# Problem: 03-02-1 | Toshiki's Note
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# Problem: 03-02-1 ​

Author:Anda Toshiki
Updated:a minute ago
Words:404
Reading:2 min

## Question ​

2 N 2O 5(g)→4 NO 2(g)+O 2(g)2 \mathrm{~N}_2 \mathrm{O}_5(g) \rightarrow 4 \mathrm{NO}_2(g)+\mathrm{O}_2(g) 2 N2​O5​(g)→4NO2​(g)+O2​(g)

The decomposition of N 2O 5(g)\mathrm{N}_2 \mathrm{O}_5(g)N2​O5​(g) is represented by the equation above. A sample of N 2O 5(g)\mathrm{N}_2 \mathrm{O}_5(g)N2​O5​(g) is monitored as it decomposes, and the concentration of N 2O 5\mathrm{N}_2 \mathrm{O}_5N2​O5​ as a function of time is recorded. The results are shown in the table below.

Time (s) | [N 2O 5]\mathrm{[N_2O_5]}[N2​O5​]
--- | ---
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 50.050.0 and 75.0 75.075.0 seconds.

## Solution ​

a A+b B→cC+dD a \mathrm{~A}+b \mathrm{~B} \rightarrow c \mathrm{C}+d \mathrm{D} a A+b B→cC+dD

the rate of reaction is defined as

rate =−1 aΔ[A]Δt=−1 bΔ[B]Δt=+1 cΔ[C]Δt=+1 dΔ[D]Δt\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} rate =−a1​ΔtΔ[A]​=−b1​ΔtΔ[B]​=+c1​ΔtΔ[C]​=+d1​ΔtΔ[D]​

Notice that the rate of change in concentration of each species is divided by its coefficient from the balanced chemical equation ( a aa, b,c b, cb,c, or d dd ). 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 N 2O 5\mathrm{N}_2 \mathrm{O}_5N2​O5​. With that in mind, let's write the reaction rate in terms of the rate of change in concentration of N 2O 5\mathrm{N}_2 \mathrm{O}_5N2​O5​ :

rate =−12Δ[N 2O 5]Δt\text { rate }=-\frac{1}{2} \frac{\Delta\left[\mathrm{N}_2 \mathrm{O}_5\right]}{\Delta t} rate =−21​ΔtΔ[N2​O5​]​

Since the coefficient for N 2O 5\mathrm{N}_2 \mathrm{O}_5N2​O5​ in the balanced equation is 2 , we divided the rate of change in concentration of N 2O 5\mathrm{N}_2 \mathrm{O}_5N2​O5​ by 2 . Additionally, since N 2O 5\mathrm{N}_2 \mathrm{O}_5N2​O5​ 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 50.050.0 and 75.0 75.075.0 seconds:

rate =−12(0.515 M−0.642 M)(75.0 s−50.0 s)=2.54×1 0−3 M s−1\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} rate ​=−21​(75.0 s−50.0 s)(0.515M−0.642M)​=2.54×10−3M s−1​

So, the average rate of the reaction between 50.0 50.050.0 and 75.0 75.075.0 seconds is 2.54×1 0−3 M s−1\bold{2.54 \times 10^{-3} \mathrm{M} \mathrm{s}^{-1}}2.54×10−3Ms−1.