Question

The table below shows recorded concentration data for the following chemical reaction:
$Rate=k\left[Fe\right(N{O}_3{)}_3{]}^x[NaOH{]}^y$
Use the data to determine the exponents $x$ and $y$ in the rate law:
$Rate=k\left[Fe\right(N{O}_3{)}_3{]}^x[NaOH{]}^y$

$\left[Fe\right(N{O}_3{)}_3]M$	$\left[NaOH\right]M$	Rate M/s
$0.100$	$0.250$	$1.3\times {10}^{-6}$
$0.100$	$0.500$	$5.2\times {10}^{-6}$
$0.200$	$0.250$	$2.6\times {10}^{-6}$
$0.300$	$0.500$	$3.9\times {10}^{-6}$

• A. $x=1,y=1$
• B. $x=1,y=2$
• C. $x=2,y=1$
• D. $x=2,y=2$

Answer 1

The correct option is A $x=1,y=1$

Rate$=R=k{\left[{Fe\left({No}_3\right)}_3\right]}^x{\left[NaOH\right]}^y$

Therefore,$1.3\times {10}^{-6}=k{\left[0.100\right]}^x{\left[0.25\right]}^y⟶\left(i\right)5.2\times {10}^{-6}=k{\left[0.1\right]}^x{\left[0.5\right]}^y⟶\left(ii\right)2.6\times {10}^{-6}=k{\left[0.2\right]}^x{\left[0.25\right]}^y⟶\left(iii\right)3.9\times {10}^{-6}=k{\left[0.3\right]}^x{\left[0.5\right]}^y⟶\left(iv\right)$

Dividing (i) by (ii)

$\frac{1.3\times {10}^{-6}}{5.2\times {10}^{-6}}={\left(\frac{0.25}{0.5}\right)}^y\therefore \frac{1}{4}={\left(\frac{1}{2}\right)}^y\therefore y=2$

Dividing (i) by (iii), $\frac{1}{2}={\left(\frac{0.1}{0.2}\right)}^x$

$\therefore x=1⟹$ Correct option is $\left(B\right)$.