# Chem Question--best answer will be awarded?

For the equilibrium

Br2(g)+Cl2(g)⇌2BrCl(g)

at 400 K, Kc = 7.0.

Part A: If 0.25 mol of Br2 and 0.55 mol of Cl2 are introduced into a 3.0-L container at 400 K, what will be the equilibrium concentration of Br2?

Part B: If 0.25 mol of Br2 and 0.55 mol of Cl2 are introduced into a 3.0-L container at 400 K, what will be the equilibrium concentration of Cl2 ?

Part C: If 0.25 mol of Br2 and 0.55 mol of Cl2 are introduced into a 3.0-L container at 400 K, what will be the equilibrium concentration of BrCl ?

### 2 Answers

- 冷眼旁觀Lv 61 month agoFavourite answer
A.

_________________ Br₂(g) __ + __ Cl₂(g) __ ⇌ __ 2BrCl(g) ___ Kc = 7

Initial (mol/L): ____ 0.25/3 ______ 0.55/3 ________ 0

Change (mol/L):_____ -y ________ -y __________ +2y

Eqm (mol/L): ___ (0.25/3) - y ___ (0.55/3 - y) ______ 2y

At equilibrium:

Kc = [BrCl]² / ([Br₂] [Cl₂])

7 {(0.25/3 - y)(0.55/3) - y} = (2y)²

0.1069 - 1.867y + 7y² = 4y²

3y² - 1.867y + 0.1069 = 0

y = [1.867 - √(1.867² - 4*3*0.1069)] / (2*3)

y = 0.0638

[Br₂] at equilibrium = {(0.25/3) - 0.0638} M = 0.0195 M

====

B.

According to Part A,

[Cl₂] at equilibrium = {(0.55/3) - 0.0638} M = 0.120 M

====

C.

According to Part A,

[BrCl] at equilibrium = 0.0638 × 2 M = 0.128 M

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- hcbiochemLv 71 month ago
There have been several similar questions asked recently. The process is the same.

Write the expression for Kc:

Kc = [BrCl]^2 / [Br2][Cl2] = 7.0

Set up an ICE table:

........................[Br2]................[Cl2].........[BrCl]

Initial.........0.25/3 = 0.0833.. 0.183..........0

Change..........-x......................-x.............+2x

Equilibrium 0.0833 -x.......0.183 -x.........2x

Kc = 7.0 = (2x)^2 / (0.0833-x)(0.183-x)

Here, again, you'll need to simplify this into a quadratic of the form ax^2+bx+c=0 and use the quadratic formula to solve for x. Once you know x, you can calculate the molar concentrations of both reactants and the product.

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