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 $Ms^{-1})$
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), $\frac{3.2 \times10^{-4}}{8.0 \times10^{-5}} = \left(2\right)^{x}$
$\left(4\right) =\left(2\right)^{x} \Rightarrow x=2$
From (2) and (3), $\frac{1.28 \times10^{-3}}{3.2 \times10^{-4}} =\left(4\right)^{y}$
$\left(4\right) =\left(4\right)^{y} \Rightarrow y = 1$
From (1) and (4), $\frac{8 \times10^{-5}}{4\times10^{-5}} = \left(2\right)^{x} \left(2\right)^{y}\left(\frac{1}{4}\right)^{x} =\left(4\right)\left(2\right)\left(\frac{1}{4}\right)^{z}$
$\Rightarrow z=1$
Thus, order w.r.t. $P = 2$
w. r. t. $Q = 1$
w. r. t. $R = 1$