at 727 degree centigrade,K=3.8*10^-5 for the dissociation of I2 into iodine atoms: I2<->2I. if the original concentration of molecular iodine is 0.20 mol/L calculate the concentration of atomic iodine at equilibrium. check to see if the hundred rule is applicable
...........I2==> 2I
I........0.20....0
C.........-x....2x
E......0.20-x....2x
K = (I)^2/(I2)
Substitute from the ICE chart and solve or x. I don't know what the hundred rule is but the answer is within 5% so you can make the approximation that 0.20-x = 0.20.
To calculate the concentration of atomic iodine at equilibrium, we can use the equilibrium constant expression and the given information.
The equilibrium constant expression for the dissociation reaction is: Kc = [I]^2 / [I2]
First, let's find the initial concentration of iodine molecules ([I2]). We are given that the original concentration of molecular iodine is 0.20 mol/L, so [I2] = 0.20 mol/L.
Substituting this value into the equilibrium constant expression gives: Kc = [I]^2 / 0.20
Next, we can rearrange the equation to solve for the concentration of atomic iodine ([I]). Rearranging gives: [I]^2 = Kc * [I2]
Now, plug in the given equilibrium constant value (K = 3.8*10^-5) and the initial concentration of iodine molecules (0.20 mol/L) into the equation:
[I]^2 = (3.8*10^-5) * 0.20
Simplifying, we get:
[I]^2 = 7.6*10^-6
To solve for [I], take the square root of both sides:
[I] = √(7.6*10^-6)
Calculating this value, [I] ≈ 0.0028 mol/L
Therefore, the concentration of atomic iodine at equilibrium is approximately 0.0028 mol/L.
Now, to check if the hundred rule is applicable, we compare the initial concentration of molecular iodine ([I2]) to the concentration of atomic iodine at equilibrium ([I]equilibrium).
If [I2] / [I]equilibrium > 100, then the hundred rule is applicable.
In our case, [I2] = 0.20 mol/L and [I]equilibrium = 0.0028 mol/L.
So, [I2] / [I]equilibrium = 0.20 / 0.0028 ≈ 71.43
Since 71.43 is not greater than 100, the hundred rule is not applicable in this scenario.