Posted by Anonymous on .
if the equilibrium concentration of nitrogen at 18degrees C is 0.256M and equil. concen. of hydrogen is 0.0184M what is the equil. concen. of ammonia?
N2 + 3 H2 <=> 2 NH3;
so K= [NH3]^2 / [N2][H2]^3
Is this correct?
Then ICE table? Giving me....
K= (2x)^2 / (0.256x)(0.0814 3x)^3
Am I on the right track?

Chem 111 
DrBob222,
You have substituted 0.256x as the equilibrium concn of N2 but the problem states it is 0.256M. You have substituted 0.01843x as the equil concn of H2 but the problem states it is 0.0184M.

Chem 111 
Anonymous,
Oh and K=.0244
so is this correct?
.0244 = (2x)^2 / (0.256)(0.0814)^3
x= 9x10^4