Homework Help: Chem 111

Posted by Anonymous on Thursday, September 8, 2011 at 12:25pm.

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 / (0.256-x)(0.0814 -3x)^3

Am I on the right track?

Chem 111 - DrBob222, Thursday, September 8, 2011 at 12:33pm
You have substituted 0.256-x as the equilibrium concn of N2 but the problem states it is 0.256M. You have substituted 0.0184-3x as the equil concn of H2 but the problem states it is 0.0184M.
Chem 111 - Anonymous, Thursday, September 8, 2011 at 6:46pm
Oh and K=.0244
so is this correct?
.0244 = (2x)^2 / (0.256)(0.0814)^3
x= 9x10^-4

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