In: Chemistry
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? M What is the equilibrium concentration of I? M |
I2(g) ⇌ 2I(g)
I 0.0461 0
C -x 2x
E 0.0461-x 2x
[I-] = 2x/2.27
[I2] = 0.0461-x/2.27
KC = [I-]^2/[I2]
3.8*10^-5 = (2x/2.27)^2/ 0.0461-x/2.27
3.8*10^-5 = 4x^2/2.27(0.0461-x)
3.8*10^-5 *(0.0461-x)*2.27 = 4x^2
x = 0.000986
[I-] = 2x/2.27 = 2*0.000986/2.27 = 0.000868M
[I2] = 0.0461-x/2.27 = 0.0461-0.000986/2.27 = 0.0451/2.27 = 0.0198M