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
0.516 m 0.689 m 4.00×10⁴ m/s
0.139 m 0.689 m m/s
0.159 m 2.24 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.
$k=\frac{rate}{[H_2][I_2]}$. Substituting $[H_2]=0.516\ M$, $[I_2]=0.689\ M$ and $rate = 4.00\times 10^{-4}\ M/s$ into the formula, we get $k=\frac{4.00\times 10^{-4}\ M/s}{0.516\ M\times0.689\ M}$.
$k=\frac{4.00\times 10^{-4}}{0.516\times0.689}\ s^{-1}\approx1.12\times 10^{-3}\ s^{-1}$.

Step2: Calculate the rate for the second row

For the second row, $[H_2]=0.139\ M$ and $[I_2]=0.689\ M$. Using the rate - law $rate = k[H_2][I_2]$, and substituting $k = 1.12\times 10^{-3}\ s^{-1}$, we have $rate=(1.12\times 10^{-3}\ s^{-1})\times0.139\ M\times0.689\ M$.
$rate=(1.12\times 10^{-3})\times0.139\times0.689\ M/s\approx1.07\times 10^{-4}\ M/s$.

Step3: Calculate the rate for the third row

For the third row, $[H_2]=0.159\ M$ and $[I_2]=2.24\ M$. Using the rate - law $rate = k[H_2][I_2]$, and substituting $k = 1.12\times 10^{-3}\ s^{-1}$, we get $rate=(1.12\times 10^{-3}\ s^{-1})\times0.159\ M\times2.24\ M$.
$rate=(1.12\times 10^{-3})\times0.159\times2.24\ M/s\approx3.97\times 10^{-4}\ M/s$.

Answer:

$[H_2]$	$[I_2]$	initial rate of reaction
$0.139\ M$	$0.689\ M$	$1.07\times 10^{-4}\ M/s$
$0.159\ M$	$2.24\ M$	$3.97\times 10^{-4}\ M/s$