Homework Answers
initially
[H2] = 10^-3
[I2] = 2*10^-3
[HI] = 0
in equilbiirum
[H2] = 10^-3 - x
[I2] = 2*10^-3 -x
[HI] = 0 + 2x
and we know
[HI] = 0 + 2x = 1.87*10^-3
x = ( 1.87*10^-3)/2 = 0.000935
then
[H2] = 10^-3 - 0.000935 = 0.000065
[I2] = 2*10^-3 -0.000935 = 0.001065
Get K
K = [HI]^2 / [H2][I2]
K = (1.87*10^-3)^2 /(0.001065*0.000065) =
K = 50.514
best answer is D
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