doc: addd new problem set for chemistry and update sidebars and titles

docs/.vitepress/config/sidebar.ts

@@ -23,7 +23,8 @@ export const sidebar: DefaultTheme.Config['sidebar'] = {
             collapsed: true,
             items: [
                 { text: 'Problem: 02-20', link: '/academic/chemistry/problems/02-20' },
-                { text: 'Problem: 03-02', link: '/academic/chemistry/problems/03-02' },
+                { text: 'Problem: 03-02-1', link: '/academic/chemistry/problems/03-02' },
+                { text: 'Problem: 03-02-2', link: '' },
             ],
         },
     ],

docs/academic/chemistry/problems/03-02-1.md

@@ -1,4 +1,4 @@
-# Problem: 03-02
+# Problem: 03-02-1
 
 ## Question
 

docs/academic/chemistry/problems/03-02-2.md

@@ -0,0 +1,27 @@
+# Problem: 03-02-2
+
+## Question
+
+The rate law for a particular reaction is rate $=k[\mathrm{XY}]^2$. In an experiment, the initial rate of the reaction is determined to be $0.16 \mathrm{~mol} /(\mathrm{L} \cdot \mathrm{s})$ when the initial concentration of $\mathrm{XY}$ is $0.40 \mathrm{~mol} / \mathrm{L}$.
+- What is the value of the rate constant, $k$, for the reaction?
+  - [ ] $0.10 \mathrm{~L} /(\mathrm{mol} \cdot \mathrm{s})$
+  - [ ] $0.40 \mathrm{~L} /(\mathrm{mol} \cdot \mathrm{s})$
+  - [x] $1.0 \mathrm{~L} /(\mathrm{mol} \cdot \mathrm{s})$
+  - [ ] $2.5 \mathrm{~L} /(\mathrm{mol} \cdot \mathrm{s})$
+
+## Solution
+
+lo tind the value of the rate constant for the reaction, let's first solve the rate law for $k$ :
+$$
+k=\frac{\text { rate }}{[\mathrm{XY}]^2}
+$$
+Next, let's plug in the initial rate and concentration given in the text:
+$$
+\begin{aligned}
+k & =\frac{0.16 \mathrm{~mol} /(\mathrm{L} \cdot \mathrm{s})}{(0.40 \mathrm{~mol} / \mathrm{L})^2} \\
+& =\frac{0.16 \mathrm{~mol} /(\mathrm{L} \cdot \mathrm{s})}{0.16 \mathrm{~mol}^2 / \mathrm{L}^2} \\
+& =1.0 \mathrm{~L} /(\mathrm{mol} \cdot \mathrm{s})
+\end{aligned}
+$$
+So, the value of the rate constant for the reaction is
+$1.0 \mathrm{~L} /(\mathrm{mol} \cdot \mathrm{s})$