The dissociation of molecular iodine into iodine atoms is represented as I2(g) ⇌ 2I(g) what are the concentrations?

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2018-02-14 12:19 pm

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 2.25−L flask at 1000 K. What are the concentrations of the gases at equilibrium?

回答 (1)

wanszeto

2018-02-14 4:06 pm

_________________ I₂(g) _______ ⇌ _____ 2I(g) _____ Kc = 3.80 × 10⁻⁵
Initial: _____ 0.0461/2.25 mol/L _________ 0 mol/L
Change: _________ -y mol/L __________ +2y mol/L
At eqm: __ (0.0461/2.25 - y) mol/L _______ 2y mol/L

At eqm: Kc = [I]² / [I₂]
3.80 × 10⁻⁵ = (2y)² / (0.0461/2.25 - y)
4y² + (3.80 × 10⁻⁵)y - (3.80 × 10⁻⁵ × 0.0461 / 2.25) = 0

Solve the above quadratic equation,
y = 4.36 × 10⁻⁴ or y = -4.46 × 10⁻⁴ (rejected)

At equilibrium:
[I₂] = {0.0641/2.25 - (4.36 × 10⁻⁴)} mol/L = 0.0201 mol/L
[I] = 2 × (4.36 × 10⁻⁴) M = 8.72 × 10⁻⁴ mol/L

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