Question

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

Experiment	$[A{]}_0=M$	$[B{]}_0=M$	Initial rate
1	$0.012$	$0.035$	$0.10$
2	$0.024$	$0.070$	$0.80$
3	$0.024$	$0.035$	$0.10$
4	$0.012$	$0.070$	$0.80$

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

Answer 1

Answer: (A)

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