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:
[tex]\frac{0.063mol}{3.7L} =0.017M[/tex]
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:
[tex]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}[/tex]
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
