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12 October, 23:08

For the reaction: SO2 (g) + NO2 (g) ⇌ SO3 (g) + NO (g), the equilibrium constant, Kc, is 18.0 at 1200 °C. If 1.0 mole of SO2 and 2.0 moles of NO2 are placed in a 2.0 L container, what concentration of SO3 will be present at equilibrium?

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  1. 12 October, 23:13
    0
    [SO3] = 0.476 M

    Explanation:

    Step 1: Data given

    Kc = 18.0 at 1200 °C

    Number of moles SO2 = 1.0 moles

    Number of moles NO2 = 2.0 moles

    Volume = 2.0 L

    Step 2: The balanced equation

    SO2 (g) + NO2 (g) ⇌ SO3 (g) + NO (g)

    Step 3: Calculate molarity

    Molarity = moles / volume

    Molarity SO2 = 1.0 mol / 2.0 L

    Molarity SO2 = 0.5 M

    Molarity NO2 = 2.0 mol / 2.0 L

    Molarity NO2 = 1.0 M

    Step 3: Initial concentrations

    [SO2] = 0.5M

    [NO2] = 1.0 M

    [SO3] = 0M

    [NO] = 0M

    Step 4: Concentration at equilibrium

    [SO2] = 0.5 - X M

    [NO2] = 1.0 - X M

    [SO3] = XM

    [NO] = XM

    Step 5: Calculate concentrations

    Kc = [SO3][NO] / [SO2][NO2]

    Kc = 18 = x² / ((0.5 - x) (1.0-x)

    x = 0.476

    [SO2] = 0.5 - 0.476 M = 0.024 M

    [NO2] = 1.0 - 0.476 M = 0.524 M

    [SO3] = XM = 0.476 M

    [NO] = XM = 0.476 M
  2. 12 October, 23:21
    0
    [SO₃] in the equilibrium, will be 0.47 M

    Explanation:

    First of all we state the equilibrium reaction:

    SO₂ (g) + NO₂ (g) ⇌ SO₃ (g) + NO (g)

    Intially we have 1 moles of sulfur dioxide and 2 moles of nitrogen dioxide.

    The stoichiometry is all 1:1.

    During the reaction x amount has reactend, then in the equilibrum I would have

    (1-x) moles of SO₂

    (2-x) moles of NO₂

    and there are formed x moles of SO₃ and x moles of NO

    The thing is that, we need molar concentrations, so all the values must be divided by 2L, the vessel volume

    Let's make the expression for Kc:

    Kc = [SO₃]. [NO] / [SO₂]. [NO₂]

    Kc = x/2. x/2 / (1-x) / 2. (2-x) / 2

    Kc = x²/4 / (1-x) (2-x) / 4 → (1-x) (2-x) = 2-x-2x+x² → 2-3x+x²

    18 = x²/4 / 2-3x+x²/4 → we cancel the 4 → 18 = x² / 2-3x+x²

    18 (2-3x+x²) = x² → 0 = 17x² - 54x + 36

    Let's solve the quadratic funtion:

    17 = a; - 54 = b; 36 = c → (-b + - √ (b²-4ac)) / 2a

    (-54 + - √ (-54² - 4. 17. 36)) / 2.36

    x1 = 0.95

    x2 = 2.22

    We take, the first value so the [SO₃] in the equilibrium, will be 0.95/2L = 0.47 M
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