Question

Which of the following rate law corresponds to the data shown for the reaction?

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

Answer 1

Answer: (A)

$\frac{\text{ Expt. } 4}{\text{ Expt. } 1}=\frac{\left[\right. 0 . 012 \left]\right.^{x} \left[\right. 0 . 070 \left]\right.^{y}}{\left[\right. 0 . 012 \left]\right.^{x} \left[\right. 0 . 035 \left]\right.^{y}}=\frac{0 . 80}{0 . 10}$
$2^{y}=8\Rightarrow y=3$
$\frac{\text{ Expt. } 3}{\text{ Expt. } 1}=\frac{\left[\right. 0 . 024 \left]\right.^{x} \left[\right. 0 . 035 \left]\right.^{3}}{\left[\right. 0 . 012 \left]\right.^{x} \left[\right. 0 . 035 \left]\right.^{3}}=\frac{0 . 10}{0 . 10}$
$2^{x}=1\Rightarrow x=0$
Rate $=k\left[\right.A\left]\right.^{0}\left[\right.B\left]\right.^{3}$ .