Question

a certain reaction is first order in h₂ and first order in i₂. use this information to complete the table below. round each of your answers to 3 significant digits.

h₂	i₂	initial rate of reaction
0.479 m	2.31 m	m/s
6.08 m	0.547 m	m/s

Explanation:

Step1: Determine the rate - law expression

Since the reaction is first - order in $H_2$ and first - order in $I_2$, the rate - law is $rate = k[H_2][I_2]$. First, find the rate constant $k$ using the first row of data. Given $[H_2]=1.44\ M$, $[I_2]=2.31\ M$ and $rate = 6.00\times 10^{5}\ M/s$. Substitute into the rate - law: $k=\frac{rate}{[H_2][I_2]}$.
$k=\frac{6.00\times 10^{5}\ M/s}{1.44\ M\times2.31\ M}\approx1.79\times 10^{5}\ M^{-1}s^{-1}$.

Step2: Calculate the rate for the second row

For the second row, $[H_2]=0.479\ M$ and $[I_2]=2.31\ M$, and $k = 1.79\times 10^{5}\ M^{-1}s^{-1}$. Using the rate - law $rate=k[H_2][I_2]$, we have $rate=(1.79\times 10^{5}\ M^{-1}s^{-1})\times0.479\ M\times2.31\ M$.
$rate=(1.79\times 10^{5})\times0.479\times2.31\ M/s\approx1.99\times 10^{5}\ M/s$.

Step3: Calculate the rate for the third row

For the third row, $[H_2]=6.08\ M$ and $[I_2]=0.547\ M$, and $k = 1.79\times 10^{5}\ M^{-1}s^{-1}$. Using the rate - law $rate = k[H_2][I_2]$, we have $rate=(1.79\times 10^{5}\ M^{-1}s^{-1})\times6.08\ M\times0.547\ M$.
$rate=(1.79\times 10^{5})\times6.08\times0.547\ M/s\approx5.95\times 10^{5}\ M/s$.

Answer:

$[H_2]$	$[I_2]$	initial rate of reaction
$0.479\ M$	$2.31\ M$	$1.99\times 10^{5}\ M/s$
$6.08\ M$	$0.547\ M$	$5.95\times 10^{5}\ M/s$