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CHEM1001 Lecture 30 Worksheet1. Calculating Equilibrium Compositions

2 2

2c

CO H

K CO H O

CO + H2O CO2 + H2

Initial 6.0 8.0 0 0

Change -x -x +x +x

Equilibrium 6.0-x 8.0-x +x +x

2.0

6.0 4.0c

x xK

x x

2

2

2

2

2.0 6.0 8.0

96.0 28.0 2.0

28.0 96.0 0

0

x x x

x x

x x

ax bx c

a = 1, b = -28.0, c = 96.0

2 2

2 2

4 28.0 28.0 4 1 9624.0

2 2

4 28.01 28.0 4 1 964.0

2 2

b b acx

a

or

b b acx

a

As we cant have 24 M CO2 and 24 M H2 (or more importantly we cant have 6.0-24 = -18

M CO), the physically reasonable answer isx = 4.0

[CO2] = [H2] =x = 4.0 M [CO] = 6.0 x = 2.0 M [H2O] = 8.0 x = 4.0 M.

(CheckKc = 4.02/(2.04.0) = 2.0.)

2. Calculating Equilibrium Compositions

[N2O4]initial = 0.0240 mole/0.372 L = 0.0645 mol L-1; [NO2]initial = 0

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N2O4 2 NO2

Initial concentrations 0.0645 0

Change in conc. -x +2x

Equilibrium concentrations 0.0645 -x +2x

2 2

2 3

2 4

24.61 10

0.0645c

NO xK

N O x

2 3

2 4 3

2 3 4

4 4.61 10 0.0645

4 2.973 10 4.61 10

4 4.61 10 2.973 10 0

x x

x x

x x

23 3 4

2 3 33

23 3 4

2 3 33

4.61 10 4.61 10 4 4 2.973 104 4.61 10 4.779 108.065 10

2 2 4 8

4.61 10 4.61 10 4 4 2.973 104 4.61 10 4.779 109.217 10

2 2 4 8

b b acx

a

or

b b acx

a

As we cant have a negative concentration of NO2, the physically reasonable answer isx =

8.065 10-3. Hence [NO2] = 2x = 0.0161 M [N2O4] = 0.0645 x = 0.0564 M.

(CheckKc = 0.01612/0.0564 = 4.60 10-3)

3. Finding Approximate Equilibrium Compositions

Find the equilibrium concentrations of all species present at 2400K in the system

N2 + O2 2 NO Kc = 2.5 10-3 (at 2400K) (1)

N2 + O2 2 NO

Initial concentrations 0.040 0.030 0

Change in conc. -x -x +2x

Equilibrium concentrations 0.040-x 0.030-x +2x

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Write out the equilibrium expression in terms ofx and then write out and solve the

quadratic expression.

2 2

3

2 2

22.5 10

0.040 0.030c

NO xK

N O x x

2 3

2 6 4 3 2

2 4 6

4 2.5 10 0.040 0.030

4 3.0 10 1.75 10 2.5 10

3.9975 1.75 10 3.0 10 0

x x x

x x x

x x

24 4 6

24

24 4 62

4

1.75 10 1.75 10 4 3.9975 3.0 1048.45 10

2 2 3.9975

1.75 10 1.75 10 4 3.9975 3.0 1048.88 10

2 2 3.9975

b b acx

a

or

b b acx

a

Choose the physically plausible (positive)x, so [NO] = 2x = 1.7 10-3 M; [N2] = 0.040 x

= 0.0392 M [O2] = 0.030 x = 0.0292 M.

The small equilibrium constant favours reactants. We expectx to be small.

2 2

3

2 2

2 3

32 7

7 4

2

2.5 100.040 0.030

4 2.5 10 0.040 0.030

2.5 10 0.040 0.0307.5 10

4

7.5 10 8.7 10

c

NO x

K N O

x

x

x

[NO] = 2x = 1.7 10-3 M; [N2] = 0.040 M [O2] = 0.030 M.

The two results are almost identical for [NO]. As x is so small, the initial concentrations ofN2 and O2 are close to their equilibrium concentrations.

The critical approximation is that little product is formed compared to the amount of

reactant present. In other words, 0.040-x 0.040, and 0.030-x 0.030.

This would not work for K > 1 as the initial and equilibrium concentrations of reactants

would be too different.