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 reaction is second - order in $\ce{N2}$ and first - order in $\ce{H2}$, so the rate - law is $r = k[\ce{N2}]^{2}[\ce{H2}]$. First, find the rate constant $k$ using the first row of data. Given $[\ce{N2}]=2.26\ M$, $[\ce{H2}]=1.08\ M$ and $r = 55.0\ M/s$. Substitute into the rate - law: $55.0=k(2.26)^{2}(1.08)$. Then $k=\frac{55.0}{(2.26)^{2}(1.08)}$.

Step2: Calculate $k$

$k=\frac{55.0}{(2.26)^{2}(1.08)}=\frac{55.0}{2.26\times2.26\times1.08}=\frac{55.0}{5.57}\approx9.87\ M^{-2}\ s^{-1}$.

Step3: Calculate the rate for the second row

For the second row, $[\ce{N2}]=2.26\ M$ and $[\ce{H2}]=0.396\ M$. Using the rate - law $r = k[\ce{N2}]^{2}[\ce{H2}]$, substitute $k = 9.87\ M^{-2}\ s^{-1}$, $[\ce{N2}]=2.26\ M$ and $[\ce{H2}]=0.396\ M$. Then $r=9.87\times(2.26)^{2}\times0.396=9.87\times5.1076\times0.396\approx20.0\ M/s$.

Step4: Calculate the rate for the third row

For the third row, $[\ce{N2}]=0.260\ M$ and $[\ce{H2}]=9.40\ M$. Using the rate - law $r = k[\ce{N2}]^{2}[\ce{H2}]$, substitute $k = 9.87\ M^{-2}\ s^{-1}$, $[\ce{N2}]=0.260\ M$ and $[\ce{H2}]=9.40\ M$. Then $r=9.87\times(0.260)^{2}\times9.40=9.87\times0.0676\times9.40\approx6.25\ M/s$.

Answer:

Second row: $20.0\ M/s$
Third row: $6.25\ M/s$