For the equilibrium below at 400 K, Kc = 7.0.

Br2(g) + Cl2(g) equilibrium reaction arrow 2 BrCl(g)
If 0.85 mol of Br2 and 0.85 mol Cl2 are introduced into a 2.0 L-container at 400. K, what will be the equilibrium concentrations of Br2, Cl2 and BrCl?
I tried to set it up and everytime i work it out i get the wrong answer

ohh I converted from grams to molariry instead of from mols i'll try it with the new calculations.

To find the equilibrium concentrations of Br2, Cl2, and BrCl, we can use the stoichiometry of the reaction and the given information. Here is a step-by-step approach:

Step 1: Write the balanced chemical equation for the reaction:
Br2(g) + Cl2(g) ⇌ 2 BrCl(g)

Step 2: Set up an ICE (Initial, Change, Equilibrium) table:
Br2(g) + Cl2(g) ⇌ 2 BrCl(g)
Initial: 0.85 mol 0.85 mol 0 mol
Change: -x -x +2x
Equilibrium:0.85 - x 0.85 - x 2x

Step 3: Use the given value of Kc to set up an expression for the equilibrium constant:
Kc = [BrCl]^2 / ([Br2]∙[Cl2])

Step 4: Substitute the equilibrium concentrations into the equilibrium constant expression:
7.0 = (2x)^2 / ([0.85 - x]∙[0.85 - x])

Step 5: Solve the equation for x:
7.0([0.85 - x]∙[0.85 - x]) = 4x^2

Step 6: Simplify and rearrange the equation:
7.0(0.7225 - 1.7x + x^2) = 4x^2
5.0575 - 11.9x + 7x^2 = 4x^2
3x^2 - 11.9x + 5.0575 = 0

Step 7: Solve the quadratic equation for x:
Using the quadratic formula, x = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 3, b = -11.9, and c = 5.0575

Step 8: Calculate the equilibrium concentrations of Br2, Cl2, and BrCl:
- Substitute the value of x into the expressions for equilibrium concentrations:
[Br2] = 0.85 - x
[Cl2] = 0.85 - x
[BrCl] = 2x

Step 9: Calculate the equilibrium concentrations:
Substitute the value of x obtained in step 7 into the expressions for equilibrium concentrations:
[Br2] = 0.85 - x
[Cl2] = 0.85 - x
[BrCl] = 2x

Step 10: Calculate the equilibrium concentrations:
[Br2] = 0.85 - x = 0.85 - (value of x)
[Cl2] = 0.85 - x = 0.85 - (value of x)
[BrCl] = 2x = 2 ∙ (value of x)

These values will give the equilibrium concentrations of Br2, Cl2, and BrCl at 400 K.

To solve this problem, we can use the principles of equilibrium and the given equilibrium constant (Kc) value. Let's break it down step by step.

Step 1: Write down the balanced equation for the reaction:
Br2(g) + Cl2(g) ⇌ 2 BrCl(g)

Step 2: Assign variables to the initial and equilibrium concentrations:
Let [Br2], [Cl2], and [BrCl] represent the equilibrium concentrations of Br2, Cl2, and BrCl, respectively, in moles per liter (M).

Step 3: Set up an ICE (Initial, Change, Equilibrium) table:
We start with initial concentrations of 0.85 mol for both Br2 and Cl2. Since the stoichiometry of the reaction tells us that 1 mol of Br2 reacts with 1 mol of Cl2 to form 2 mol of BrCl, the initial concentration of BrCl will be zero.

Initial Change Equilibrium
Br2 0.85 M -x 0.85 - x
Cl2 0.85 M -x 0.85 - x
BrCl 0 M +2x 2x

Step 4: Set up the expression for the equilibrium constant:
Kc = ([BrCl]²) / ([Br2] * [Cl2])
Kc = (2x)² / ((0.85 - x) * (0.85 - x))
Kc = 4x² / ((0.85 - x)²)

Step 5: Use the given equilibrium constant (Kc) value:
Kc = 7.0

Step 6: Set up the equation and solve for x:
7.0 = 4x² / ((0.85 - x)²)

To find the equilibrium concentrations, you'll need to solve this equation for x. You can either simplify it and solve using algebraic methods or use numerical methods, such as trial and error or a numerical solver. Once you find the value of x, you can substitute it back into the ICE table to get the equilibrium concentrations of Br2, Cl2, and BrCl.

Note: It's important to double-check your calculations along the way, as any small errors can lead to incorrect answers.

Why don't you show your work and let us find the error?

M = 0.85 mols/2L = 0.425M
........Br2 + Cl2 ==> 2BrCl
I.....0.425..0.425.....0
C.......-x.....-x......2x
E.....0.425-x.0.425-x..2x

Go from there.