Post a New Question

Chem 111

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

Am I on the right track?

  • Chem 111 - ,

    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 - ,

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

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question