Question

The results given in the below table were obtained during kinetic studies of the following reaction:
$3A+2B\to C+D$

Experiment	$[A{]}_0$ $\left(mol{L}^{-1}\right)$	$[B{]}_0$ $\left(mol{L}^{-1}\right)$	Intial rate $\left(mol{L}^{-1}mi{n}^{-1}\right)$
$I$	$0.2$	$0.2$	$9\times {10}^{-3}$
$II$	$0.2$	$0.4$	$3.6\times {10}^{-2}$
$III$	$0.4$	$0.2$	$1.8\times {10}^{-2}$
$IV$	$X$	$0.4$	$1.08\times {10}^{-1}$
$V$	$0.6$	$Y$	$4.32\times {10}^{-1}$

Choose the correct option:

• A. $X$ is $0.6$
• B. $Y$ is $1.0$
• C. $X$ is $0.4$
• D. $Y$ is $0.3$

Answer 1

The correct option is A $X$ is $0.6$

Let the rate of reaction be :
$R=k\left[A{]}^a\right[B{]}^b$
Solution:
From $I$ and $II$, by initial rate method :
${\left(\frac{0.2}{0.4}\right)}^b=\frac{9\times {10}^{-3}}{3.6\times {10}^{-2}}$

${\left(\frac{1}{2}\right)}^b=\frac{1}{4}$

$b=2$
Rate of reaction w.r.t B is 2

From $I$ and $III$, by initial rate method :
${\left(\frac{0.2}{0.4}\right)}^a=\frac{9\times {10}^{-3}}{1.8\times {10}^{-2}}$
${\left(\frac{1}{2}\right)}^a=\frac{1}{2}$
$a=1$
Rate of reaction w.r.t A is 1
The rate of reaction is is :
$R=k\left[A\right][B{]}^2$
Now ,
From $II$ and $IV$, by initial rate method :
${\left(\frac{0.2}{X}\right)}^1=\frac{3.6\times {10}^{-2}}{1.08\times {10}^{-1}}X=0.6$

From $II$ and $V$, by initial rate method :
${\left(\frac{0.2}{0.6}\right)}^1\times {\left(\frac{0.4}{Y}\right)}^2=\frac{3.6\times {10}^{-2}}{4.32\times {10}^{-1}}\frac{1}{3}\times {\left(\frac{0.4}{Y}\right)}^2=\frac{1}{12}Y=0.8$

Theory:
Initial Rate Method
The method involves finding of initial rate of reaction by taking known concentrations of different reactants.
Involves comparison of different initial rates of a reaction by varying the concentration of one of the reactants while other reactants are kept constant
For $aA+bB+cC\to \text{Products}$
$\text{Rate}=k\left[A{]}^p\right[B{]}^q[C{]}^r$
p, q, and r are the order of reaction with respect to
A, B, and C, respectively.
Concentration of A is changed, keeping concentrations of
B and C same as before.
Two different initial concentrations of $A,[A_0{]}_1,[A_0{]}_2$ are taken .

$\text{Rate}=k\left[A{]}^p\right[B{]}^q[C{]}^r$
The initial rates of the reaction are determined as
$r_1=k^{\prime }[A_0{]}_1^p$

$r_2=k^{\prime }[A_0{]}_2^p$

$k^{\prime }=k\left[B{]}^q\right[C{]}^r$

$\frac{r_1}{r_2}={\left(\frac{[A_0{]}_1}{[A_0{]}_2}\right)}^p$

p can be calculated by measuring $r_1,r_2,[A_0{]}_1,[A_0{]}_2$ values.
Following the same method, q and r can also be calculated