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Difficulty: Very hardEquilibrium Constant Expression and Calculations
At a constant temperature, dinitrogen tetroxide gas, N2O4(g)\text{N}_2\text{O}_4(g), dissociates into nitrogen dioxide gas, NO2(g)\text{NO}_2(g), in a closed vessel according to the balanced equation:
N2O4(g)⇌2NO2(g)\text{N}_2\text{O}_4(g) \rightleftharpoons 2\text{NO}_2(g)
If the total equilibrium pressure of the system is 2.0 atm2.0\text{ atm} and the equilibrium mole fraction of N2O4(g)\text{N}_2\text{O}_4(g) is 0.200.20, what is the numerical value and unit of the equilibrium constant, KpK_p, for this reaction?
  1. 6.4 atm6.4\text{ atm}Answer
  2. B
    3.2 atm3.2\text{ atm}
  3. C
    4.0 atm4.0\text{ atm}
  4. D
    0.16 atm−10.16\text{ atm}^{-1}

Answer

The numerical value of the equilibrium constant KpK_p is 6.4 atm6.4\text{ atm}.
To find KpK_p, first calculate the partial pressure of each gas using Dalton's Law (Pi=χiPtotalP_i = \chi_i P_{\text{total}}). The mole fraction of NO2\text{NO}_2 is 1.00−0.20=0.801.00 - 0.20 = 0.80. Thus, PN2O4=0.20×2.0 atm=0.40 atmP_{\text{N}_2\text{O}_4} = 0.20 \times 2.0\text{ atm} = 0.40\text{ atm} and PNO2=0.80×2.0 atm=1.60 atmP_{\text{NO}_2} = 0.80 \times 2.0\text{ atm} = 1.60\text{ atm}. Substituting into Kp=(PNO2)2PN2O4K_p = \frac{(P_{\text{NO}_2})^2}{P_{\text{N}_2\text{O}_4}} gives (1.60)20.40=6.4 atm\frac{(1.60)^2}{0.40} = 6.4\text{ atm}.

Step-by-Step Solution

1
Determine the equilibrium mole fractions of all gaseous species.
Mole fraction of N2O4(g)\text{N}_2\text{O}_4(g), χN2O4=0.20\chi_{\text{N}_2\text{O}_4} = 0.20. Since the sum of mole fractions equals 1.00, the mole fraction of NO2(g)\text{NO}_2(g) is χNO2=1.00−0.20=0.80\chi_{\text{NO}_2} = 1.00 - 0.20 = 0.80.
The sum of all mole fractions in a mixture must equal 1.
2
Calculate the equilibrium partial pressures of each gas.
PN2O4=0.20×2.0 atm=0.40 atmP_{\text{N}_2\text{O}_4} = 0.20 \times 2.0\text{ atm} = 0.40\text{ atm}, and PNO2=0.80×2.0 atm=1.60 atmP_{\text{NO}_2} = 0.80 \times 2.0\text{ atm} = 1.60\text{ atm}.
Dalton's Law states that partial pressure equals mole fraction multiplied by total pressure (Pi=χiPtotalP_i = \chi_i P_{\text{total}}).
3
Write the KpK_p expression and calculate its value and unit.
Kp=(PNO2)2PN2O4=(1.60 atm)20.40 atm=2.56 atm20.40 atm=6.4 atmK_p = \frac{(P_{\text{NO}_2})^2}{P_{\text{N}_2\text{O}_4}} = \frac{(1.60\text{ atm})^2}{0.40\text{ atm}} = \frac{2.56\text{ atm}^2}{0.40\text{ atm}} = 6.4\text{ atm}.
The equilibrium constant KpK_p uses partial pressures raised to the power of their respective stoichiometric coefficients.

Key Concept

Equilibrium Constant in Terms of Partial Pressures (KpK_p)
Estimated Time:2m 30s
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Topics

Equilibrium Constant Expression and Calculations