Ok so I am given H2 + I2 = 2HI .500 mol H2 1.00 mole I2 placed in a 10 L vessel

Question

Ok so I am given H2 + I2 = 2HI
.500 mol H2
1.00 mole I2
placed in a 10 L vessel what is the mole fraction of HI in mixture when equilibrium is reached at 205C
I have an appendix that provides deltaG,H ect
from this I conclude that I have a reaction that has 16.208 KJ/mol
an equilibrium constain of K= 1.44x10^-3
I have a K2 = 6.14x10^-3

and delta G of 16.208. What is the mole fraction of this reaction?

Explanation / Answer

we only require Keq for finding mol fraction suppose x mole of H2 reacted => x mol of I2 reacted and 2x mol of HI formed at equilibrium x^2/[(0.5-x)(1-x) = 1.44*10^-3 solving we get x=0.025 so mol fraction = 2x/(1.5-2x+2x) =0.033

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