At $$ 400 \mathrm{~K}, \mathrm{~N}_{2} \mathrm{O}_{4} $$ decomposes into $$ \mathrm{NO}_{2} $$ in the gas phase: $$ \mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{~g}) $$ In an experiment, 0.328 g of of $$ \mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g}) $$ were placed in a 0.500 L evacuated flask at 400 K , and allowed to come to equilibrium, the final pressure was 0.331 atm. What were the partial pressures of $$ \mathrm{N}_{2} \mathrm{O}_{4} $$ and $$ \mathrm{NO}_{2} $$, and what is the equilibrium constant? $$ \mathrm{P}_{\mathrm{N} 2 \mathrm{O} 4}=0.135 $$ $$ \mathrm{P}_{\mathrm{NO} 2}=0.196 $$ atm $$ \mathrm{~K}_{\mathrm{c}}= $$ i eTextbook and Media

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1. **Calculate the initial moles of $$ \mathrm{N_2O_4} $$:**  
   The molar mass of $$ \mathrm{N_2O_4} $$ is  
   $$
   M = 2(14.01) + 4(16.00) \approx 28.02 + 64.00 = 92.02 \, \mathrm{g/mol}.
   $$  
   Given 0.328 g of $$ \mathrm{N_2O_4} $$, the initial number of moles is  
   $$
   n_0 = \frac{0.328 \, \mathrm{g}}{92.02 \, \mathrm{g/mol}} \approx 0.003565 \, \mathrm{mol}.
   $$

2. **Set up the reaction and the ICE table:**  
   The reaction is  
   $$
   \mathrm{N_2O_4 (g)} \rightleftharpoons 2\,\mathrm{NO_2 (g)}.
   $$  
   Let $$ x $$ be the number of moles of $$ \mathrm{N_2O_4} $$ that decompose at equilibrium. Then

| Species      | Initial (mol) | Change (mol)    | Equilibrium (mol)     |
   |--------------|---------------|-----------------|-----------------------|
   | $$ \mathrm{N_2O_4} $$ | $$ 0.003565 $$   | $$ -x $$          | $$ 0.003565 - x $$     |
   | $$ \mathrm{NO_2} $$   | 0             | $$ +2x $$         | $$ 2x $$               |

3. **Find the total moles at equilibrium using the ideal gas law:**  
   The total pressure at equilibrium is $$ P_{\text{total}} = 0.331 \, \mathrm{atm} $$, the volume is $$ V = 0.500 \, \mathrm{L} $$, and the temperature is $$ T = 400 \, \mathrm{K} $$. Using the ideal gas law  
   $$
   P V = n_{\text{total}} R T,
   $$
   we solve for the total moles $$ n_{\text{total}} $$:  
   $$
   n_{\text{total}} = \frac{P V}{R T} = \frac{0.331 \times 0.500}{0.08206 \times 400}.
   $$
   Compute the denominator:  
   $$
   0.08206 \times 400 = 32.824.
   $$
   Thus,  
   $$
   n_{\text{total}} \approx \frac{0.1655}{32.824} \approx 0.00504 \, \mathrm{mol}.
   $$

4. **Relate the number of moles at equilibrium:**  
   The total moles at equilibrium also equal  
   $$
   n_{\text{total}} = (0.003565 - x) + 2x = 0.003565 + x.
   $$
   Equate this with the value from the ideal gas law:  
   $$
   0.003565 + x = 0.00504.
   $$
   Solve for $$ x $$:  
   $$
   x = 0.00504 - 0.003565 \approx 0.001475 \, \mathrm{mol}.
   $$

5. **Calculate the equilibrium moles of each gas:**  
   $$
   n(\mathrm{N_2O_4}) = 0.003565 - x \approx 0.003565 - 0.001475 = 0.00209 \, \mathrm{mol},
   $$
   $$
   n(\mathrm{NO_2}) = 2x \approx 2 \times 0.001475 = 0.00295 \, \mathrm{mol}.
   $$

6. **Determine the partial pressures:**  
   The mole fractions are  
   $$
   \chi_{\mathrm{N_2O_4}} = \frac{0.00209}{0.00504} \approx 0.414, \quad \chi_{\mathrm{NO_2}} = \frac{0.00295}{0.00504} \approx 0.586.
   $$
   Their partial pressures are given by  
   $$
   P_i = \chi_i \times P_{\text{total}}.
   $$
   Thus,  
   $$
   P_{\mathrm{N_2O_4}} \approx 0.414 \times 0.331 \approx 0.137 \, \mathrm{atm},
   $$
   $$
   P_{\mathrm{NO_2}} \approx 0.586 \times 0.331 \approx 0.194 \, \mathrm{atm}.
   $$
   These values are in good agreement with the given values $$ P_{\mathrm{N_2O_4}} \approx 0.135 \, \mathrm{atm} $$ and $$ P_{\mathrm{NO_2}} \approx 0.196 \, \mathrm{atm} $$.

7. **Calculate the equilibrium constant $$ K_c $$:**  
   First, find the concentrations by dividing the moles by the volume of the flask, $$ V = 0.500 \, \mathrm{L} $$:
   $$
   [\mathrm{N_2O_4}] = \frac{0.00209}{0.500} = 0.00418 \, \mathrm{M},
   $$
   $$
   [\mathrm{NO_2}] = \frac{0.00295}{0.500} = 0.00590 \, \mathrm{M}.
   $$
   The equilibrium constant for the reaction  
   $$
   \mathrm{N_2O_4} \rightleftharpoons 2\, \mathrm{NO_2}
   $$
   is  
   $$
   K_c = \frac{[\mathrm{NO_2}]^2}{[\mathrm{N_2O_4}]}.
   $$
   Substitute the equilibrium concentrations:
   $$
   K_c = \frac{(0.00590)^2}{0.00418} = \frac{0.00003481}{0.00418} \approx 0.00832.
   $$

Thus, the results are:

- $$ P_{\mathrm{N_2O_4}} \approx 0.135 \, \mathrm{atm} $$,
- $$ P_{\mathrm{NO_2}} \approx 0.196 \, \mathrm{atm} $$,
- $$ K_c \approx 8.3 \times 10^{-3}. $$
At equilibrium: - $$ \mathrm{N_2O_4} $$ partial pressure is approximately 0.135 atm, - $$ \mathrm{NO_2} $$ partial pressure is approximately 0.196 atm, - The equilibrium constant $$ K_c $$ is approximately 0.0083.

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