Chemistry test tomorrow and I'm lost, please help!?

D

Dr.A

New member
H2 + I2 = 2 HI
initial concentration
4.2 .. .5.0 .. 1.0
change
-x .. ..-x .. . +2x

4.2 - x = 0.6
x = 3.6
2x = 7.2

[H2] = 0.6
[I2]= 5.0 - 3.6= 1.4 M
[HI]= 7.2 M

Kc = (7.2)^2 / 0.6 x 1.4 = 61.7
 
Initially the concentrations of [H2]= 4.2 M ,[I2] = 5.0 M and [HI] = 1.0 M are in an sealed vessel at 500K. At equilibrium, the concentration of [H2]= 0.6 M. What is the EQUILIBRIUM concentration for [H2],[I2]and [HI]? H2(g) + I2(g) 2 HI(g) [HI] =__M

[H2] =___M

[I2] =___M
 
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