## Grade 12 Chemistry - Help with Equilibrium Constant problem!!!

Hi, i was wondering if anyone could help me with this problem because im not sure how to start or go about it.

The equilibrium constant for the reaction of hydrogen gas and iodine gas to form hdrogen iodide is 57.00 at 700K. If 1.00mole of each reactant is put into a 10.0L reaction vessel at 700K, what are the concentrations of each of the 3 gases at equilibrium? how many moles of each gas are present?

I have tried to solve the concentrations of each of the gases by using the quadratic formula.. but it just doesnt seem right.. if anyone could help anytime soon.. i'd rreally appreciate it, Thanks a lot!

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### Hi! Well, you said you

Hi! Well, you said you needed help anytime soon and it's been a week now, so i don't know if you still need this. Sorry...

The reaction described is:
H2(g) + I2(g) «--» 2 HI(g) [Kc = 57.00 at 700K]

If V = 10.0 L, the concentrations of the 3 gases at the beginning of the reaction are:
[H2]=[I2]=[HI]= 1.00/10.0 = 0.100 mol/dm^3

At the equilibrium, the concentrations should be:
[H2]=[I2]= 0.100-x mol/dm^3
[HI]= 0.100+2x mol/dm^3

Using the equilibrium constant, we determine x:
Kc = [HI]^2/([H2][I2])
57 = (0.1+2x)^2/(0.1-x)^2
57(0.01-0.2x+x^2)-(0.01+0.4x+4x^2) = 0
53x^2-11.8x+0.56 = 0
x = 0.154 v x = 0.067

The x value 0.154 is irrelevant (0.154>0.100).

Therefore, the concentrations at the equilibrium are:
[H2]=[I2]= 0.100-0.067 = 0.033 mol/dm^3
[HI]= 0.100+2(0.067) = 0.234 mol/dm^3

And the moles:
n(H2)=n(I2)= 0.033*10.0 = 0.33 mol
n(HI)= 0.158*10.0 = 2.34 mol

Hope this helps!

### At the equilibrium, the

At the equilibrium, the concentrations should be:
[H2]=[I2]= 0.100-x mol/dm^3
[HI]= 0.100+2x mol/dm^3

[HI] should = 0+2x mol/dm^2 since the question specifies that only 1 mol of the REACTANTS were put in the 10L container no?

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