July 09, 2020, 05:27:56 PM
Forum Rules: Read This Before Posting


Topic: calculating moles at equilibrium,quadratic eqautions  (Read 7158 times)

0 Members and 1 Guest are viewing this topic.

Offline zcbteo6

  • New Member
  • **
  • Posts: 3
  • Mole Snacks: +0/-1
calculating moles at equilibrium,quadratic eqautions
« on: April 14, 2009, 05:51:49 AM »
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

Offline AWK

  • Retired Staff
  • Sr. Member
  • *
  • Posts: 7390
  • Mole Snacks: +516/-86
  • Gender: Male
Re: calculating moles at equilibrium,quadratic eqautions
« Reply #1 on: April 14, 2009, 06:04:20 AM »
It should be:
470= (10+ 2x)2 /  (2-x) (2-x)
AWK

Offline zcbteo6

  • New Member
  • **
  • Posts: 3
  • Mole Snacks: +0/-1
Re: calculating moles at equilibrium,quadratic eqautions
« Reply #2 on: April 14, 2009, 06:50:32 AM »
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

Offline zcbteo6

  • New Member
  • **
  • Posts: 3
  • Mole Snacks: +0/-1
Re: calculating moles at equilibrium,quadratic eqautions
« Reply #3 on: April 14, 2009, 03:57:23 PM »
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)

Offline AWK

  • Retired Staff
  • Sr. Member
  • *
  • Posts: 7390
  • Mole Snacks: +516/-86
  • Gender: Male
Re: calculating moles at equilibrium,quadratic eqautions
« Reply #4 on: April 17, 2009, 07:11:22 AM »
470(2-x)(2-x) = (10+2x)2
AWK

Sponsored Links