Calculate ΔG∘rxn at 298 K for the following reaction: I2(g)+Br2(g)⇌2IBr(g)Kp=436

Question

Calculate ΔG∘rxn at 298 K for the following reaction:
I2(g)+Br2(g)⇌2IBr(g)Kp=436

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Dulcie 5 years 2021-07-30T07:43:32+00:00 1 Answers 64 views 0

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  1. thucuc
    0
    2021-07-30T07:45:05+00:00
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    Answer:

    ΔG°rxn = -15,1 kJ/mol

    Explanation:

    ΔG°rxn = -RTlnK = -(8.314 J/mol K)(298 K)×ln436

    ΔG°rxn = -1.51×10^4 J/mol = -15.1 kJ/mol

    an exergonic reaction, it makes sense, since K > 1

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Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )