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Calculate equilibrium concentrations A closed system initially containin 1.000 x 10-3 M H2(g) and 2.000 x 10-3 M 12(9) at 448 °C is allowed to form HI and reach equilibrium. Analysis of the equilibrium mixture shows that the concentration of HI is 1.87 x Calculate K, at 448 °C for the reaction. a) 935 b) 27000 c) 4.00 d) 50.5
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Answer #1

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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