Question

Find the rate law that corresponds to the data shown for the following reaction?

Experiment	$\left[A_0\right]=M$	$\left[B_0\right]=M$	Initial rate
1	0.012	0.035	0.10
2	0.024	0.070	0.80
3	0.024	0.035	0.10
4	0.012	0.070	0.080

• A. $\text{Rate}=\left[A{]}^2\right[B{]}^2$
• B. $\text{Rate}=k[B{]}^4$
• C. $\text{Rate}=k\left[A\right][B{]}^3$
• D. $\text{Rate}=k\left[A{]}^0\right[B{]}^3$

Answer 1

The correct option is D $\text{Rate}=k\left[A{]}^0\right[B{]}^3$

$\frac{Expt.4}{Expt1}=\frac{\left[0.012{]}^x\right[0.070{]}^y}{\left[0.012{]}^x\right[0.035{]}^y}=\frac{0.80}{0.10}$
$\Rightarrow 2^y=8\therefore y=3$
$\frac{Expt.3}{Expt.1}=\frac{\left[0.024{]}^x\right[0.035]}{\left[0.012{]}^x\right[0.035{]}^y}=\frac{0.10}{0.10}$
$\Rightarrow 2^x=1;\therefore x=0$
$\therefore Rate=k\left[A{]}^0\right[B{]}^3$