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
2.26 m	0.396 m	m/s
0.260 m	9.40 m	m/s

Explanation:

Step1: Determine the rate - law expression

The rate - law for the reaction is $r = k[\text{N}_2]^2[\text{H}_2]$. First, find the rate constant $k$ using the first set of data. Given $[\text{N}_2]=2.26\ M$, $[\text{H}_2]=1.08\ M$ and $r = 55.0\ M/s$.

$$k=\frac{r}{[\text{N}_2]^2[\text{H}_2]}=\frac{55.0}{(2.26)^2\times1.08}$$

$$k=\frac{55.0}{2.26^2\times1.08}=\frac{55.0}{5.55}\approx9.91\ M^{-2}s^{-1}$$

Step2: Calculate the rate for the second row

For the second row, $[\text{N}_2]=2.26\ M$ and $[\text{H}_2]=0.396\ M$. Using the rate - law $r = k[\text{N}_2]^2[\text{H}_2]$ and $k = 9.91\ M^{-2}s^{-1}$.

$$r=9.91\times(2.26)^2\times0.396$$

$$r=9.91\times5.1076\times0.396$$

$$r=9.91\times2.0226\approx20.0\ M/s$$

Step3: Calculate the rate for the third row

For the third row, $[\text{N}_2]=0.260\ M$ and $[\text{H}_2]=9.40\ M$. Using the rate - law $r = k[\text{N}_2]^2[\text{H}_2]$ and $k = 9.91\ M^{-2}s^{-1}$.

$$r=9.91\times(0.260)^2\times9.40$$

$$r=9.91\times0.0676\times9.40$$

$$r=9.91\times0.63544\approx6.30\ M/s$$

Answer:

$[\text{N}_2]$	$[\text{H}_2]$	initial rate of reaction
$2.26\ M$	$0.396\ M$	$20.0\ M/s$
$0.260\ M$	$9.40\ M$	$6.30\ M/s$