Question

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 0.0454 mol of I2 in a 2.35−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

Answer 1

initial concentration I₂ (molarity) = moles / volume = (0.0454 mol / 2.35 L ) = 0.01932 M

now using ICE table at equilibrium find concentration

Equilibrium reaction: I2 (g) <----------> 2I (g)
Initial molarity : ...........(0.01932 M)..............................0
Change in molarity : ........( - x )................................. (+ 2x)
Molarity at equilib. : ........ (0.01932 - x) .....................(+ 2x)

now find concentration at equilibrium using equilibrium expression

Kc = [ I ]² / [ I₂ ]
3.80 x 10^-5 = [2x]² / [0.01932 - x ] :- { we neglect x because x << 0.01932 }

3.80x10^-5 = [2x]² / [0.01932]

0.073416 x 10^-5 = 4x²  

x² = 0.018354 x10^-5

x = 0.4284 x10^-3 M = 0.0004284 M

concentration at equilibrium

I2 = (0.01932 - x) = (0.01932 - 0.0004284) = 0.01889 M

I = (2x) = 0.0004284x2 = 0.0008568 M