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PIT_PIT [208]
3 years ago
6

The equilibrium constant, Kc, for the following reaction is 77.5 at 600 K. CO(g) Cl2(g) COCl2(g) Calculate the equilibrium conce

ntrations of reactant and products when 0.555 moles of CO and 0.555 moles of Cl2 are introduced into a 1.00 L vessel at 600 K.
Chemistry
1 answer:
Margarita [4]3 years ago
6 0

Answer:

[CO] = 0.078M

[Cl2] = 0.078M

[COCl2] = 0.477M

Explanation:

Based on the reaction:

CO(g) Cl2(g) ⇄ COCl2(g)

<em>Where equilibrium constant, kc, is:</em>

kc = 77.5 = [COCl2] / [CO] [Cl2]

[] represents the equilibrium concentration of each gas. The initial concentration of each gas is:

[CO] = 0.555mol/1.00L = 0.555M

[Cl2] = 0.555M

And equilibrium concentrations are:

[CO] = 0.555M - x

[Cl2] = 0.555M - x

[COCl2] = x

<em>Where x is reaction coordinate</em>

Replacing in kc expression:

77.5 = [x] / [0.555M - x] [0.555M - x]

77.5 = x / 0.308025 - 1.11 x + x²

23.8719 - 86.025 x + 77.5 x² = x

23.8719 - 87.025 x + 77.5 x² = 0

x = 0.477M. Right answer

x = 0.646M. False answer. Produce negative concentrations

Replacing:

<h3>[CO] = 0.555M - 0.477M = 0.078M</h3><h3>[Cl2] = 0.078M</h3><h3>[COCl2] = 0.477M</h3>

And those concentrations are the equilibrium concentrations

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