Question

When 36.8g N₂O₄ (g) is introduced into a 1.0-litre flask at 27°C. The following equilibrium reaction occurs :

N₂O₄ (g) ⇌ 2NO₂ (g); Kₚ = 0.164 atm.

(a) Calculate K of the equilibrium reaction. (in 10⁻³ M)

(b) What are the number of millimoles of N₂O₄ at equilibrium?

(c) What is the total pressure of gases (in atm) in the flask at equilibrium?

(d) What is the percent dissociation of N₂O₄?

Answer 1

(a) $K_c = 6.66 \times 10^{-3}$ M (or 6.66 in 10⁻³ M) (b) 375 mmol (c) 10.5 atm (d) 6.25%

Answer 2

1. Calculate Initial Moles and Concentration of N₂O₄:

   • Molar mass of N₂O₄ = 92.02 g/mol.
   • Initial moles of N₂O₄ = $36.8 \text{ g} / 92.02 \text{ g/mol} \approx 0.400 \text{ mol}$.
   • Initial concentration of N₂O₄, $[N₂O₄]_0 = 0.400 \text{ M}$.

2. Calculate $K_c$ (Part a):

   • $\Delta n_g = 1$. $T = 27^\circ\text{C} = 300 \text{ K}$. $R = 0.0821 \text{ L atm/mol K}$.
   • $K_c = \frac{K_p}{(RT)^{\Delta n_g}} = \frac{0.164}{(0.0821 \times 300)^1} \approx 0.00666 \text{ M} = 6.66 \times 10^{-3} \text{ M}$.

3. Set up ICE Table and Calculate Equilibrium Concentrations:

   • Reaction: N₂O₄ (g) ⇌ 2NO₂ (g)
   • Initial (M): 0.400 0
   • Change (M): -x +2x
   • Equilibrium (M): 0.400 - x 2x
   • $K_c = \frac{[NO₂]^2}{[N₂O₄]} \implies 0.00666 = \frac{(2x)^2}{0.400 - x}$.
   • $4x^2 + 0.00666x - 0.002664 = 0$.
   • Solving for $x \approx 0.0250 \text{ M}$.

4. Calculate Millimoles of N₂O₄ at Equilibrium (Part b):

   • Equilibrium concentration of N₂O₄ = $0.400 \text{ M} - 0.0250 \text{ M} = 0.375 \text{ M}$.
   • Moles of N₂O₄ = $0.375 \text{ mol/L} \times 1.0 \text{ L} = 0.375 \text{ mol}$.
   • Millimoles of N₂O₄ = $0.375 \text{ mol} \times 1000 \text{ mmol/mol} = 375 \text{ mmol}$.

5. Calculate Percent Dissociation of N₂O₄ (Part d):

   • Degree of dissociation, $\alpha = \frac{x}{[N₂O₄]_0} = \frac{0.0250 \text{ M}}{0.400 \text{ M}} = 0.0625$.
   • Percent dissociation = $0.0625 \times 100\% = 6.25\%$.

6. Calculate Total Pressure at Equilibrium (Part c):

   • Equilibrium concentration of NO₂ = $2x = 2 \times 0.0250 \text{ M} = 0.0500 \text{ M}$.
   • Moles of NO₂ = $0.0500 \text{ mol}$.
   • Total moles = Moles of N₂O₄ + Moles of NO₂ = $0.375 \text{ mol} + 0.0500 \text{ mol} = 0.425 \text{ mol}$.
   • $P_{total} = \frac{n_{total}RT}{V} = \frac{0.425 \text{ mol} \times 0.0821 \text{ L atm/mol K} \times 300 \text{ K}}{1.0 \text{ L}} \approx 10.5 \text{ atm}$.