At a certain temperature, the reaction H2(g)+ Cl(g)=2HCl(g)has Kc = 5 x 10^8. If a reaction mixture at equilibrium contains 0.48 M of H2 and 0.23 M of Cl2, what is the concentration of HCl?

I was wondering if I had to rearrange the equilibrium constant equation to work out the concentration of HCl (product).

Thanks for your help.

H2 + Cl2 ==> 2HCl

Kc = (HCl)^2/(H2)(Cl2)
You know Kc, (H2) and (Cl2). The only unknown is (HCl). Substitute and solve for that.

To find the concentration of HCl at equilibrium, you can use the equilibrium constant expression. The equilibrium constant expression for the given reaction, H2(g) + Cl2(g) ⇌ 2HCl(g), is written as:

Kc = [HCl]^2 / ([H2] * [Cl2])

Since you have the values for the concentrations of H2 and Cl2, you can substitute these values into the expression and solve for the concentration of HCl.

Substituting [H2] = 0.48 M and [Cl2] = 0.23 M into the expression, you have:

Kc = [HCl]^2 / (0.48 * 0.23)

Now, rearrange the equation to solve for [HCl]:

[HCl]^2 = Kc * (0.48 * 0.23)

[HCl] = √(Kc * (0.48 * 0.23))

Plugging in the given value for Kc (5 x 10^8), you can calculate the concentration of HCl at equilibrium.