chem 12

posted by George

for the equilibrium
2HI(g)--- H2(g) + I2 k eq = 8.0
2.0 mol of HI are placed in a 4.0 L container, and the system is allowed to reach equilibrium. Calculate the equilibrium concentration of all three gases.
Inital
2HI 0.5mol/l
H2 =0
I2=0
change
2HI= -2x
H2=+x
I2=+x
equilibrium
2HI = 0.5-2x
H2= x
I2=x
keq= H2 x I2/ HI^2
8.0= (x) (x)/0.5-2x
taking sqr root of both sides
2.828=x/0.5-2x
0.2154=x
so 0.2154 mol/L of H2 and I2 at equilibrium and
2HI = 2x=2(0.2154)= 0.4308
0.5-0.4308=0.069
2HI=0.069 mol/L
If you could look this over to see if I have done this right. I appreciate any help you can give. Thank you for taking the time to offer your help. Again it is greatly appreciated.

1. bobpursley

keq= H2 x I2/ HI^2
8.0= (x) (x)/0.5-2x

what happened to the square in the denominator?

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