A solution initially contains 0.1 M of I2 and 0.1 M of I-. The following reaction occurs and the equilibrium concentration for I2 is 0.01 M. What is the Kc for the reaction?
I2 (aq) + I- (aq) ---> I3- (aq)
Given : initial concentration of I2 = 0.1 M
initial concentration of I- = 0.1 M
equilibrium concentration of I2 = 0.01 M
| ICE Table | I2 | I- | I3- |
| Initial | 0.1 M | 0.1 M | 0 |
| Change | -x | -x | +x |
| Equilibrium | 0.1 M - x | 0.1 M - x | +x |
equilibrium concentration of I2 = 0.1 M - x = 0.01 M
x = 0.1 M - 0.01 M
x = 0.09 M
equilibrium concentration of I- = 0.1 M - x
equilibrium concentration of I- = 0.1 M - 0.09 M
equilibrium concentration of I- = 0.01 M
equilibrium concentration of I3- = x
equilibrium concentration of I3- = 0.09 M
concentration equilibrium constant, Kc = [I3-]eq / [I2]eq[I-]eq
Kc = (0.09 M) / (0.01 M) * (0.01 M)
Kc = 900
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