The reaction N2O4(g) -->2NO2(g) has Kc = 0.133 at 25 °C. What is the NO2 concentration in a 5.00 liter flask if it contains 0.250 mol of N2O4 at equilibrium

I know I set something equal to Kc?

(NO2) at equilibrium = 0.250 mols/5 L = 0.05

Kc = (NO2)^2/(N2O4) = 0.133

(NO2)2 = x^2
(N2O4 = 0.05
Solve for x

To solve this problem, we can use the equilibrium expression. The equilibrium expression for the given reaction is:

Kc = [NO2]^2 / [N2O4]

We are given the value of Kc as 0.133, the initial moles of N2O4 as 0.250 mol, and the volume of the flask as 5.00 L.

At equilibrium, let's assume x mol of N2O4 has reacted to form 2x mol of NO2.

Substituting these values into the equilibrium expression, we get:

0.133 = (2x)^2 / (0.250 - x)

Now we need to solve this equation for x. To simplify the equation, we can multiply both sides by (0.250 - x) to eliminate the denominator:

0.133 * (0.250 - x) = (2x)^2

0.03325 - 0.133x = 4x^2

Rearranging the equation:

4x^2 + 0.133x - 0.03325 = 0

Now we can solve this quadratic equation using the quadratic formula:

x = (-b ± sqrt(b^2 - 4ac)) / 2a

Here, a = 4, b = 0.133, and c = -0.03325. By substituting these values into the quadratic formula, we can calculate the value of x. Remember that we are interested in the positive value of x, as negative values do not have physical meaning in this context.

Once we find the value of x, we can calculate the concentration of NO2 in the 5.00 L flask by multiplying x by 2 (since 2x mol of NO2 is formed for every x mol of N2O4 reacted).