T (K) [N2] [H2] [NH3] [Kc]
575 0.110M − 0.134M 9.6
Find [H2] at 575 K.
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...
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 ?
To find the concentration of [H2] at 575 K, we can use the equilibrium constant expression and the given values for [N2], [NH3], and Kc.
The balanced chemical equation for the reaction is:
N2 + 3H2 ↔ 2NH3
The equilibrium constant expression, Kc, for this reaction is:
Kc = [NH3]^2 / ([N2] * [H2]^3)
Given:
[N2] = 0.110 M
[NH3] = 0.134 M
Kc = 9.6
We need to find [H2] at 575 K.
Let's rearrange the equilibrium constant expression to solve for [H2]:
[H2]^3 = [NH3]^2 / (Kc * [N2])
Now substitute the given values into the equation:
[H2]^3 = (0.134 M)^2 / (9.6 * 0.110 M)
Simplify the expression:
[H2]^3 = 0.0190845 M / 1.056
[H2]^3 = 0.018072
To find [H2], we need to take the cube root of both sides:
[H2] = (0.018072)^(1/3)
[H2] ≈ 0.276 M
Therefore, [H2] at 575 K is approximately 0.276 M.