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13 November, 22:33

A mixture of 3.00 mol of Cl2 and 2.10 mol I2 is initially placed into a rigid 1.00-L container at 350 °C. When the mixture has come to equilibrium, the concentration of ICl is 2.32 M. What is the equilibrium constant for this reaction at 350 °C? Cl2 (g) + I2 (g) ⇌ 2 ICl (g)

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  1. 13 November, 23:50
    0
    Kc = [ICl]² / [Cl₂]. [I₂] →2.32² / 1.84. 0.94 = 3.11

    Explanation:

    Let's propose the equilibrium reaction:

    Cl₂ (g) + I₂ (g) ⇄ 2ICl (g)

    Initial 3 m 2.10m -

    React x x 2x

    X amount has reacted of chlorine and iodine, so by stoichiometry, we made 2X of ICl.

    As we have this molar concentration we can determine the x → 2x = 2.32. Then x = 2.32 / 2 = 1.16

    Eq (3 m - 1.16) (2.10-1.16) 2.32

    Molar concentrations in the equilibrium are: [Cl₂] = 1.84M, [I₂] = 0.94M

    Let's make the expression for Kc:

    Kc = [ICl]² / [Cl₂]. [I₂] →2.32² / 1.84. 0.94 = 3.11
  2. 13 November, 23:57
    0
    The equilibrium constant is 3.11

    Explanation:

    Step 1: Data given

    Number of moles Cl2 = 3.00 moles

    Number of moles I2 = 2.10 moles

    Volume = 1.00L

    Temperature = 350 °C

    Concentration of ICl at the equilibrium = 2.32 M

    Step 2: The balanced equation

    Cl2 + I2 → 2ICl

    Step 3: Calculate initial concentrations

    [Cl2] = 3.00 moles / 1L = 3.00 M

    [I2] = 2.10 moles / 1L = 2.10 M

    [ICl] = 0M

    Step 4: Calculate concentration at equilibrium

    [Cl2] = 3.00 - X M

    [I2] = 2.10 - X M

    [ICl] = 2X = 2.32 M

    X = 2.32 / 2 = 1.16 M

    [Cl2] = 3.00 - 1.16 M = 1.84 M

    [I2] = 2.10 - 1.16 M = 0.94 M

    [ICl] = 2X = 2.32 M

    Step 5: Calculate Kc

    Kc = [ICl]²/[Cl2][I2]

    Kc = 2.32² / (1.84*0.94)

    Kc = 3.11

    The equilibrium constant is 3.11
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