Question

21.Be sure to answer all parts. The dissociation of molecular iodine into iodine atoms is represented as I2(g) ⇌ 2I(g) At 1000 K, the equilibrium constant Kc for the reaction is 3.80 ×10−5. Suppose you start with 0.0461 mol of I2 in a 2.27−L flask at 1000 K. What are the concentrations of the gases at equilibrium? What is the equilibrium concentration of I2? What is the equilibrium concentration of I?

Answer 1

initial [I2] = number of mol / volume
= 0.0461 mol / 2.27 L
= 0.0203 M

I2 (g)   <---> 2I (g)
0.0203          0      (initial)
0.0203-x        2x      (at equilibrium)

Kc= [I]^2 / [I2]
3.80*10^-5 = (2x)^2 / (0.0203-x)
since Kc is small, x can be ignored as compared to 0.0203
Above expression thus becomes,
3.80*10^-5 = (2x)^2 / (0.0203)
(2x)^2 = 7.714*10^-7
x = 4.39*10^-4 M

[I2] = 0.0203-x = 0.0203 - 4.39*10^-4 = 0.0199 M
[I] = 2x= 2*4.39*10^-4 = 8.78*10^-4 M