Keq Probs

CHEMICAL EQUILIBRIUM

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Calculation of the Equilibrium Constant 1.00 mol SO2 and 1.00 mol O2 are confined at 1000 K in a 1.00-liter container. At equilibrium, 0.925 mol SO3 has been formed. Calculate Kc for the reaction 2SO2(g) + O2(g)  2SO3(g) at 1000 K. Solution: 2SO2(g) + O2(g)  2SO3(g) Initial moles 1.00 1.00 0 Change - 2y -y + 2y Moles at equil. 1.00 – 2y 1.00 – y 0 + 2y = 0.925 mol 

n SO3 = 2y = 0.925 mol n SO2 = 1.00 – 2y, 1.00 – 0.925 = 0.075 mol n O2 = 1.00 – y, 1.00 – 0.46 = 0.54 mol 2

2SO2(g) + O2(g)  2SO3(g) K

= [SO3]2 　 / 　 [SO2]2 [O2]

K

= [0.925]2 　 / 　 [0.075]2 [0.54]

K

= 282

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２ ) 3.00 moles of pure SO3 are introduced into an 8.00-liter container at 1105 K. At equilibrium 0.58 mole of O2 has been formed. Calculate Kc for the reaction: 2 SO3(g)  2 SO2(g) + O2(g) at 1105 K.  Answer:

2 SO3(g)  2 SO2(g) + O2(g) Initial: 3.00 Change: - 2x Equilibrium: 3.00-2x

0 + 2x 0+2x

0 + 1x 0+1x= 0.58 mole 4

n O2 = 0 + x = 0.58 mole n SO3 = 3 – 2x, 3 – 1.16 = 1.84 mole n SO2 = 0 + 2x = 1.16 mole [O2] = 0.58 　 / 　 8 = 0.0725 mole/liter [SO3] = 1.84 　 / 　 8 = 0.23 M [SO2] = 1.16 　 / 　 8 = 0.145 M K = [SO2]2 [O2] 　 / 　 [SO3]2 K = [0.145]2 [0.0725] 　 / 　 [0.23]2 -2

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3) A 1-liter container is filled with 0.50 mol of HI at 448oC. The value of the equilibrium constant for the reaction: H2(g) + I2(g)  2 HI(g) at this temperature is 50.5. What are the concentrations of H2, I2 and HI in the vessel at equilibrium? Answer:

Given: K = 50.5 H2(g) + I2(g)  2 HI(g)

Initial: 0 Change: x Equilibrium: 0+x

0 x 0+x

0.50 mol - 2x 0.50-2x 6

= [HI]2 　 / 　 [H2] [I2]

K

 50.5  (x x

= [0.5 – 2x]2 　 / 　 [x] [x]

= - 0.0979 mol)

or

(x = 0.0549 mol)

= 0.0549 mol H2(g) + I2(g)  2 HI(g)

Equilibrium: 0+x  [H2]

0+x

and [I2] = 0.0549 M

0.50-2x [HI] = 0.39 M 7

Checking: H2(g) + I2(g)  2 HI(g) Equilibrium: 0+x 0+x 0.50-2x  [H2]

and [I2] = 0.0549 M

[HI] = 0.39 M

K

= [HI]2 　 / 　 [H2] [I2]

K

= [0.39]2 　 / 　 [0.0549] [0.0549]

K

= 50.5

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4) For the reaction: H2(g) + I2(g)  2 HI(g) ; the equilibrium constant at 420oC is 54.3. Suppose that the initial concentrations of H2 and I2 forming HI are 0.00623 M, 0.00414 M, and 0.0224 M respectively, what are the concentrations of these species at equilibrium? Answer: H2(g) + I2(g)  2 HI(g) Initial: 0.00623 0.00414 0.0224 Change: - x -x + 2x Equil.: 0.00623-x 0.00414-x 0.0224+2x 9

Solution:

54.3 = [0.0224 + 2x]2 [0.00623 – x] [0.00414 –x]

x = 0.0114 M or

x = 0.00156 M

x = 0.00156 M [H2] = 0.00467 [I2] = 0.00258 M Checking: K = [HI]2 [H2] [I2]

[HI] = 0.0255 M

= [0.0255]2 [0.00467] [0.00258] K = 53.97 K = 54

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5) At 1600 K, the equilibrium constant (Kc) for the reaction: Br2(g)  2 Br(g) is 1.04 X10-3. The amount of 0.10 mol Br2 is confined in a 1.0-liter container and heated to 1600 K. Calculate (a) the concentration of Br atoms present at equilibrium and (b) the percentage of the initial Br2 that is dissociated into atoms.

%

dissociation = quantity dissociated

X 100

original quantity 11

 Solution:

Initial: Change: Equili.:

K = 1.04 X10-3 Br2(g)  2 Br(g) 0.1 -y 0.1 – y

0 + 2y 0 + 2y

1.04 X10-3 = [2y]2 [0.1-y] y = 4.97 x10-3 M (a) 2y = 9.94 x10-3 M (b) % dissociation = 4.97 x10-3 M

X 100

0.10 M = 4.97% or 5%

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