At 25 oC, Kc = 0.145 for the following reaction in the solvent CCl4.

2BrCl <----> Br2 + Cl2

A solution was prepared with the following initial concentrations: [BrCl] = 0.0482 M, [Br2]= 0.0307 M, and [Cl2]= 0.0277 M.

What will their equilibrium concentrations be?
How do i know when the change in concentration is +x or -x?

I tried-2BrCl<-->Br2 Cl2

.0482 .0307 .0277
-2x +x +x

and reverse too but it's wrong. Any ideas?

You must have made a math error.

........2BrCl ==> Br2 + Cl2
I.......0.0482..0.0307..0.0277

First decide which direction the reaction will move? How? Qrxn.
Q = (0.0307)(0.0277)/(0.0482) = 0.0176 and compare this with Kc = 0.145. Q < K which means the fraction is too low in the numerator and too high in the denominator and that means the rxn must go to the right.
C.........-2x.......x.......x
E......0.0482-2x....x.......x

and solve for x. and 0.0482-2x.If you still have a problem post your work and I'll find the error.

You are wrong!!!!!!

To solve this problem, you can use the equilibrium expression and the given value of Kc to determine the equilibrium concentrations.

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

Given:
Kc = 0.145
[BrCl] = 0.0482 M
[Br2] = 0.0307 M
[Cl2] = 0.0277 M

Let's assume that at equilibrium, the change in concentration for each species is -x.

So, the equilibrium concentrations can be expressed as follows:
[BrCl] = 0.0482 - x
[Br2] = 0.0307 + x
[Cl2] = 0.0277 + x

Substituting these values into the equilibrium expression gives:
Kc = ([Br2] * [Cl2]) / [BrCl]^2
0.145 = (0.0307 + x) * (0.0277 + x) / (0.0482 - x)^2

Next, you need to solve this equation to find the value of x. This can be done by rearranging the equation and solving for x using algebraic methods or by using numerical methods such as a calculator or computer software.

Once you find the value of x, you can substitute it back into the equilibrium concentrations equation to find the equilibrium concentrations of each species at 25°C.

To determine the equilibrium concentrations, we need to use the equilibrium expression and initial concentrations. In this case, the equilibrium expression is Kc = [Br2][Cl2]/[BrCl]^2.

Let's assign a variable for the change in concentration of Br2, Cl2, and BrCl. Let's assume that the change in concentration of Br2 is "x", the change in concentration of Cl2 is also "x", and the change in concentration of BrCl is "-2x" because the stoichiometric coefficient of BrCl is 2 in the balanced equation.

Initial concentrations:
[BrCl] = 0.0482 M
[Br2] = 0.0307 M
[Cl2] = 0.0277 M

At equilibrium:
[BrCl] = 0.0482 - 2x
[Br2] = 0.0307 + x
[Cl2] = 0.0277 + x

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

Kc = [Br2][Cl2]/[BrCl]^2
0.145 = (0.0307 + x)(0.0277 + x)/(0.0482 - 2x)^2

This is a quadratic equation that can be solved to find the value of x. Once you solve for x, you can plug the value back into the equilibrium concentrations expressions to determine the equilibrium concentrations for [BrCl], [Br2], and [Cl2].

Regarding the sign of the change in concentration, it depends on the direction of the reaction. In your case, the reaction is given as 2BrCl <----> Br2 + Cl2. If you assume the reaction is proceeding from left to right, then the change in concentration of Br2 and Cl2 would be positive and the change in concentration of BrCl would be negative. If you assume the opposite, where the reaction is proceeding in the reverse direction, then the signs would be reversed accordingly.

Therefore, the signs of the changes in concentration depend on the direction of the reaction and need to be determined based on the given information or experimental conditions.