doc: add new problems set for chemistry and update titles of sidebar

docs/.vitepress/config/sidebar.ts

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

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

@@ -3,19 +3,23 @@
 ## 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] $1.0 \mathrm{~L} /(\mathrm{mol} \cdot \mathrm{s})$
-  - [ ] $2.5 \mathrm{~L} /(\mathrm{mol} \cdot \mathrm{s})$
+
+-   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:
+
 $$
 \begin{aligned}
 k & =\frac{0.16 \mathrm{~mol} /(\mathrm{L} \cdot \mathrm{s})}{(0.40 \mathrm{~mol} / \mathrm{L})^2} \\
@@ -23,5 +27,6 @@ k & =\frac{0.16 \mathrm{~mol} /(\mathrm{L} \cdot \mathrm{s})}{(0.40 \mathrm{~mol
 & =1.0 \mathrm{~L} /(\mathrm{mol} \cdot \mathrm{s})
 \end{aligned}
 $$
+
 So, the value of the rate constant for the reaction is
-$1.0 \mathrm{~L} /(\mathrm{mol} \cdot \mathrm{s})$
+$1.0 \mathrm{~L} /(\mathrm{mol} \cdot \mathrm{s})$

docs/academic/chemistry/problems/03-02-3.md

@@ -0,0 +1,23 @@
+# Problem 03-02-3
+
+## Question
+
+$$
+2 \mathrm{X}(g)+2 \mathrm{Y}(g) \rightarrow \mathrm{Q}(g)+2 \mathrm{R}(g)
+$$
+
+The reaction represented above is found to be second order with respect to $\mathrm{X}$ and first order with respect to $\mathrm{Y}$.
+
+What happens to the rate of the reaction when $[\mathrm{X}]$ is halved and $[\mathrm{Y}]$ is doubled?
+
+-   [ ] It increases by a factor of 4.
+-   [x] It decreases by a factor of 2.
+-   [ ] It decreases by a factor of 4.
+-   [ ] It does not change.
+
+## Solution
+
+To solve this problem, let's first write out the rate law for the reaction. According to the text, the reaction is second order with respect to $\mathrm{X}$ and first order with respect to $\mathrm{Y}$, so the rate law is rate $=k[\mathrm{X}]^2[\mathrm{Y}]$
+
+Now, let's think about how the rate changes when $[\mathrm{X}]$ is halved and $[\mathrm{Y}]$ is doubled. Since $[\mathrm{X}]$ is raised to the second power in the rate law, halving $[\mathrm{X}]$ decreases the reaction rate by a factor of 4 . Similarly, since $[\mathrm{Y}]$ is raised to the first power, doubling $[\mathrm{Y}]$ increases the reaction rate by a factor of 2 . Combined, these changes result in the reaction rate decreasing by a factor of 2 overall.
+So, when $[\mathrm{X}]$ is halved and $[\mathrm{Y}]$ is doubled, the rate of the reaction decreases by a factor of 2 .