GChem Question...Help??

[jdoc04]

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N2 + 3H2 <----> 2NH3
If 8 moles of N2 and 8 moles of H2 are placed in a 2L flask & allowed to come to equilibrium. At equilibrium, 2 moles of NH3 are formed. Calculate Keq.

I know I need to setup the Initial, Change, Equilibrium value chart. I don't understand Dr. Romano's explanation.

Anyone???????

 

[Shunwei]

N2 + 3H2 <----> 2NH3
If 8 moles of N2 and 8 moles of H2 are placed in a 2L flask & allowed to come to equilibrium. At equilibrium, 2 moles of NH3 are formed. Calculate Keq.

I know I need to setup the Initial, Change, Equilibrium value chart. I don't understand Dr. Romano's explanation.

Anyone???????

You have to first convert the given moles into concentration, M, and the volume here is a given. So for both starting material you got 4 M, and that is what you would put for the Initial line in the ICE chart. Then, for the Change line, take into account the stoichiometric ratios, so for every two NH3 formed (+2x), you would lose one N2 (-x) and three H2(-3x). That would be the second line. Then, to get the equalibrium line, simply add the first "I" and "C" lines. So you would get 4 - x, 4 - 3x, and +2x. But here you are given the final product concentrations, 2 moles of NH3 in 2L; this translates to 1 M NH3 = + 2x, with x being 0.5 M.

So now you have x and then you can use it to backsolve for the equalibrium concentrations of 4 - x and 4 - 3x to get your answer. Hope this helps.

 

[jdoc04]

You have to first convert the given moles into concentration, M, and the volume here is a given. So for both starting material you got 4 M, and that is what you would put for the Initial line in the ICE chart. Then, for the Change line, take into account the stoichiometric ratios, so for every two NH3 formed (+2x), you would lose one N2 (-x) and three H2(-3x). That would be the second line. Then, to get the equalibrium line, simply add the first "I" and "C" lines. So you would get 4 - x, 4 - 3x, and +2x. But here you are given the final product concentrations, 2 moles of NH3 in 2L; this translates to 1 M NH3 = + 2x, with x being 0.5 M.

So now you have x and then you can use it to backsolve for the equalibrium concentrations of 4 - x and 4 - 3x to get your answer. Hope this helps.

Thanks so much!!

 

[Tejv24]

i had the same problem. thanks for futher elaboration.