In a study of halogen bond strength 0.50mol of iodine was heated in 2.5litre vessel and the following reaction occurred. Calculate the concentrations of the iodine and iodide at equilibrium at 2000k. Given that Kc=0.209?
What reaction occurred?
To calculate the concentrations of iodine and iodide at equilibrium, we need to use the given equilibrium constant, Kc. The equation for the reaction is:
I2(g) ⇌ 2I(g)
The equilibrium constant expression for this reaction can be written as:
Kc = [I(g)]^2 / [I2(g)]
Given that Kc = 0.209, we can rewrite the equation as:
0.209 = [I(g)]^2 / [I2(g)]
Now, let's determine the initial moles of iodine and the volume of the vessel to calculate the initial concentrations:
Number of moles of I2 = 0.50 mol
Volume of the vessel = 2.5 L
The initial concentration of I2 can be calculated as:
Initial concentration of I2 = Number of moles of I2 / Volume of vessel
= 0.50 mol / 2.5 L
= 0.20 M
Let's assume the equilibrium concentrations of iodine and iodide are x M and 2x M, respectively, where x is the change in concentration.
Substituting these values into the equilibrium constant expression, we have:
0.209 = (2x)^2 / x
Simplifying, we get:
0.209 = 4x^2 / x
0.209 = 4x
Solving for x, we find:
x = 0.209 / 4
x = 0.05225 M
Therefore, the concentration of iodine at equilibrium is:
[I(g)] = x = 0.05225 M
And the concentration of iodide at equilibrium is:
[I2(g)] = 2x = 2 * 0.05225 M = 0.1045 M