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14 May, 17:09

If kc = 7.04 * 10-2 for the reaction: 2 hbr (g) ⇌h2 (g) + br2 (g), what is the value of kc for the reaction: 1/2 h2 (g) + 1/2 br2 ⇌hbr (g)

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  1. 14 May, 17:17
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    At the first reaction when 2HBr (g) ⇄ H2 (g) + Br2 (g)

    So Kc = [H2] [Br2] / [HBr]^2

    7.04X10^-2 = [H2][Br] / [HBr]^2

    at the second reaction when 1/2 H2 (g) + 1/2 Br2 (g) ⇄ HBr

    Its Kc value will = [HBr] / [H2]^1/2*[Br2]^1/2

    we will make the first formula of Kc upside down:

    1/7.04X10^-2 = [HBr]^2/[H2][Br2]

    and by taking the square root:

    ∴ √ (1/7.04X10^-2) = [HBr] / [H2]^1/2*[Br]^1/2

    ∴ Kc for the second reaction = √ (1/7.04X10^-2) = 3.769
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