b.) The concentration of I2 is increased to 1.5 M, disrupting equilibrium. Calculate the new ratio of products to reactants with this higher concentration of iodine. Assume that the reaction has not yet regained equilibrium.
the reaction is I2 + Cl2 = 2ICl pls help
DrBob222
answered
Part a is just a matter of plugging into the Kc expression and solve for Kc.
Part b appears to me nothing more than adding 1.5 to the I2 equilibrium amount and recalculating Kc (which would now be called Q).
Step-by-Step Bot
answered
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.
Explain Bot
answered
The equilibrium constant expression for this reaction is:
Kc = [ICl]^2 / ([I2] * [Cl2])
a.) Using the given molar concentrations at equilibrium:
[I2] = 0.50 M
[Cl2] = 0.60 M
[ICl] = 5.0 M
Plugging these values into the equilibrium constant expression, we have:
Kc = (5.0 M)^2 / (0.50 M * 0.60 M)
Kc = 25.0 M^2 / 0.30 M^2
Kc = 83.33 M^-1
Therefore, the equilibrium constant (Kc) for this reaction is 83.33 M^-1.
b.) If the concentration of I2 is increased to 1.5 M, we can calculate the new ratio of products to reactants using the new concentration.
Let's call the new concentration of I2 [I2]' = 1.5 M.
Assuming that the reaction has not yet regained equilibrium, we can use the reaction stoichiometry to calculate the new concentrations of ICl and Cl2.
From the balanced equation: I2 + Cl2 ⇌ 2ICl
Since the reaction is not at equilibrium, we don't know the exact concentrations of the new reactants and products. However, let's assume that the reaction has proceeded to some extent, with the formation of a small amount of ICl.
Let's assume x M of ICl has formed. This means that the concentrations of I2 and Cl2 have decreased by x M.
[I2]' = [I2] - x = 1.5 M - x
[Cl2]' = [Cl2] - x = 0.60 M - x
[ICl]' = [ICl] + 2x = 5.0 M + 2x
Now we can write the new equilibrium constant expression:
Kc' = [ICl]'^2 / ([I2]' * [Cl2]')
Plugging in the new concentrations, we have:
Kc' = ([ICl] + 2x)^2 / ([I2] - x * [Cl2] - x)
To determine the new ratio of products to reactants, we need to find the value of x. This can be done by using the initial equilibrium constant (Kc) and rearranging the equation with the given concentrations:
Kc = [ICl]^2 / ([I2] * [Cl2])
Simplifying this expression, we have:
83.33 M^-1 = (5.0 M)^2 / (0.50 M * 0.60 M)
Now solve for x using this equation:
Kc' = ([ICl] + 2x)^2 / ([I2]' * [Cl2]')
83.33 M^-1 = (5.0 M + 2x)^2 / ((1.5 M - x) * (0.60 M - x))
Solving this equation will give you the new ratio of products to reactants with the increased concentration of iodine.
