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1 September, 11:08

a flask was filled with SO2 at a partial pressure of 0.409 atm and O2 at a partial pressure of 0.601 atm. The following gas-phase reaction was allowed to reach equilibrium: 2SO2 + O2 2SO3 At equilibrium, the partial pressure of SO2 was 0.297 atm. Calculate the equilibrium partial pressure of O2.

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  1. 1 September, 13:38
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    The equilibrium partial pressure of O2 is 0.545 atm

    Explanation:

    Step 1: Data given

    Partial pressure of SO2 = 0.409 atm

    Partial pressure of O2 = 0.601 atm

    At equilibrium, the partial pressure of SO2 was 0.297 atm.

    Step 2: The balanced equation

    2SO2 + O2 ⇆ 2SO3

    Step 3: The initial pressure

    pSO2 = 0.409 atm

    pO2 = 0.601 atm

    pSO3 = 0 atm

    Step 4: Calculate the pressure at the equilibrium

    pSO2 = 0.409 - 2X atm

    pO2 = 0.601 - X atm

    pSO3 = 2X

    pSO2 = 0.409 - 2X atm = 0.297

    X = 0.056 atm

    pO2 = 0.601 - 0.056 = 0.545 atm

    pSO3 = 2*0.056 = 0.112 atm

    Step 5: Calculate Kp

    Kp = (pSO3) ² / ((pO2) * (pSO2) ²)

    Kp = (0.112²) / (0.545 * 0.297²)

    Kp = 0.261

    The equilibrium partial pressure of O2 is 0.545 atm
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