Question

The following results have been obtained during the kinetic study of the reaction
$A+B⟶C+D$
$\begin{array}{cccc}\text{Experiment}& \text{[A]}& \text{[B]}& \text{Initial rate of}\\ & & & \text{formation of}\\ & & & \left(Mse{c}^{-1}\right)\\ \text{I}& 0.1& 0.1& 6.0\times {10}^{-3}\\ \text{II}& 0.3& 0.2& 7.2\times {10}^{-2}\\ \text{III}& 0.3& 0.4& 2.88\times {10}^{-1}\\ \text{IV}& 0.4& 0.1& 2.40\times {10}^{-2}\end{array}$
Determine the rate law of the above reaction.

• A. Rate law $=k\left[A{]}^2\right[B]$
• B. Rate law $=k\left[A\right]\left[B\right]$
• C. Rate law $=k\left[A{]}^2\right[B{]}^2$
• D. Rate law $=k\left[A\right][B{]}^2$

Answer 1

The correct option is D Rate law $=k\left[A\right][B{]}^2$

$r_1=6\times {10}^{-3}=k\left[A{]}^m\right[B{]}^n$
$6\times {10}^{-3}=k\left[0.1{]}^m\right[0.1{]}^n$ ....(1)
$r_2=7.2\times {10}^{-2}=k\left[0.3{]}^m\right[0.2{]}^n$ ....(2)
$r_3=2.88\times {10}^{-1}=k\left[0.3{]}^m\right[0.4{]}^n$ .....(3)
$r_4=2.40\times {10}^{-2}=k\left[0.4{]}^m\right[0.1{]}^n$ ....(4)
equation (ii)/(iii) $\Rightarrow \frac{2.88}{0.72}=4=(\frac{0.4}{0.2}{)}^n\Rightarrow (2{)}^n=4\Rightarrow n\Rightarrow 2$
doing (iv)/(i) $\frac{2.40}{0.6}=(4{)}^m\Rightarrow m=1$
rate law = - $\frac{d\left[A\right]}{dt}=k\left[A\right][B{]}^2$