Question

. If 2.00 mol of H2 and 1.00 mol of I2 come to equilibrium at this...

. If 2.00 mol of H2 and 1.00 mol of I2 come to equilibrium at this temperature (458ºC), how many moles of HI will be in the final mixture? (First, consider this question: Why is the volume of the container not needed for this problem?)

H2(g) + I2(g)  2HI(g) Kc = 50.3 at 458ºC

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Answer #1

The volume is not needed since there in no net change in the number of moles. 2 moles of reactants gives 2 moles of product.

You can assume Volume as V and solve. V will cancel out and you would not need its value

The reaction is,

Reaction: H2 + I2 ----> 2 HI

Initial:       2 .... 1...........0

Change: -X.... -X.........+2X

Final:    2-X.....1-X.........2X

Kc = [HI]^2 / [H2] [I2]

=> 50.3 = (2X)^2 / (2-X) (1-X)

=> 50.3 = 4X^2 / (X^2 - 3X + 2)

=> 50.3 X^2 - 150.9 X + 100.6 = 4X^2

=> 46.3 X^2 - 150.9 X + 100.6 = 0

Solving the above quadratic we get,

X = 0.935 M

Moles of HI = 2X = 2 x 0.935 = 1.87 moles

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