a. At a certain temperature, Kc is 4.13 x 10 -2 for the equilibrium:2 IBr (g) I2 (g) + Br2 (g)Assume that equilibrium is established at the above temperature by adding only IBr (g) to the reactionflask. What are the concentrations of I2 (g) and Br2 (g) in equilibrium with 0.0124 moles/liter of IBr(g) ?

To solve this problem, we can use the equilibrium expression and the given value of Kc to determine the concentrations of I2 (g) and Br2 (g) in equilibrium with 0.0124 moles/liter of IBr (g).

The equilibrium expression for the given reaction is: Kc = [I2] * [Br2] / [IBr]^2

We are given the value of Kc as 4.13 x 10^-2 and the initial concentration of IBr as 0.0124 moles/liter.

Let's assume the equilibrium concentrations of I2 and Br2 as x moles/liter. Since the stoichiometric coefficient of IBr is 2, the concentration of IBr in equilibrium will be (0.0124 - 2x) moles/liter.

Now, we can substitute these values into the equilibrium expression and solve for x:

4.13 x 10^-2 = (x) * (x) / (0.0124 - 2x)^2

Simplifying the equation, we get:

4.13 x 10^-2 = x^2 / (0.0124 - 2x)^2

Cross-multiplying, we have:

4.13 x 10^-2 * (0.0124 - 2x)^2 = x^2

Expanding and rearranging the equation, we get:

0.0539 - 0.0826x + 0.004x^2 = x^2

0.004x^2 + 0.0826x - 0.0539 = 0

Now, we can solve this quadratic equation using the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / 2a

Plugging in the values, we get:

x = (-0.0826 ± √(0.0826^2 - 4 * 0.004 * -0.0539)) / (2 * 0.004)

Simplifying further, we get:

x ≈ 0.019 moles/liter or x ≈ 0.001 moles/liter

Since the concentration of I2 and Br2 cannot be negative, we discard the negative value. Therefore, the concentration of I2 (g) in equilibrium is approximately 0.019 moles/liter and the concentration of Br2 (g) in equilibrium is approximately 0.001 moles/liter.