Question

a certain reaction is first order in n2 and second order in h2. use this information to complete the table below. round each of your answers to 3 significant digits.
n2, 0.237 m, 0.865 m, 0.865 m, h2, 7.29 m, 0.643 m, 2.00 m, initial rate of reaction, 7.00×10^4 m/s, □ m/s, □ m/s

Explanation:

Step1: Write the rate - law expression

The rate - law for the reaction is $rate = k[N_2][H_2]^2$. First, we need to find the rate constant $k$ using the given data. Let's use the first set of data: $[N_2]=0.865M$, $[H_2]=2.00M$, and $rate = 7.00\times10^{-4}M/s$.
Substitute into the rate - law: $7.00\times10^{-4}=k\times0.865\times(2.00)^2$.
Solve for $k$:
\[

$$\begin{align*} k&=\frac{7.00\times 10^{-4}}{0.865\times4.00}\\ &=\frac{7.00\times 10^{-4}}{3.46}\\ &\approx2.02\times 10^{-4}M^{-2}s^{-1} \end{align*}$$

\]

Step2: Calculate the rate for the second set of concentrations

For $[N_2]=0.237M$ and $[H_2]=7.29M$, use the rate - law $rate = k[N_2][H_2]^2$.
Substitute $k = 2.02\times 10^{-4}M^{-2}s^{-1}$, $[N_2]=0.237M$, and $[H_2]=7.29M$ into the rate - law:
\[

$$\begin{align*} rate&=(2.02\times 10^{-4})\times0.237\times(7.29)^2\\ &=(2.02\times 10^{-4})\times0.237\times53.1441\\ &=(2.02\times 10^{-4})\times12.6951517\\ &\approx2.56\times 10^{-3}M/s \end{align*}$$

\]

Step3: Calculate the rate for the third set of concentrations

For $[N_2]=0.865M$ and $[H_2]=0.643M$, use the rate - law $rate = k[N_2][H_2]^2$.
Substitute $k = 2.02\times 10^{-4}M^{-2}s^{-1}$, $[N_2]=0.865M$, and $[H_2]=0.643M$ into the rate - law:
\[

$$\begin{align*} rate&=(2.02\times 10^{-4})\times0.865\times(0.643)^2\\ &=(2.02\times 10^{-4})\times0.865\times0.413449\\ &=(2.02\times 10^{-4})\times0.357633485\\ &\approx7.22\times 10^{-5}M/s \end{align*}$$

\]

Answer:

For the first blank: $2.56\times 10^{-3}M/s$
For the second blank: $7.22\times 10^{-5}M/s$