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
2.03 m 1.22 m 3.00 m/s
2.03 m 3.67 m m/s
8.54 m 0.290 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 $r = k[H_2][I_2]$. First, find the rate constant $k$ using the first row of data. Given $[H_2]=2.03\ M$, $[I_2]=1.22\ M$ and $r = 3.00\ M/s$. Then $k=\frac{r}{[H_2][I_2]}$.
$k=\frac{3.00\ M/s}{2.03\ M\times1.22\ M}\approx1.21\ M^{-1}s^{-1}$

Step2: Calculate the rate for the second row

For the second row, $[H_2]=2.03\ M$ and $[I_2]=3.67\ M$. Using the rate - law $r = k[H_2][I_2]$, substitute $k = 1.21\ M^{-1}s^{-1}$, $[H_2]=2.03\ M$ and $[I_2]=3.67\ M$.
$r=1.21\ M^{-1}s^{-1}\times2.03\ M\times3.67\ M\approx9.00\ M/s$

Step3: Calculate the rate for the third row

For the third row, $[H_2]=8.54\ M$ and $[I_2]=0.290\ M$. Using the rate - law $r = k[H_2][I_2]$, substitute $k = 1.21\ M^{-1}s^{-1}$, $[H_2]=8.54\ M$ and $[I_2]=0.290\ M$.
$r=1.21\ M^{-1}s^{-1}\times8.54\ M\times0.290\ M\approx3.00\ M/s$

Answer:

9.00 M/s
3.00 M/s