the way i set it up was
9.6=(.134)^2/(.110)(x)^3
just don't know if it is right or how to solve for x...
T (K) [N2] [H2] [NH3] [Kc]
575 0.110M − 0.134M 9.6
Find [H2] at 575 K.
3 answers
If these are equibrium concentrations that is right.
To solve for x do
(0.110)(x^3)(9.6) = (0.134)^2
x^3 = (0.134)^2/9.6*0.110 = ?
Then x = cube root of ?
(0.110)(x^3)(9.6) = (0.134)^2
x^3 = (0.134)^2/9.6*0.110 = ?
Then x = cube root of ?
1 answer