chemical reaction is an equation that links the initial or forward reaction rate with the concentrations or pressures of the reactants and constant parameters.