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3 January, 02:43

A mixture of 0.600 mol of bromine and 1.600 mol of iodine is placed into a rigid 1.000-L container at 350°C. Br2 (g) + I2 (g) 2IBr (g) When the mixture has come to equilibrium, the concentration of iodine monobromide is 1.190 M. What is the equilibrium constant for this reaction at 350°C?

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  1. 3 January, 06:12
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    The answer to your question is K = 1.48

    Explanation:

    Balanced chemical reaction

    Br₂ + I₂ ⇒ 2IBr

    Process

    1. - Calculate the concentration of each of the reactants

    [Br₂] = 0.6 mol / 1 L

    = 0.6 M

    [I₂] = 1.6 mol / 1 L

    = 1.6 M

    [IBr] = 1.19 M

    2. - Write the equation of the Equilibrium constant

    K = [IBr]² / [Br₂][I₂]

    -Substitution

    K = [1.19]² / [0.6][1.6]

    - Simplification

    K = 1.4161 / 0.96

    - Result

    K = 1.48
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