For the reaction below at a particular temperature, the concentrations at equilibrium were observed to be: [f2] = 2.1× 10–2 mol/l and [f] = 2.0 × 10–4mol/l. calculate the value of the equilibrium constant from these data. (the units are deleted.)

Kc = 1.9x10^-6

Explanation:

You are missing the reaction, however, let me write the reaction for you:

F2(g) <> 2F(g)

Now, remember that for an equilibrium reaction, that means that a reaction is not complete at all, we have a value of equilibrium constant, usually called Kc.

This value of Kc is the result of concentrations of products and reactants:

Kc = [Products] / [Reactants]

However, you should note that only substances and compounds in gaseous phase and aquous contribute to the equilibrium. This is because solid and liquid concentrations values are constants and near to 1.

Now for a general reaction like this one:

aA(g) + bB(g) <> cC(g) + dD(g)

The expression of Kc would be:

Kc = [C]^c * [D]^d / [A]^a * [B]^b

The letters a, b, c, d are the coefficients that balance the equation and they should be considered in the equilibrium, that is why they are elevated.

Now, with this clear, let's apply these concepts to the reaction here:

F2(g) <> 2F(g)

The expression of Kc will be:

Kc = [F]² / [F2]

Replacing the given values in the above expression, we can solve for Kc:

Kc = (2x10^-4)² / (2.1x10^-2)

Kc = 1.9x10^-6

This is the value of Kc and have no units.

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