calculating keq

[spoog74]

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problem #202 in dat destroyer 2011

the final equilibrium goes;

4-x + 4 -3x = 1

how does that equal 2x=1 ?

Basically this is the question

Consider teh following N2+3H2 >< 2NH3

8 moles of N2 and 8 Moles of H2 are placed in a 2 liter flask and allowed to come to equilibrium. At equilibrium, 2 moles of NH3 are formed. Calculate keq

and they give you the answer in terms of the equation you should setup....

 

[casper22]

Oh that 2x+1 comes from the C and E (ICE table) of the NH3..

 

[spoog74]

Oh that 2x+1 comes from the C and E (ICE table) of the NH3..

can you elaborate? Im trying to figure out how those equations give 2x+1

 

[casper22]

I dont have the book with me...but the final equation is not solved to get 2x=1...instead if you see the ICE table in the solution, the last column is for NH3... the initial concentration (I) of which is 0, change (C) = +2x and E = 2/2, since 2 moles of NH3 is formed and is placed in 2 lit flask..

thus 2x = 1

There is no way the final equation will give you 2x = 1

 

[Jab1113]

quick question while we are talking about the ICE tables, when do you know to use them??

this is the one thing I never understood

 

[spoog74]

I dont have the book with me...but the final equation is not solved to get 2x=1...instead if you see the ICE table in the solution, the last column is for NH3... the initial concentration (I) of which is 0, change (C) = +2x and E = 2/2, since 2 moles of NH3 is formed and is placed in 2 lit flask..

thus 2x = 1

There is no way the final equation will give you 2x = 1

dude i am SO sorry if i am annoying you, but i understand EVERYTHING up untiil you wrote "thus 2x=1" ...

What does 2 moles of NH3 forming in a 2 L flask have to do with 2x=1? I get that E = 2/2 and i get that the change is +2x... But where does 2x=1 come in?

 

[casper22]

dude i am SO sorry if i am annoying you, but i understand EVERYTHING up untiil you wrote "thus 2x=1" ...

What does 2 moles of NH3 forming in a 2 L flask have to do with 2x=1? I get that E = 2/2 and i get that the change is +2x... But where does 2x=1 come in?

You get E = 2/2 from the fact that 2 moles of NH3 forming in a 2 L flask...thus E = 2/2 = 1....now instead of setting up the final equation that you posted in your initial post, just work with the NH3 column...since your I is 0, C = 2x and E = 1...so just working with that entire column, you can solve for x by the equation 2x+1...

 

[casper22]

quick question while we are talking about the ICE tables, when do you know to use them??

this is the one thing I never understood

I want to say, to calculate Keq..basically solve equilibrium problems...

 

[spoog74]

You get E = 2/2 from the fact that 2 moles of NH3 forming in a 2 L flask...thus E = 2/2 = 1....now instead of setting up the final equation that you posted in your initial post, just work with the NH3 column...since your I is 0, C = 2x and E = 1...so just working with that entire column, you can solve for x by the equation 2x+1...

ok i get it now. But one more question;

How do you know when to solve for x in the second column or in the last one? I remember another problem that had the same setup solving for x in the last column......

Whats the difference and why do it this way in this problem?

 

[casper22]

ok i get it now. But one more question;

How do you know when to solve for x in the second column or in the last one? I remember another problem that had the same setup solving for x in the last column......

Whats the difference and why do it this way in this problem?

The other problem that you are talking about does not tell you how much the final amount is formed...and in that case the when you right the final equation, both sides will have x...