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?
2 answers
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.
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.
2 answers