Ncert Exercises & Solutions

4.10 In a reaction between A and B, the initial rate of reaction (r0) was measured for different initial concentrations of A and B as given below:

The order of the reaction with respect to A and B would be:
1. The order w.r.t A is 0.5 and w.r.t B is zero
2. The order w.r.t A is 1 and w.r.t B is 0.5
3. The order w.r.t A is 2 and w.r.t B is 1
4. The order w.r.t A is 1.5 and w.r.t B is zero

NEETprep Answer:

Let the order of the reaction with respect to A be x and with respect to B be y.
Therefore,

$r_{0} = K {[A]}^{x} {[B]}^{y}$
$5.07 \times 10^{- 5} = K {[0.20]}^{x} {[0.30]}^{x} (i)$
$5.07 \times 10^{- 5} = K {[0.20]}^{x} {[0.10]}^{x} (i i)$
$1.43 \times 10^{- 4} = K {[0.40]}^{x} {[0.05]}^{x} (i i i)$

Dividing equation (i) by (ii), we obtain

$\frac{5.07 \times 10^{- 5}}{5.07 \times 10^{- 5}} = \frac{K {[0.20]}^{x} {[0.30]}^{y}}{K {[0.20]}^{x} {[0.10]}^{y}}$
$\Rightarrow 1 = \frac{{[0.30]}^{y}}{[0.10]}$
$\Rightarrow {(\frac{0.30}{0.10})}^{0} {(\frac{0.30}{0.10})}^{y}$
$\Rightarrow y = 0$

Dividing equation (iii) by (ii), we obtain

$\frac{1.43 \times 10^{- 4}}{5.07 \times 10^{- 5}} = \frac{K {[0.40]}^{N} [0.05]}{K {[0.20]}^{N} {[0.30]}^{y}}$
$\Rightarrow \frac{1.43 \times 10^{- 4}}{5.07 \times 10^{- 5}} = \frac{[0.40] z}{{[0.20]}^{x}} = \frac{[0.40]}{{[0.20]}^{x}} [\begin{matrix} S i n c e y = \\ {[0.05]}^{y} = {[0.30]}^{y} = 1 \end{matrix}]$
$\Rightarrow 2.821 = 2^{N}$
$\Rightarrow \log 2.821 = x \log 2 (T a l k i n g \log b o t h s i d e s)$
$\Rightarrow x = \frac{\log 2.821}{\log 2}$
$= 1.496$
$= 1.5 (a p p r o x i m a t e l y)$
$H e n c e, t h e o r d e r o f t h e r e a c t i o n w i t h r e s p e c t t o A i s 1.5 a n d w i t h r e s p e c t t o B i s z e r o$

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