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!