Question

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

n₂	h₂	initial rate of reaction
2.20m	7.12m	55.0m/s
0.517m	1.92m	0.221m/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 $H_2$

Compare the first and second rows where $[N_2]$ is constant ($2.20M$). Let the rate - law be $rate = k[N_2]^m[H_2]^n$.
$\frac{rate_2}{rate_1}=\frac{k[N_2]_2^m[H_2]_2^n}{k[N_2]_1^m[H_2]_1^n}$. Since $[N_2]_1 = [N_2]_2$, we have $\frac{55.0M/s}{4.00M/s}=\frac{[7.12M]^n}{[1.92M]^n}$, $\frac{55.0}{4.00}=(\frac{7.12}{1.92})^n$, $13.75=(3.71)^n$. Taking the logarithm of both sides, $n=\log_{3.71}13.75\approx 2$.

Step2: Determine the order with respect to $N_2$

Compare the first and third rows where $[H_2]$ is constant ($1.92M$).
$\frac{rate_3}{rate_1}=\frac{k[N_2]_3^m[H_2]_3^n}{k[N_2]_1^m[H_2]_1^n}$. Since $[H_2]_1 = [H_2]_3$, we have $\frac{0.221M/s}{4.00M/s}=\frac{[0.517M]^m}{[2.20M]^m}$, $\frac{0.221}{4.00}=(\frac{0.517}{2.20})^m$, $0.05525=(0.235)^m$. Taking the logarithm of both sides, $m = \log_{0.235}0.05525\approx 1$.

Step3: Write the rate - law

The rate - law is $rate=k[N_2][H_2]^2$.

Step4: Calculate the rate constant $k$

Using the first row of data ($[N_2]=2.20M$, $[H_2]=1.92M$, $rate = 4.00M/s$) and the rate - law $rate=k[N_2][H_2]^2$, we can solve for $k$.
$k=\frac{rate}{[N_2][H_2]^2}=\frac{4.00M/s}{2.20M\times(1.92M)^2}=\frac{4.00}{2.20\times3.6864}M^{-2}s^{-1}\approx0.49M^{-2}s^{-1}$

Answer:

rate = $k[N_2][H_2]^2$
k = $0.49M^{-2}s^{-1}$