Chemical Kinetics - Result Question 8
8. For the reaction, $2 A+B \rightarrow C$, the values of initial rate at different reactant concentrations are given in the table below. The rate law for the reaction is
(2019 Main, 8 April I)
| $[\boldsymbol{A}]\left(mol L^{-1}\right)$ | $[\boldsymbol{B}]\left(mol L^{-1}\right)$ | Initial rate $\left(mol L^{-1} s^{-1}\right)$ |
|---|---|---|
| 0.05 | 0.05 | 0.045 |
| 0.10 | 0.05 | 0.090 |
| 0.20 | 0.10 | 0.72 |
(a) rate $=k[A][B]^{2}$
(b) rate $=k[A]^{2}[B]^{2}$
(c) rate $=k[A][B]$
(d) rate $=k[A]^{2}[B]$
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Answer:
Correct Answer: 8. (a)
Solution:
- Let the rate equation be $k[A]^{x}[B]^{y}$
From Ist values,
$ 0.045=k[0.05]^x[0.05]^y ……(i) $
From 2nd values,
$ 0.090=k[0.10]^x[0.05]^y……(ii) $
From 3rd values,
$ 0.72=k[0.20]^x[0.10]^y……(iii) $
On dividing equations (i) by (ii), we get
$ \begin{aligned} \frac{0.045}{0.09} & =\left[\frac{0.05}{0.10}\right]^x \\ {\left[\frac{0.05}{0.10}\right]^1 } & =\left[\frac{0.05}{0.10}\right]^x \\ \therefore x & =1 \end{aligned} $
Similarly on dividing Eq. (ii) by (iii) we get
$ \begin{aligned} & \frac{0.09}{0.72}=\left[\frac{0.1}{0.2}\right]^x\left[\frac{0.05}{0.10}\right]^y \\ & \frac{0.01}{0.08}=\frac{0.1}{0.2}\left[\frac{0.05}{0.1}\right]^y \\ & 0.25=\left[\frac{0.05}{0.10}\right]^y \\ & 0.25=[0.5]^\gamma \\ & {[0.5]^2=[0.5]^y} \\ & \therefore y =2 \\ & \end{aligned} $
Hence, the rate law for the reaction
$ \text { Rate }=k[A][B]^2 $
BETA
