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24 May, 13:18

Kc for the reaction N2O4 2NO2 is 0.619 at 45 degrees C If 50.0g of N2O4 is introduced into an empty 2.10L container, what are the partial pressures of NO2 and N2O4 after equilibrium has been achieved at 45 degrees C?

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  1. 24 May, 14:29
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    p (N2O4) = 0.318 atm

    p (NO2) = 7.17 atm

    Explanation:

    Step 1: Data given

    Kc = 0.619

    Temperature = 45.0 °C

    Mass of N2O4 = 50.0 grams

    Volume = 2.10 L

    Molar mass N2O4 = 92.01 g/mol

    Step 2: The balanced equation

    N2O4 ⇔ 2NO2

    Step 3: Calculate moles N2O4

    Moles N2O4 = 50.0 grams / 92.01 g/mol

    Moles N2O4 = 0.543 moles

    Step 4: The initial concentration

    [N2O4] = 0.543 moles/2.10 L = 0.259 M

    [NO2] = 0 M

    Step 5: Calculate concentration at the equilibrium

    For 1 mol N2O4 we'll have 2 moles NO2

    [N2O4] = (0.259 - x) M

    [NO2] = 2x

    Step 6: Calculate Kc

    Kc = 0.619 = [NO2]² / [N2O4]

    0.619 = (2x) ² / (0.259-x)

    0.619 = 4x² / (0.259 - x)

    x = 0.1373

    Step 7: Calculate concentrations

    [N2O4] = (0.259 - x) M = 0.1217 M

    [NO2] = 2x = 0.2746 M

    Step 8: The moles

    Moles = molarity * volume

    Moles N2O4 = 0.1217 M * 2.10 = 0.0256 moles

    Moles NO2 = 0.2746 M * 2.10 = 0.577 moles

    Step 9: Calculate partial pressure

    p*V = n*R*T

    ⇒ with p = the partial pressure

    ⇒ with V = the volume = 2.10 L

    ⇒ with n = the number of moles

    ⇒ with R = the gas constant = 0.08206 L*atm/mol*K

    ⇒ with T = the temperature = 45 °C = 318 K

    p = (nRT) / V

    p (N2O4) = (0.0256 * 0.08206 * 318) / 2.10

    p (N2O4) = 0.318 atm

    p (NO2) = (0.577 * 0.08206 * 318) / 2.10

    p (NO2) = 7.17 atm
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