a.) The equation for the reaction is: I2 + Cl2 ⇌ 2ICl
To find the equilibrium constant (Kc), you need to set up the expression for Kc using the concentrations of the reactants and products at equilibrium.
Kc = [ICl]^2 / ([I2] ∙ [Cl2])
Given the molar concentrations at equilibrium:
[I2] = 0.50 M
[Cl2] = 0.60 M
[ICl] = 5.0 M
Substituting these values into the equation for Kc:
Kc = (5.0 M)^2 / (0.50 M ∙ 0.60 M)
Kc = 25.0 M^2 / 0.30 M
Kc = 83.3 (rounded to three significant figures)
Therefore, the equilibrium constant (Kc) for this reaction is approximately 83.3.
b.) If the concentration of I2 is increased to 1.5 M, it will disrupt the equilibrium. However, you can calculate the new ratio of products to reactants by using the initial concentrations and the stoichiometry of the reaction.
The initial concentration of I2 was 0.50 M, and it has been increased to 1.5 M, which is three times the initial concentration.
According to the balanced equation: I2 + Cl2 ⇌ 2ICl
Since the stoichiometry of I2 is 1 in the balanced equation, the new concentration of Cl2 can be calculated as follows:
[Cl2]new = [Cl2]initial = 0.60 M
The new concentration of ICl can also be determined by using the stoichiometry:
[ICl]new = 2 ∙ [I2]new
[ICl]new = 2 ∙ 1.5 M
[ICl]new = 3.0 M
Therefore, the new ratio of products to reactants with the higher concentration of iodine is:
[ICl]new / ([I2]new ∙ [Cl2]new)
= 3.0 M / (1.5 M ∙ 0.60 M)
= 3.0 M / 0.9 M
= 3.33 (rounded to two decimal places)
So, the new ratio of products to reactants, with the increased concentration of I2, is approximately 3.33.