Question

If you take the following chemical equation:

Br_2(g)   --->   2Br_(g)

The K_c is 8.08 x 10 ^-4 for the temperature of 1756K. If the initial concentration of Br_2(g) was 3.2M, what are the equilibrium concentrations? If after you get to equilibrium you add an extra 1.3M of Br₂, what is the Q value and what would be the new equilibrium concentrations?

Answer 1

Br2 =============== 2Br+   

inicially 3,2M 0

equilibrium 3,2-X X

Kc= [Br+]² / [Br2] = X² / (3,2-X)

We can assume that 3,2-X = 3,2, so we can simplify the equation

Kc=  X² / 3,2 -----------X= (Kc*3,2)^1/2 = 0,051

So in the equilibrium we have that

[Br+] = 0,051M

[Br2] = 3,2-0,051= 3,15M, this certify our assumption that 3,2-X = 3,2.

Now when we add [Br2] we can calculate Q using inicial concentrations, Q that represent a constant when the reaction hasn't got to the equilibrium,

Br2 =============== 2Br+   

inicially 3,2M +1,3M 0,051

equilibrium 3,2 +1,3 -X 0,051+ X

Q=  [Br+]² / [Br2] = 0,051² / (3,2+1,3)

Q=  0,051² / (3,2+1,3) = 5,84*10-4

Because Q its minor that K, the equilibrium will be desplaced to the producto to get to the equilibrium again.

Br2 =============== 2Br+   

New Equilibrium 3,2 +1,3 -X ---------------- 0,051+ X

Kc = (0,051+ X)² /4,45 -X

We assume that 4,45 -X = 4,45, so

Kc= (0,051+ X)² /4,45 ------------- X = (Kc*4,45)^1/2 -0,051 ----- X= 0,009M

[Br+] = 0,051M + 0,009M = 0,060M

[Br2] = 4,45 --0,009= 4,44M