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15 December, 01:22

Methanol and oxygen react to form carbon dioxide and water, like this: 2CH3OH (l) + 3O2 (g) → 2CO2 (g) + 4H2O (g) At a certain temperature, a chemist finds that a 9.3L reaction vessel containing a mixture of methanol, oxygen, carbon dioxide, and water at equilibrium has the following composition:compound CH3OH O2 CO2 H2Oamount 1.47g 1.56g 2.28g 3.33gCalculate the value of the equilibrium constant Kc for this reaction. Round your answer to 2 significant digits.

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  1. 15 December, 03:40
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    The value of the equilibrium constant Kc for this reaction is 3.34*10^-5

    Explanation:

    Step 1: Data given

    Volume of the reaction vessel = 9.3 L

    Mass of CH3OH at equilibrium = 1.47 grams

    Mass of O2 at equilibrium = 1.56 grams

    Mass of CO2 at equilibrium = 2.28 grams

    Mass of H2O at equilibrium = 3.33 grams

    Molar mass CH3OH = 32.04 g/mol

    Molar mass of O2 = 32.0 g/mol

    Molar mass of CO2 = 44.01 g/mol

    Molar mass of H2O = 18.02 g/mol

    Step 2: The balanced equation

    2CH3OH (l) + 3O2 (g) → 2CO2 (g) + 4H2O (g)

    Step 3: Calulate moles

    Moles = mass / molar mass

    Moles CH3OH = 1.47 grams / 32.04 g/mol

    Moles CH3OH = 0.0459 moles

    Moles O2 = 1.56 grams / 32.0 g/mol

    Moles O2 = 0.0489 moles

    Moles CO2 = 2.28 grams / 44.01 g/mol

    Moles CO2 = 0.0518 moles

    Moles H2O = 3.33 grams / 18.02 g/mol

    Moles H2O = 0.185 moles

    Step 4: Calculate molarity

    Molarity = moles / volume

    [CH3OH] = 0.0459 moles / 9.3 L

    [CH3OH] = 0.00494 M

    [O2] = 0.0489 moles / 9.3 L

    [O2] = 0.00526 M

    [CO2] = 0.0518 moles / 9.3 L

    [CO2] = 0.00557 M

    [H2O] = 0.185 moles / 9.3 L

    [H2O] = 0.0199 M

    Step 5: Calculate Kc

    Since liquids and solids do not affect the equilibrium, CH3OH will not take part

    Kc = [CO2]²[H2O]^4 / [O2]³

    Kc = (0.00557² * 0.0199^4) / 0.00526³

    Kc = 3.34*10^-5

    The value of the equilibrium constant Kc for this reaction is 3.34*10^-5
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