Question

a For the reaction H20g) Br2 (g) 2HBr(g) Kp 3.6 x 104 at 1496 K. What is the value of Kp for the following reaction at 1496 K? HBr(g) H2 (g) h Br20g)
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Answer #1

When an equation is reversed in direction the new Keq is reciprocal of the old one.

When an equationis divided by a factor x , then the K' = K1/x

and if we multiply the equation by x, then K' = Kx

Thus

K for the given new equation ( which is reversed and divided by 2)

K' = 1/K1/2

   = 1/(3.6x104)1/2

   = 5.27x10-3

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