https://file.aiquickdraw.com/mathbot/file/1730142087311bfwriwpz.jpg

Let's work through the questions step-by-step. 

### Given:
- **Reaction:** \( 2 \, \text{NO}_2 (g) \rightleftharpoons \text{N}_2\text{O}_4 (g) \)  
- **\(\Delta H\)** = –57.2 kJ/mol (exothermic reaction)
- **Initial NO\(_2\) amount:** 2.00 mol
- **Container volume:** 500 cm³ = 0.500 L
- **At equilibrium:** 0.65 mol of N\(_2\)O\(_4\) has formed.

---

### **7.1. How much NO\(_2\) was used to produce 0.65 mol N\(_2\)O\(_4\)?**

From the balanced reaction:
\[ 2 \, \text{NO}_2 \rightarrow \text{N}_2\text{O}_4 \]  
2 moles of NO\(_2\) produce 1 mole of N\(_2\)O\(_4\).

Thus, to form 0.65 mol of N\(_2\)O\(_4\):
\[
\text{Moles of NO}_2 \, \text{used} = 2 \times 0.65 = 1.30 \, \text{mol}
\]

---

### **7.2. How much NO\(_2\) is left in the container at equilibrium?**

Initially, there were 2.00 mol of NO\(_2\). At equilibrium, 1.30 mol was used to form N\(_2\)O\(_4\), so:
\[
\text{Remaining NO}_2 = 2.00 - 1.30 = 0.70 \, \text{mol}
\]

---

### **7.3. Calculate the concentration at equilibrium.**

- **7.3.1.** **Concentration of NO\(_2\):**  
\[
\text{Concentration of NO}_2 = \frac{0.70 \, \text{mol}}{0.500 \, \text{L}} = 1.40 \, \text{mol/L}
\]

- **7.3.2.** **Concentration of N\(_2\)O\(_4\):**  
\[
\text{Concentration of N}_2\text{O}_4 = \frac{0.65 \, \text{mol}}{0.500 \, \text{L}} = 1.30 \, \text{mol/L}
\]

---

### **7.4. Calculate the equilibrium constant \(K_c\) at 200°C.**

For the reaction:
\[ 2 \, \text{NO}_2 (g) \rightleftharpoons \text{N}_2\text{O}_4 (g) \]
The equilibrium expression is:
\[
K_c = \frac{[\text{N}_2\text{O}_4]}{[\text{NO}_2]^2}
\]
Substitute the equilibrium concentrations:
\[
K_c = \frac{1.30}{(1.40)^2} = \frac{1.30}{1.96} \approx 0.663
\]

---

### **7.5. Will an increase in temperature affect \(K_c\)?**  
Since the reaction is **exothermic** (negative \(\Delta H\)), increasing the temperature will shift the equilibrium to the **left** (more NO\(_2\) is produced). As a result, \(K_c\) will **decrease**.

---

### **7.6. What will a decreased volume of the container affect in terms of color?**  
Decreasing the volume will increase the pressure, and the equilibrium will shift to the side with fewer gas molecules (Le Chatelier's principle).  
- **Reactants:** 2 moles of NO\(_2\)  
- **Product:** 1 mole of N\(_2\)O\(_4\)  

Since the equilibrium will shift towards N\(_2\)O\(_4\) (colourless), the mixture will become **less red-brown** (more colourless).

---

## Summary of Answers:
1. **7.1:** 1.30 mol of NO\(_2\) was used.
2. **7.2:** 0.70 mol of NO\(_2\) is left at equilibrium.
3. **7.3.1:** Concentration of NO\(_2\) = 1.40 mol/L.  
   **7.3.2:** Concentration of N\(_2\)O\(_4\) = 1.30 mol/L.
4. **7.4:** \(K_c = 0.663\).
5. **7.5:** **Decrease** in \(K_c\) with increasing temperature.
6. **7.6:** Mixture will become **more colourless** with decreased volume.

---

Do you need further details or have any questions?

### 5 Related Questions:
1. What happens to the equilibrium if additional NO\(_2\) is introduced to the container?
2. How would a catalyst affect the equilibrium concentrations?
3. How does pressure influence the equilibrium if temperature is kept constant?
4. What is the significance of \(K_c\) values being greater or less than 1?
5. How would the equilibrium concentrations change if the initial amount of NO\(_2\) was lower?

### Tip:
Le Chatelier’s principle is a powerful tool to predict how changes in conditions will shift chemical equilibrium. Always identify which side of the reaction has fewer molecules to understand pressure effects!

Dinitrogen tetroxide (N2O4), a colourless gas, is formed when two molecules of red-brown nitrogen dioxide (NO2) gas bond together, reaching chemical equilibrium according to the equation: 2NO2(g) ⇌ N2O4(g) Δ = -57.2 kJ/mol. Initially, 2.00 mol of NO2 gas was sealed in a 500 cm³ container. At equilibrium, it was found that 0.65 mol of N2O4 had been formed as a result of NO2 reacting. Answer the following questions: 7.1. How much NO2 was used to produce 0.65 mol of N2O4? 7.2. How much of the NO2 is left in the container at equilibrium? 7.3 Calculate the concentration of NO2 and N2O4 at equilibrium. 7.4 Calculate the equilibrium constant (Kc) for the reaction at 200°C. 7.5 How will an increase in temperature affect the Kc value? 7.6 What will a decreased volume of the container affect the color in the container? Only write MORE RED-BROWN, MORE COLOURLESS, or REMAIN THE SAME.

Solve a detailed equilibrium problem involving NO2 and N2O4 at 200°C. Learn how to calculate equilibrium concentrations, the equilibrium constant (Kc), and predict changes due to temperature and volume. This chemistry problem is suitable for students in Grades 11-12.

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