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The equilibrium constant Kc for the reaction, COCl₂(g)⇌CO(g)+Cl₂(g) is 0.08 at a certain temperature. If the equilibrium concentration of COCl₂ is 0.10M, what are the equilibrium concentration of CO and Cl₂?

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Final answer:

To find the equilibrium concentrations of CO and Cl₂ when COCl₂ dissociates with Kc = 0.08, we set up an ICE table and solve the quadratic equation resulting from the equilibrium constant expression. Both CO and Cl₂ have identical equilibrium concentrations of 0.0894 M.

Step-by-step explanation:

To calculate the equilibrium concentrations of CO and Cl₂ when the equilibrium concentration of COCl₂ is given as 0.10M, and the equilibrium constant (Kc) for the reaction COCl₂(g) ⇌ CO(g) + Cl₂(g) is 0.08 at a certain temperature, we can set up an ICE (Initial, Change, Equilibrium) table.

We're given Kc and the equilibrium concentration of COCl₂, but not the initial concentrations of CO and Cl₂. We can assume that these start at 0 since the COCl₂ is dissociating into them. At equilibrium, the concentration of COCl₂ decreases by some amount x, and the concentrations of CO and Cl₂ each increase by x. Therefore, the equilibrium concentrations of CO and Cl₂ are both x M.

The equilibrium expression for this reaction is:

Kc = [CO][Cl₂] / [COCl₂]

Substituting the known values into this expression, we get:

0.08 = (x)(x) / 0.10

Solving for x, we have:

x² = 0.08 * 0.10

x² = 0.008

x = √0.008

x = 0.0894 M

Since x is the change in concentration for CO and Cl₂, and they were both initially 0 M, the equilibrium concentrations of CO and Cl₂ will both be 0.0894 M.

Using these values, we can now confirm that the change is small enough to be neglected. Since the initial concentration was 0.10M and the change x is around 0.09, which is almost the same as the initial concentration, we can see that this change is not negligible.

That means our assumption of negligible x would lead to an inaccurate result and we need to solve the equilibrium concentrations using the quadratic formula, which confirms x = 0.0894 M as a legitimate solution.

User Andre M
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