2010 destroyer

[recyrb]

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consider the following: N2 + 3H2 <--> 2NH3, if 8 moles of N2 and 8 moles of H2 are placed in a 2 liter flask and allowed to come to quilibrium. At equilibrium, 2 moles of NH3 are formed. Calculate the Keq.

gen chem --on questions 192

how do you figure that 2x = 1
and x = .5??

 

[gtk06]

do products over reactants with the coefficients placed accordingly then substitute the moles for each of the molecules given. then solve it...correct me if im wrong

 

[recyrb]

consider the following: N2 + 3H2 <--> 2NH3, if 8 moles of N2 and 8 moles of H2 are placed in a 2 liter flask and allowed to come to quilibrium. At equilibrium, 2 moles of NH3 are formed. Calculate the Keq.

gen chem --on questions 192

how do you figure that 2x = 1
and x = .5??

no, because its in two liters, the quantities are all divided by 2, which i understand, but it shows the 4-x 4 - 3x and then some thing to do with the addition of NH3 so +2x but divided by two, and then x = .5 or some chit, anyone ??

 

[recyrb]

bumpp

 

[FROGGBUSTER]

Make an ICE table.

Initially, you have 4M N2 & 4/3M H2 in the flask. You know this has to react to form NH3 since there is initially no NH3 in the flask.

Try to do the rest of the problem and post back if you need help.

 

[recyrb]

yea i see the ice table, but i dont see what to do form

4-x 4-3x = (0+2x) this is the step i am confused about, how does x = .5

 

[rizzler]

yea i see the ice table, but i dont see what to do form

4-x 4-3x = (0+2x) this is the step i am confused about, how does x = .5

So the mole ratio's are 1:3:2 and you're told you have two liters. You are told that two moles of NH3 are formed and you are told it is in a two liter flask. 2moles/2liters = 1 M

When solving products over reactants the product (NH3) is +2x(the amount created) = 1 (to the known mole concentration you calculated above)
2x=1 (the confusing part because orgoman didn't explain it =.=")
working backwards essentially to get X you can solve the rest of the problem. It took me a while to understand this and I hope it helps.

 

[recyrb]

Yes, thats great thanks!