Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: zcbteo6 on April 14, 2009, 05:51:49 AM
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ive been stuck on this for a while now
What will be the equilibrium number of moles of the following different species in the reaction below? The equilibrium constant, K, is 470 and the starting number of moles are given below.
h2 + 12 ---> 2HI
Formation of HI from elements
H2: 2
I2: 2
HI: 10
so the change for h2 and 12 will be 2-x...and for HI it would be 10+ 2x
so i thought
470= 10+ 2x(squared) / (2-x) (2-x)
and then i cross multiplied etc to get
470x (squared) - 1880 + 1880x= 10 + 4x
i did my quadratic eqaution and got
470x(squared) - 1876x + 1870=0
im hopeless as alegbra,and quadratic eqautions
i got the wrong answer when i did this
because the right answer for the moles at equilibrium is
HI: 10.6364 moles
H2: 1.6828 moles
I2: 1.6828 moles
so can anyone pleas help,and attempt this..because i think my algrebra has gone wrong..evidently
much appreciated
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It should be:
470= (10+ 2x)2 / (2-x) (2-x)
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thank you
ok so i re-did it with doing (10+ 2x) squared
and so i got as ax(squared) + bx + c=0
470x (squared) - 1880x + 1880 = 100=40 + 4x (squaured)
so 466x (squared)- 1840 + 1780 x=0
but then i did the quadratic equation and its still wrong
can anyone go through with me how to get the correct quadratic formula..as this is where im going wrong
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i know how to calculate quadratic eqautions, i just dont know how to rearrange into a quadratic form
470 = (10+2x) squared/ (2-x)(2-x)
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470(2-x)(2-x) = (10+2x)2