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April 19, 2014

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?

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