For the equilibrium Br2 + Cl2 goes to 2BrCl at 400 K ,Kc = 7.0. 0.30 of Br2 and 0.30 of Cl2 are introduced into a 1.0 L- container at 400 K.a) What will be the equilibrium concentration of Br2? b) What will be the equilibrium concentration of Cl2 ? c) What will be the equilibrium concentration of BrCl?

0.30 WHAT of Br2 and 0.30 WHAT of Cl2 are introduced........ I suspect you meant moles and the following answer is based on that assumption.

0.30 moles/1 L = 0.30M for (Br2) and (Cl2) initially. Set up and ICE chart.

............Br2 + Cl2 ==> 2BrCl
initial...0.30M...0.30M.....0
change......-x....-x........2x
equil.....0.30-x..0.30-x....2x

Substitute the equilibrium values into the Kc expression and solve for x. From that you can calculate (Br2), (Cl2) and (BrCl) at equilibrium.

To determine the equilibrium concentrations of Br2, Cl2, and BrCl, we can use the equilibrium expression and the given values.

The equilibrium expression for the reaction Br2 + Cl2 ⇌ 2BrCl is as follows:
Kc = [BrCl]^2 / [Br2] x [Cl2]

Now let's solve the problem step by step:

a) To find the equilibrium concentration of Br2:

First, let's define the initial concentrations of Br2, Cl2, and BrCl:
[Br2]initial = 0.30 M
[Cl2]initial = 0.30 M
[BrCl]initial = 0 M (since no BrCl is initially present)

Using the equilibrium expression, we can set up the following equation:
Kc = [BrCl]^2 / [Br2] x [Cl2]

Since we know the value of Kc (7.0) and the initial concentrations of Br2 and Cl2, we can substitute these values into the equation to solve for [BrCl]:
7.0 = ([BrCl]eq)^2 / (0.30 M) x (0.30 M)

Simplifying the equation:
7.0 = ([BrCl]eq)^2 / 0.09

Rearranging the equation to solve for [BrCl]eq:
([BrCl]eq)^2 = 0.09 x 7.0
[BrCl]eq = √(0.09 x 7.0)
[BrCl]eq = √0.63
[BrCl]eq ≈ 0.79 M

So the equilibrium concentration of Br2 ([Br2]eq) is approximately 0.30 M.

b) To find the equilibrium concentration of Cl2:

Since the balanced equation shows that 1 mole of Cl2 reacts with 2 moles of BrCl, and assuming the reaction goes to completion, the concentration of Cl2 will decrease by twice the concentration of BrCl formed. Therefore, the equilibrium concentration of Cl2 ([Cl2]eq) is given by:
[Cl2]eq = [Cl2]initial - 2 x [BrCl]eq
[Cl2]eq = 0.30 M - 2 x 0.79 M
[Cl2]eq = 0.30 M - 1.58 M
[Cl2]eq ≈ -1.28 M

However, negative concentrations are not physically possible, so the equilibrium concentration of Cl2 should be zero (0 M) since it is entirely consumed in the reaction.

c) To find the equilibrium concentration of BrCl:

The equilibrium concentration of BrCl ([BrCl]eq) was already calculated in part a) and found to be approximately 0.79 M.

Therefore, the equilibrium concentrations are:
[Br2]eq ≈ 0.30 M
[Cl2]eq = 0 M
[BrCl]eq ≈ 0.79 M

To answer these questions, we need to use the given equilibrium constant (Kc) and apply the principles of stoichiometry.

First, let's define the initial concentrations of the reactants:

[Br2] = 0.30 M
[Cl2] = 0.30 M

a) To find the equilibrium concentration of Br2, we will use the equation for the equilibrium constant (Kc):

Kc = [BrCl]^2 / ([Br2] * [Cl2])

We know Kc is 7.0, and we want to find [Br2]. Therefore, we can rearrange the equation and solve for [Br2]:

[Br2] = [BrCl]^2 / (Kc * [Cl2])

Now, we need to find the equilibrium concentration of BrCl, so let's solve for [BrCl]:

[BrCl] = sqrt(Kc * [Br2] * [Cl2])

Substituting the values we know:
[BrCl] = sqrt(7.0 * 0.30 * 0.30)

b) To find the equilibrium concentration of Cl2, we can use the stoichiometry of the balanced equation (1:1) and the fact that the concentration of Cl2 will decrease by x. Thus, we have:

[Cl2] = 0.30 - x

c) Finally, to find the equilibrium concentration of BrCl, we can use the stoichiometry of the balanced equation (2:1) and the fact that the concentration of BrCl will increase by 2x. Hence, we have:

[BrCl] = 2x

By substituting the values we have:

[BrCl] = 2 * x
[BrCl] = 2 * sqrt(7.0 * 0.30 * 0.30)

Therefore, the equilibrium concentration of Br2, Cl2, and BrCl can be determined by substituting the appropriate values into the respective equilibrium expressions.