From 38b6455b2b63dcc6b2c33611f60c4c9adeed4c76 Mon Sep 17 00:00:00 2001
From: andatoshiki <101481353+andatoshiki@users.noreply.github.com>
Date: Thu, 2 Mar 2023 13:48:35 +0800
Subject: [PATCH] doc: addd new problem set for chemistry

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 docs/academic/chemistry/problems/3-2.md | 43 +++++++++++++++++++++++++
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+# Problem: 3-2
+
+## 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 $\bold{2.54 \times 10^{-3} \mathrm{M} \mathrm{s}^{-1}}$.
\ No newline at end of file