Question

missed this? watch kcv: the rate law for a chemical reaction, we: determining the order and rate constant of a reaction. read section 15.3. you can click on the review link to access the section in your e - text. experimental data for the reaction 2no₂(g)+f₂(g)→2no₂f(g) are given in the table below. no₂ (m) f₂ (m) initial rate (m/s) 0.100 0.100 0.026 0.200 0.100 0.051 0.200 0.200 0.103 0.400 0.400 0.411 part a enter an expression for the reaction rate law. express your answer in terms of k, no₂, and f₂. view available hint(s) rate =

Explanation:

Step1: Determine order with respect to $\ce{NO2}$

Compare first and second experiments. When $[\ce{F2}]$ is constant and $[\ce{NO2}]$ is doubled from $0.100\ M$ to $0.200\ M$, the rate doubles from $0.026\ M/s$ to $0.051\ M/s$. So the reaction is first - order with respect to $\ce{NO2}$.

Step2: Determine order with respect to $\ce{F2}$

Compare second and third experiments. When $[\ce{NO2}]$ is constant and $[\ce{F2}]$ is doubled from $0.100\ M$ to $0.200\ M$, the rate doubles from $0.051\ M/s$ to $0.103\ M/s$. So the reaction is first - order with respect to $\ce{F2}$.

Step3: Write the rate law

The general rate law is rate = $k[\ce{NO2}]^m[\ce{F2}]^n$, where $m$ and $n$ are the orders with respect to $\ce{NO2}$ and $\ce{F2}$ respectively. Since $m = 1$ and $n = 1$, the rate law is rate = $k[\ce{NO2}][\ce{F2}]$.

Answer:

rate = $k[\ce{NO2}][\ce{F2}]$