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11 May, 14:36

For the equilibrium PCl5 (g) PCl3 (g) + Cl2 (g), Kc = 2.0 * 101 at 240°C. If pure PCl5 is placed in a 1.00-L container and allowed to come to equilibrium, and the equilibrium concentration of PCl3 (g) is 0.27 M, what is the equilibrium concentration of PCl5 (g) ?

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  1. 11 May, 15:50
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    the equilibrium concentration of [PCl₅] is 3.64*10⁻³ M

    Explanation:

    for the reaction

    PCl₅ (g) → PCl₃ (g) + Cl₂ (g)

    where

    Kc = [PCl₃]*[Cl₂]/[PCl₅] = 2.0*10¹ M = 20 M

    and [A] denote concentrations of A

    if initially the mixture is pure PCl₅, then it will dissociate according to the reaction and since always one mole of PCl₃ (g) is generated with one mole of Cl₂ (g), the total number of moles of both at the end is the same → they have the same concentration → [PCl₃ (g) ] = [Cl₂]=0.27 M

    therefore

    Kc = [PCl₃]*[Cl₂]/[PCl₅] = 0.27 M * 0.27 M / [PCl₅] = 20 M

    [PCl₅] = 0.27 M * 0.27 M / 20 M = 3.64*10⁻³ M

    [PCl₅] = 3.64*10⁻³ M

    the equilibrium concentration of [PCl₅] is 3.64*10⁻³ M
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