Question

For the reaction, $P+Q+R\to S$
Experimental data for the measured initial rates in given below.

Expt.	(P)	(Q)	(R)	Intial rate (In $M{s}^{-1})$
1	0.2 M	0.5 M	0.4 M	$8.0\times 10^{-5}$
2.	0.4 M	0.5 M	0.4 M	$3.2\times 10^{-4}$
3.	0.4 M	2.0 M	0.4 M	$1.28\times 10^{-3}$
4.	0.1 M	0.2 M	1.6 M	$4.0\times 10^{-8}$

The order of the reaction with respect of $P,Q$ and $R$ respectively is

• A. $4,1,3$
• B. $2,2,1$
• C. $1,2,1$
• D. $2,1,1$

Answer 1

Answer: (D)

By rate law, Rate$=k{\left[P\right]}^x{\left[Q\right]}^y{\left[R\right]}^z$
$x=$ order w.r.t. $P$
$y=$ order w.r.t. $Q$
$z=$order w.r.t. $R$
(1) $8.0\times 10^{-5}=k{\left(0.2\right)}^x{\left(0.5\right)}^y{\left(0.4\right)}^z$
(2) $3.2\times 10^{-4}=k{\left(0.4\right)}^x{\left(0.5\right)}^y{\left(0.4\right)}^z$
(3) $1.28\times 10^{-3}=k{\left(0.4\right)}^x{\left(2.0\right)}^y{\left(0.4\right)}^z$
(4) $4.0\times 10^{-5}=k{\left(0.1\right)}^x{\left(0.25\right)}^y{\left(0.6\right)}^z$
From (1) and (2), $\frac{3.2\times 10^{-4}}{8.0\times 10^{-5}}={\left(2\right)}^x$
$\left(4\right)={\left(2\right)}^x\Rightarrow x=2$
From (2) and (3), $\frac{1.28\times 10^{-3}}{3.2\times 10^{-4}}={\left(4\right)}^y$
$\left(4\right)={\left(4\right)}^y\Rightarrow y=1$
From (1) and (4), $\frac{8\times 10^{-5}}{4\times 10^{-5}}={\left(2\right)}^x{\left(2\right)}^y{\left(\frac{1}{4}\right)}^x=\left(4\right)\left(2\right){\left(\frac{1}{4}\right)}^z$
$\Rightarrow z=1$
Thus, order w.r.t. $P=2$
w. r. t. $Q=1$
w. r. t. $R=1$