# Problem: 03-02-1

## Question

$$
2 \mathrm{~N}_2 \mathrm{O}_5(g) \rightarrow 4 \mathrm{NO}_2(g)+\mathrm{O}_2(g)
$$

The decomposition of $\mathrm{N}_2 \mathrm{O}_5(g)$ is represented by the equation above. A sample of $\mathrm{N}_2 \mathrm{O}_5(g)$ is monitored as it decomposes, and the concentration of $\mathrm{N}_2 \mathrm{O}_5$ as a function of time is recorded. The results are shown in the table below.

| Time (s) | $\mathrm{[N_2O_5]}$ |
| -------- | ------------------- |
| 0        | 1.000               |
| 25.0     | 0.801               |
| 50.0     | 0.642               |
| 75.0     | 0.515               |

Calculate the average rate of the reaction between $50.0$ and $75.0$ seconds.

## Solution

$$
a \mathrm{~A}+b \mathrm{~B} \rightarrow c \mathrm{C}+d \mathrm{D}
$$

the rate of reaction is defined as

$$
\color{cyan}\text { rate }=-\frac{1}{a} \frac{\Delta[\mathrm{A}]}{\Delta t}=-\frac{1}{b} \frac{\Delta[\mathrm{B}]}{\Delta t}=+\frac{1}{c} \frac{\Delta[\mathrm{C}]}{\Delta t}=+\frac{1}{d} \frac{\Delta[\mathrm{D}]}{\Delta t}
$$

Notice that the rate of change in concentration of each species is divided by its coefficient from the balanced chemical equation ( $a$, $b, c$, or $d$ ). This ensures that the calculated reaction rate is the same no matter which reactant or product is monitored for changes in concentration.
In this case, the monitored species was $\mathrm{N}_2 \mathrm{O}_5$. With that in mind, let's write the reaction rate in terms of the rate of change in concentration of $\mathrm{N}_2 \mathrm{O}_5$ :

$$
\text { rate }=-\frac{1}{2} \frac{\Delta\left[\mathrm{N}_2 \mathrm{O}_5\right]}{\Delta t}
$$

Since the coefficient for $\mathrm{N}_2 \mathrm{O}_5$ in the balanced equation is 2 , we divided the rate of change in concentration of $\mathrm{N}_2 \mathrm{O}_5$ by 2 .
Additionally, since $\mathrm{N}_2 \mathrm{O}_5$ is being consumed in the reaction, we included a negative sign in front of the expression.

Now, let's plug in the information from the table to calculate the average reaction rate between $50.0$ and $75.0$ seconds:

$$
\begin{aligned}
\text { rate } & =-\frac{1}{2} \frac{(0.515 M-0.642 M)}{(75.0 \mathrm{~s}-50.0 \mathrm{~s})} \\
& =2.54 \times 10^{-3} M \mathrm{~s}^{-1}
\end{aligned}
$$

So, the average rate of the reaction between $50.0$ and $75.0$ seconds is $\bold{2.54 \times 10^{-3} \mathrm{M} \mathrm{s}^{-1}}$.
