Question

some measurements of the initial rate of a certain reaction are given in the table below.

n₂	h₂	initial rate of reaction
2.18m	0.444m	0.810m/s
9.43m	0.444m	15.2m/s
2.18m	0.813m	2.72m/s

use this information to write a rate law for this reaction, and calculate the value of the rate constant k. round your value for the rate constant to 2 significant digits. also be sure your answer has the correct unit symbol.
rate = k
k =

Explanation:

Step1: Determine the order with respect to $N_2$

Compare the first and second rows where $[H_2]$ is constant. $\frac{[N_2]_2}{[N_2]_1}=\frac{9.43M}{2.18M}\approx4.326$, and $\frac{\text{rate}_2}{\text{rate}_1}=\frac{15.2M/s}{0.810M/s}\approx18.77$. Since $\frac{\text{rate}_2}{\text{rate}_1}=(\frac{[N_2]_2}{[N_2]_1})^n$, $18.77=(4.326)^n$. Taking the logarithm of both sides, $n = \frac{\log(18.77)}{\log(4.326)}\approx2$.

Step2: Determine the order with respect to $H_2$

Compare the first and third rows where $[N_2]$ is constant. $\frac{[H_2]_3}{[H_2]_1}=\frac{0.813M}{0.444M}\approx1.83$, and $\frac{\text{rate}_3}{\text{rate}_1}=\frac{2.72M/s}{0.810M/s}\approx3.36$. Since $\frac{\text{rate}_3}{\text{rate}_1}=(\frac{[H_2]_3}{[H_2]_1})^m$, $3.36=(1.83)^m$. Taking the logarithm of both sides, $m=\frac{\log(3.36)}{\log(1.83)}\approx2$.

Step3: Write the rate - law

The rate - law is $\text{rate}=k[N_2]^2[H_2]^2$.

Step4: Calculate the rate constant $k$

Using the first row of data, $0.810M/s = k(2.18M)^2(0.444M)^2$. First, calculate $(2.18M)^2(0.444M)^2=(2.18)^2\times(0.444)^2M^4\approx0.887M^4$. Then $k=\frac{0.810M/s}{0.887M^4}\approx0.91M^{-3}s^{-1}$.

Answer:

rate = $k[N_2]^2[H_2]^2$
k = $0.91M^{-3}s^{-1}$