Given the concentrations, calculate the equilibrium constant for this reaction:
I2(g) + Cl2(g)---> 2ICl(g)
At equilibrium, the molar concentrations for reactants and products are found to be I2 = 0.50M, Cl2 = 0.60M, and ICl = 5.0M. What is the equilibrium constant (Kc) for this reaction?
this question is also based on the reaction above. The concentration of I2(g) 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. Pls. do help it's a Homework given by the teacher before discussing about the Kinetics and equilibrium. thanks
........I2(g) + Cl2(g)---> 2ICl(g)
E.......0.50....0.60.........5.0
Kc = (ICl)^2/(I2)(Cl2)
Substitute and solve for Kc which I estimate to be 83 but you need to do it more accurately.
I'm a little confused by part 2. Apparently you are NOT asked to calculate the new equilibrium concns although that could be done rather easily. So if we increase I2 to 1.5 (not + 1.5), the ratio will be (5.0)^2/(1.5)(0.60) = about 28. I suspect your instructor wants you to see how Kc < or > Qc and which way the equilibrium will shift.
To calculate the equilibrium constant (Kc), you need to use the molar concentrations of the reactants and products at equilibrium. The general equation for the equilibrium constant is:
Kc = [products]^(coefficients of products) / [reactants]^(coefficients of reactants)
In this case, the molar concentrations of the reactants and products at equilibrium are:
[I2] = 0.50 M
[Cl2] = 0.60 M
[ICl] = 5.0 M
Since the coefficient for I2 and Cl2 is 1, and the coefficient for ICl is 2, we can substitute the values into the equation:
Kc = ([ICl]^2) / ([I2] * [Cl2])
Plugging in the values:
Kc = (5.0 M)^2 / (0.50 M * 0.60 M)
Kc = 25 M^2 / 0.30 M^2
Kc = 83.33
Therefore, the equilibrium constant (Kc) for this reaction is approximately 83.33.
To calculate the equilibrium constant (Kc) for this reaction, you need to use the concentrations of the reactants and products at equilibrium. The equilibrium constant expression for this reaction can be written as:
Kc = [ICl]^2 / ([I2] * [Cl2])
Given concentrations at equilibrium:
[I2] = 0.50 M
[Cl2] = 0.60 M
[ICl] = 5.0 M
Substituting these values into the equilibrium constant expression:
Kc = (5.0)^2 / (0.50 * 0.60)
= 25 / 0.30
= 83.33
Therefore, the equilibrium constant (Kc) for this reaction is approximately 83.33.