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 and 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)

In the beginning, you add 0.0401M of both [I₂] [I⁻]. When the reaction reach the equilibrium, xM of both [I₂] [I⁻] is consumed producing xM of [I₃⁻]. That is written as:

[I₃⁻] = X

[I₂] = 0.0401M - X

[I⁻] = 0.0401M - X

X is known as reaction coordinate.

Replacing in Kc:

0.25 = [I₂] [I⁻] / [I₃⁻]

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

0.25X = 0.00160801 - 0.0802X + X²

0 = 0.00160801 - 0.3302X + X²

Solving for X:

X = 0.0049M → Right solution

X = 0.3252M → False solution. Produce negative concentrations

Replacing, equilibrium concentrations will be:

[I₃⁻] = X

[I₂] = 0.0401M - X

[I⁻] = 0.0401M - X

[I₃⁻] = 0.0049M

[I₂] = 0.0352M

[I⁻] = 0.0352M

andromache andromache · Answer 2 · 2022-04-07T14:25:51+02:00

The equilibrium concentration (in molarity) of [I⁻] should be considered as the 0.0352M.

Calculation of the equilibrium concentration:

Since

I₃⁻(aq) ⇄ I₂(aq) + I⁻(aq)

Here Kc should be defined

Kc = 0.25 = [I₂] [I⁻] / [I₃⁻]

Also, The system finished the equilibrium at the time when the ratio [I₂] [I⁻] / [I₃⁻] is equivalent to 0.25.

Also,

[I₃⁻] = X

[I₂] = 0.0401M - X

[I⁻] = 0.0401M - X

Also,

0.25 = [I₂] [I⁻] / [I₃⁻]

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

0.25X = 0.00160801 - 0.0802X + X²

0 = 0.00160801 - 0.3302X + X²

Now

X = 0.0049M → Right solution

X = 0.3252M → False solution

Now equilibrium concentrations will be:

[I₃⁻] = X

[I₂] = 0.0401M - X

[I⁻] = 0.0401M - X

[I₃⁻] = 0.0049M

[I₂] = 0.0352M

[I⁻] = 0.0352M

Hence, The equilibrium concentration (in molarity) of [I⁻] should be considered as the 0.0352M.

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