Chemical Equilibrium, finding concentration

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SUMMARY

The discussion focuses on calculating the concentration of hydrogen iodide (HI) in a chemical equilibrium reaction involving hydrogen (H2) and iodine (I2) at a temperature of 731K with an equilibrium constant (K) of 49.0. Given 0.171 mol of H2 and 1.571 mol of I2 in a 1.04 L vessel, the concentrations of H2 and I2 are calculated by dividing the moles by the volume. The equilibrium expression K = [HI]² / ([H2][I2]) is utilized to solve for the concentration of HI.

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parwana
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H2(g) + I2(g) = 2HI(g)

T = 731K
K = 49.0


If at equilibrium 0.171 mol H2 and 1.571 mol I2 are present in a 1.04 L vessel, what is the concentration of HI in the gaseous mixture in moles/liter ?

Ok I tried doing this question by setting up a mole ratio first and then finding moles for HI, but I don't get it. HELP.
 
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So we know that

[tex]K=\frac{[HI]^{2}}{[H_{2}][I_{2}]}[/tex]
that is
[tex]49=\frac{[HI]^{2}}{[H_{2}][I_{2}]}[/tex]

now for the concentrations of H2 and I2, we take the number of moles and divide by 1.04, the size of the container.

Then we can put these two numbers in for

and [I2] in the above equation and solve for the concentration of HI.

hope it helped

 
ChemRookie had a similar question in the high school HW section, of which I attempted an answer.
 

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