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Topic: Equilibrium Concentration  (Read 3956 times)

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Offline JessMaksut

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Equilibrium Concentration
« on: March 04, 2009, 09:38:30 PM »
At 430 degrees C, the following reaction

H2 (g) + I2 (g)  ::equil:: 2HI
has an equilibrium constant of 54.3. If 0.765 mol HI are placed in 1.00 L container what would the equilibrium concentrations be?


H2 (g) + I2 (g)  ::equil:: 2HI
-x        -x                   +2x
0           0                   +2(0.765)

K = [HI]2 / [H2][I2]

54.3 = (2(0.765))2 / (-x)(-x)

Take square root:

7.37 = 2(0.765) / (-x)(-x)

...Then I don't know where to go from here. :(
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Offline Astrokel

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Re: Equilibrium Concentration
« Reply #1 on: March 04, 2009, 11:25:59 PM »
x is not 0.765. Initially, you have 0.765mol of HI.
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Offline frenchy

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Re: Equilibrium Concentration
« Reply #2 on: March 05, 2009, 02:13:03 AM »
Should you not be left with only one -x after taking the square root of both side of your equation?
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