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Lemur [1.5K]
3 years ago
11

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:
lutik1710 [3]3 years ago
0 0

Answer:

[I⁻] = 0.0352M

Explanation:

Based on the equilibrium:

I₃⁻(aq) ⇄ I₂(aq) + I⁻(aq)

Kc is defined as:

Kc = 0.25 = [I₂] [I⁻] / [I₃⁻]

<em>The system reaches the equilbrium when the ratio</em> [I₂] [I⁻] / [I₃⁻] <em>is equal to 0.25</em>

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

<em>X is known as reaction coordinate.</em>

<em />

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

<h3>[I⁻] = 0.0352M</h3>

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