Question

a certain reaction is second order in n₂ and first order in h₂. use this information to complete the table below. round each of your answers to 3 significant digits.

n₂	h₂	initial rate of reaction
2.43 m	1.74 m	m/s
0.810 m	3.06 m	m/s

Explanation:

Step1: Write the rate - law expression

The rate - law for the reaction is $rate = k[N_2]^2[H_2]$. First, find the rate constant $k$ using the first set of data.
Given $[N_2]=2.43\ M$, $[H_2]=1.02\ M$, and $rate = 0.0796\ M/s$.
$0.0796\ M/s=k(2.43\ M)^2(1.02\ M)$
$k=\frac{0.0796\ M/s}{(2.43\ M)^2(1.02\ M)}$
$k=\frac{0.0796}{2.43^{2}\times1.02}\ s^{-1}M^{-2}$
$k=\frac{0.0796}{6.007338}\ s^{-1}M^{-2}\approx0.01325\ s^{-1}M^{-2}$

Step2: Calculate the rate for the second row

For the second row, $[N_2]=2.43\ M$, $[H_2]=1.74\ M$, and $k = 0.01325\ s^{-1}M^{-2}$.
$rate=k[N_2]^2[H_2]$
$rate=(0.01325\ s^{-1}M^{-2})(2.43\ M)^2(1.74\ M)$
$rate = 0.01325\times6.007338\times1.74\ M/s$
$rate=0.01325\times10.452768\ M/s\approx0.138\ M/s$

Step3: Calculate the rate for the third row

For the third row, $[N_2]=0.810\ M$, $[H_2]=3.06\ M$, and $k = 0.01325\ s^{-1}M^{-2}$.
$rate=k[N_2]^2[H_2]$
$rate=(0.01325\ s^{-1}M^{-2})(0.810\ M)^2(3.06\ M)$
$rate=0.01325\times0.6561\times3.06\ M/s$
$rate=0.01325\times1.907666\ M/s\approx0.0253\ M/s$

Answer:

$[N_2]$	$[H_2]$	initial rate of reaction
$2.43\ M$	$1.74\ M$	$0.138\ M/s$
$0.810\ M$	$3.06\ M$	$0.0253\ M/s$