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14 November, 14:21

If one starts with pure NO2 (g) at a pressure of 0.500 atm, the total pressure inside the reaction vessel when 2NO2 (g) 2NO (g) + O2 (g) reaches equilibrium is 0.674 atm. Calculate the equilibrium partial pressure of NO2.

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  1. 14 November, 14:30
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    The partial pressure of NO2 = 0.152 atm

    Explanation:

    Step 1: Data given

    Pressure NO2 = 0.500 atm

    Total pressure at equilibrium = 0.674 atm

    Step 2: The balanced equation

    2NO2 (g) → 2NO (g) + O2 (g)

    Step 3: The initial pressure

    pNO2 = 0.500 atm

    pNO = 0 atm

    p O2 = 0 atm

    Step 4: Calculate pressure at the equilibrium

    For 2 moles NO2 we'll have 2 moles NO and 1 mol O2

    pNO2 = 0.500 - 2x atm

    pNO = 2x atm

    pO2 = xatm

    The total pressure = p (total) = p (NO2) + p (NO) + p (O2)

    p (total) = (0.500 - 2x) + 2x + x = 0.674 atm

    0.500 + x = 0.674 atm

    x = 0.174 atm

    This means the partial pressure of NO2 = 0.500 - 2*0.174 = 0.152 atm
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