[Hgrigsby]

ok here goes

For the following equilibrium, Kc=8.5x10^-3 at 150 degrees C
  2Ibr(g)<====>I2(g)+Br2(g)

a)WHat is the value of Kp for this reaction at 150 degrees C?
b) If 0.025 mol of IBr is placed in a 2.0L container, what is the molar concentration of IBr at equilibrium?

For a) I got 1, it makes me uneasy to get such an exact number.
For b) I wanted to make an ICE table to tabulate all my numbers but I think all I have is the initial concentration of IBr.  Does the Kc value given equal the Change?

Thanks

[Donaldson Tan]

regarding question 1,

Kc = [I2][Br2]/[IBr]²

assuming ideal gas behavior, [ PV = nRT => P = (n/V)RT ]
P_I2 = [I2]RT
P_Br2 = [Br2]RT
P_IBr = [IBr]RT

(substitute the above terms in the Kp equation)
Kp = P_I2*P_Br2/P_IBr² = [I2][Br2]/[IBr]² = Kc

regarding question 2,
to achieve equilibrium, X amount of I2 forms. This means the amount of Br2 formed must be X, then the amount of IBr at equilibrium is (initial concentration - X). Hope the hint helps!

[Hgrigsby]

Oh thank you thank you thank you!!!!!!!!!! I got it!

[sdekivit]

ok here goes

For the following equilibrium, Kc=8.5x10^-3 at 150 degrees C
  2Ibr(g)<====>I2(g)+Br2(g)

a)WHat is the value of Kp for this reaction at 150 degrees C?
b) If 0.025 mol of IBr is placed in a 2.0L container, what is the molar concentration of IBr at equilibrium?

For a) I got 1, it makes me uneasy to get such an exact number.
For b) I wanted to make an ICE table to tabulate all my numbers but I think all I have is the initial concentration of IBr.  Does the Kc value given equal the Change?

Thanks

Kc can be converted into Kp using the following relation:

Kp = Kc * (RT)^delta n with delta n = number of gasmolecules on the right - gas molecules on the left.

--> so: delta n = 0, so Kp = Kc

2) Kc = [I2][Br2] / [IBr]^2. Since all the molecules are in the same volume, we can replace concentration by mols.

--> no say that the amount of moles IBr in equilibrium is 1,25 - x and there is formed x mol I2 and Br2, you must simply solve the following equation for x:

Kc = x^2 / (1,25 - x)