Consider the reaction: 2 IF5 (g) + I4F2 (g) ⇋ 3 I2 (g) + 6 F2 (g)
6.0 mol of IF5 and 8.0 mol of I4F2 are placed in a 5.0 L container. At equilibrium, 6.0 mol of I4F2 are present. Calculate K for this reaction.
This is the same type of problem you posted earlier; i.e., the SO2 + O2 ==> SO3 problem. This one is done the same way.
I saw the follow up to the SO2 problem. Where is the confusion. If we work one in detail as I did you should be able to follow.
To calculate the equilibrium constant (K) for the given reaction, we need to use the equilibrium concentrations of the reactants and products. We also need to recognize the stoichiometry of the balanced chemical equation.
First, let's define the initial concentrations as follows:
Initial concentration of IF5 = 6.0 mol / 5.0 L = 1.2 M
Initial concentration of I4F2 = 8.0 mol / 5.0 L = 1.6 M
Initial concentration of I2 = 0 M (since it is not present initially)
Initial concentration of F2 = 0 M (since it is not present initially)
At equilibrium, the concentration of I4F2 is given as 6.0 mol / 5.0 L = 1.2 M. Since the volume does not change, we can assume that the total volume of the container remains constant.
Using the stoichiometry from the balanced equation, we can calculate the equilibrium concentrations of I2 and F2:
I2: The coefficient in front of I2 is 3. The change in concentration of I2 is 3 times the change in concentration of I4F2.
Change in concentration of I2 = 3 * (initial concentration of I4F2 - equilibrium concentration of I4F2) = 3 * (1.6 M - 1.2 M) = 1.2 M
Equilibrium concentration of I2 = 0 M (initial concentration) + 1.2 M (change in concentration) = 1.2 M
F2: The coefficient in front of F2 is 6. The change in concentration of F2 is 6 times the change in concentration of I4F2.
Change in concentration of F2 = 6 * (initial concentration of I4F2 - equilibrium concentration of I4F2) = 6 * (1.6 M - 1.2 M) = 2.4 M
Equilibrium concentration of F2 = 0 M (initial concentration) + 2.4 M (change in concentration) = 2.4 M
Now, we have the equilibrium concentrations of all the species:
[IF5] = 1.2 M
[I4F2] = 1.2 M
[I2] = 1.2 M
[F2] = 2.4 M
Finally, we can calculate the equilibrium constant K:
K = ([I2]^3 * [F2]^6) / ([IF5]^2 * [I4F2])
= (1.2 M)^3 * (2.4 M)^6 / (1.2 M)^2 * (1.2 M)
= 5.76 * 298.5984 / 1.44
≈ 119.07
Therefore, the equilibrium constant (K) for this reaction is approximately 119.07.
To calculate the equilibrium constant (K) for the given reaction, we need to use the balanced equation and the concentrations or partial pressures of the reactants and products at equilibrium.
The balanced equation for the reaction is:
2 IF5 (g) + I4F2 (g) ⇋ 3 I2 (g) + 6 F2 (g)
At equilibrium, the equilibrium concentrations of the reactants and products are as follows:
[IF5] = 6.0 mol / 5.0 L = 1.2 M
[I4F2] = 6.0 mol / 5.0 L = 1.2 M
[I2] = 0.0 mol / 5.0 L = 0.0 M (because there are no I2 molecules initially)
[F2] = 6.0 mol / 5.0 L = 1.2 M
Now, we can use these values to calculate the equilibrium constant (K). The general expression for K for the given reaction is:
K = ([I2]^3 * [F2]^6) / ([IF5]^2 * [I4F2])
Substituting the given equilibrium concentrations into the equation:
K = (0.0^3 * 1.2^6) / (1.2^2 * 1.2) = 0.0 / 1.44 = 0.0
Therefore, the equilibrium constant (K) for this reaction is 0.0.