Question

Consider the following equilibrium at 970 K for the dissociation of molecular iodine into atoms of iodine. I2(g) equilibrium reaction arrow 2 I(g); Kc = 1.35 ✕ 10−3 Suppose this reaction is initiated in a 3.7 L container with 0.063 mol I2 at 970 K. Calculate the concentrations of I2 and I at equilibrium.

Answer 1

Answer:

[ I₂] = 0.015 M

[I] = 0.0044 M

Explanation:

Let's consider the following reaction.

I₂(g) ⇄ 2 I(g)

The initial concentration of I₂ is:

$\frac{0.063mol}{3.7L} =0.017M$

To find the concentrations at equilibrium we use an ICE Chart. We identify 3 stages (Initial, Change and Equilibrium) and complete each row with the concentration or change in concentration.

      I₂(g) ⇄ 2 I(g)

I     0.017       0

C      -x        +2x

E   0.017-x     2x

The equilibrium constant (Kc) is:

$Kc=1.35 \times 10^{-3} =\frac{[I]^{2} }{[I_{2}]} =\frac{(2x)^{2} }{(0.017-x)} \\4x^{2} +1.35 \times 10^{-3}x - 2.3 \times 10^{-5}$

x₁ = 0.0022 and x₂ = -0.0025. We take the positive value because it is the one with physical meaning.

[ I₂] = 0.017-x = 0.017-0.0022 = 0.015 M

[I] = 2x = 2(0.0022) = 0.0044 M