The initial concentrations of I2 and I− in the reaction below are each 0.0401 M. If the initial concentration of I−3 is 0.0 M an

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The initial concentrations of I2 and I− in the reaction below are each 0.0401 M. If the initial concentration of I−3 is 0.0 M an

d the equilibrium constant is Kc=0.25 under certain conditions, what is the equilibrium concentration (in molarity) of I−? I−3(aq)↽−−⇀I2(aq)+I−(aq)

Chemistry

1 answer:

Anarel [2.9K] · Answer 1 · 2026-03-01T16:56:06+03:00

Solution:

[I⁻] = 0.0352M

Explanation:

Referring to the equilibrium:

I₃⁻(aq) ⇄ I₂(aq) + I⁻(aq)

Kc is defined by the equation:

Kc = 0.25 = [I₂] [I⁻] / [I₃⁻]

Equilibrium is established when the proportion [I₂] [I⁻] / [I₃⁻] matches 0.25

Initially, we have 0.0401M for both [I₂] and [I⁻]. When the reaction stabilizes, xM of both [I₂] and [I⁻] gets consumed, leading to xM of [I₃⁻]. We represent this as:

[I₃⁻] = X

[I₂] = 0.0401M - X

[I⁻] = 0.0401M - X

X is referred to as the reaction coordinate.

Substituting into Kc gives:

0.25 = [I₂] [I⁻] / [I₃⁻]

0.25 = [0.0401M - X] [0.0401M - X] / [X]

0.25X = 0.00160801 - 0.0802X + X²

Thus, we have 0 = 0.00160801 - 0.3302X + X².

Solving for X yields:

X = 0.0049M → Valid solution

X = 0.3252M → Invalid solution due to negative concentrations.

The equilibrium concentrations calculate to:

[I₃⁻] = X

[I₂] = 0.0401M - X

[I⁻] = 0.0401M - X

[I₃⁻] = 0.0049M

[I₂] = 0.0352M