OC destoryer question about Keq

[vDDmaniaC]

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#146 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 equilibrium. At equilibrium, amoles of NH3 are formed. Calculate Keq.

Why can't we use a simple formula [NH3]^2 / [N]^2[H2]^2 here?

The solution uses the table technique that I am not very familiar.
Is there any reason why we can't just put the M/L directly into the above formula?

 

[Contach]

#146 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 equilibrium. At equilibrium, amoles of NH3 are formed. Calculate Keq.

Why can't we use a simple formula [NH3]^2 / [N]^2[H2]^2 here?

The solution uses the table technique that I am not very familiar.
Is there any reason why we can't just put the M/L directly into the above formula?

what they are giving you are initial concentrations..
at equilibrium there will be less of the reactants than what you had initially. Since keq is dependent on equilibrium concentrations, you have to find the eq concentrations. Thus you cannot just plug the numbers they give you into the 'simple formula' you speak of.

you would only be able to put the numbers into the formula if the question gave you equilibrium concentrations or reactants and products.

does that help?

 

[vDDmaniaC]

So what you says is that the 8 moles of N2 and 8 moles of H2 are placed in a 2 liter flask is not in the equilibrium condition yet?