For the equilibrium system:

H2 + F2 <----> 2HF Kc = 230
If 5.80 mol of hydrogen gas and 5.80 mole are added to a 1.00 L flask what will be the concentration of HF at equilibrium?

Set up an ICE chart, substitute into Kc, and solve. Post your work if you get stuck.

To find the concentration of HF at equilibrium, we need to calculate the equilibrium concentrations of H2 and F2 first. We are given that the initial concentration of H2 is 5.80 mol and the initial concentration of F2 is also 5.80 mol.

Since the equation gives a 1:1 mole ratio between H2 and F2, the equilibrium concentration of H2 and F2 will be x, where x is the change in concentration from the initial concentration. Therefore, the equilibrium concentration of H2 and F2 will both be 5.80 - x.

Now, let's use the equilibrium expression Kc = [HF]^2 / ([H2] * [F2]) to find the equilibrium concentration of HF.

Given:
Initial concentration of H2 = 5.80 M
Initial concentration of F2 = 5.80 M
Equilibrium concentration of H2 = 5.80 - x
Equilibrium concentration of F2 = 5.80 - x
Equilibrium concentration of HF = ?

We can substitute these values into the equilibrium expression, Kc = [HF]^2 / ([H2] * [F2]):

230 = [HF]^2 / ((5.80 - x) * (5.80 - x))

Note that the equilibrium concentration of HF is not given, so we will denote it as x as well.

Now, we can solve for x. Rearranging the equation, we have:

230 * (5.80 - x)^2 = [HF]^2

Taking the square root of both sides, we get:

sqrt(230 * (5.80 - x)^2) = |HF|

Since the concentration cannot be negative, we can ignore the absolute value symbol.

sqrt(230 * (5.80 - x)^2) = HF

So, the equilibrium concentration of HF at equilibrium is sqrt(230 * (5.80 - x)^2).