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Chemistry

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For the following reaction, Kp = 3.5 104 at 1495 K.

H2(g) + Br2(g) 2 HBr(g)

What is the value of Kp for the following reactions at 1495 K?
(a) HBr(g) 1/2 H2(g) + 1/2 Br2(g)
(b) 2 HBr(g) H2(g) + Br2(g)
(c) 1/2 H2(g) + 1/2 Br2(g) HBr(g)

  • Chemistry - ,

    For 1/2 the reaction, new k is sqrt Kp.
    For 2 x reaction, new k is K^2p
    For the reverse and 1/2 new k is 1/sqrt Kp.

  • Chemistry - ,

    2.33

  • Chemistry - ,

    a) K'p=(Kp)^{-1/2}=5.3*10^-3

    b)K''p=(Kp)^{-1}=2.77*10^-5

    c)K'''p=(Kp)^{1/2}=189.7

    - Reversing the direction of a reaction gives the inverse of the equilibrium constant.
    - When multiplying by a constant, raise K to the power of that constant

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