Iodine and iodide both start at 1.000 x 10^-3 M and iodine settles at 6.61 x 10^-4 M as I3- forms. What is Kc?

1. Set the change in iodine as minus x, so iodide also falls by x and triiodide rises by x. 2. Solve for x from the iodine row: 1.000 x 10^-3 minus x = 6.61 x 10^-4. 3. That gives x = 3.39 x 10^-4 M, which is the triiodide concentration at equilibrium. 4. Iodide fell by the same x, so it also sits at 6.61 x 10^-4 M. 5. Write the expression: Kc = [I3-] divided by ([I2] x [I-]). 6. Denominator: 6.61 x 10^-4 x 6.61 x 10^-4 = 4.37 x 10^-7. 7. Divide: 3.39 x 10^-4 / 4.37 x 10^-7 = 776. 8. The value 0.339 is x alone, and 1.29 comes from dividing x by one iodide term only. _Source: OpenStax Chemistry (CC BY 4.0), Ch 13 "Fundamental Equilibrium Concepts", section 13.4 Equilibrium Calculations_