the ice table is as follows: h 2 i 2 hi 601. the ice table is as follows: h 2 i 2 hi 602

1

The ICE table is as follows:

H2 I2 HI

M0.00623l3.50mol0.0218

M0.00414l3.50mol0.0145

M0.0224l3.50mol0.0783

2


H2 I2 HI

M0.00623l3.50mol0.0218

M0.00414l3.50mol0.0145

y- y- y2M0.0224l3.50mol0.0783

3


H2 I2 HI

0.00623 – y 0.00414 – y 0.0224 + 2y

M0.00623l3.50mol0.0218

M0.00414l3.50mol0.0145

y- y- y2M0.0224l3.50mol0.0783

4


H2 I2 HI

0.00623 – y 0.00414 – y 0.0224 + 2y

M0.00623l3.50mol0.0218

M0.00414l3.50mol0.0145

y- y- y2M0.0224l3.50mol0.0783

]][I[H[HI]K

22

2

c

5

Now think about any possible simplification.y)-y)(0.00414-(0.00623

y)2(0.0224K2

c

6

Now think about any possible simplification.There is none!

y)-y)(0.00414-(0.00623y)2(0.0224K

2

c

7

Math Aside: The quadratic equation

8


A quadratic equations looks like:

9



0 c x b xa 2

10



The solutions (there are two of them) take the form

0 c x b xa 2

11




0 c x b xa 2

a 24acb b- x

2

12




Note that in an equilibrium problem, only one of

the solutions will generate a physically acceptable solution to the problem.

0 c x b xa 2

a 24acb b- x

2

13

We now need to solve

y)-y)(0.00414-(0.00623y)2(0.022454.3

2

14


2y)2(0.0224y)-y)(0.00414-(0.0062354.3 y)-y)(0.00414-(0.00623

y)2(0.022454.32

15


2y)2(0.0224y)-y)(0.00414-(0.0062354.3

)y4 y-8.96x10 -10 x (5.02 224 )y y-1.037x10 - -(2.58x1054.3 225

y)-y)(0.00414-(0.00623y)2(0.022454.3

2

16


2y)2(0.0224y)-y)(0.00414-(0.0062354.3

)y4 y-8.96x10 -10 x (5.02 224 0-8.99x10 y0.653y50.3 42

)y y-1.037x10 - -(2.58x1054.3 225

y)-y)(0.00414-(0.00623y)2(0.022454.3

2

17


Now solve the quadratic equation using

with a = 50.3, b = -0.653, and c = 8.99x10-4.

2y)2(0.0224y)-y)(0.00414-(0.0062354.3

)y4 y-8.96x10 -10 x (5.02 224 0-8.99x10 y0.653y50.3 42

)y y-1.037x10 - -(2.58x1054.3 225

y)-y)(0.00414-(0.00623y)2(0.022454.3

2

a 24acb b- y

2

18

(50.3) 2)-99x104(50.3)(8.0.653 0.653 y

42

19

y = 0.0114 M or y = 0.00157 M.

(50.3) 2)-99x104(50.3)(8.0.653 0.653 y

42

20

y = 0.0114 M or y = 0.00157 M.

The first answer is physically impossible.

(50.3) 2)-99x104(50.3)(8.0.653 0.653 y

42

21

y = 0.0114 M or y = 0.00157 M.

The first answer is physically impossible. Why?

(50.3) 2)-99x104(50.3)(8.0.653 0.653 y

42

22

y = 0.0114 M or y = 0.00157 M.

The first answer is physically impossible. Why? The concentration of I2 = 0.00414 – y

= -0.00726 M which is physically impossible. We cannot have a

negative concentration.

(50.3) 2)-99x104(50.3)(8.0.653 0.653 y

42

23

y = 0.0114 M or y = 0.00157 M.

The first answer is physically impossible. Why? The concentration of I2 = 0.00414 – y

= -0.00726 M which is physically impossible. We cannot have a

negative concentration. So the solution we seek is y = 0.00157 M.

(50.3) 2)-99x104(50.3)(8.0.653 0.653 y

42

24

At equilibrium:

25

At equilibrium: [H2] = 0.00623 – 0.00157 = 0.00466 M

26


[I2] = 0.00414 – 0.00157 = 0.00257 M

27


[I2] = 0.00414 – 0.00157 = 0.00257 M

[HI] = 0.0224 + 2x 0.00157 = 0.0255 M

28


[I2] = 0.00414 – 0.00157 = 0.00257 M

[HI] = 0.0224 + 2x 0.00157 = 0.0255 M

Check:

]][I[H[HI]K

22

2

c

29


[I2] = 0.00414 – 0.00157 = 0.00257 M

[HI] = 0.0224 + 2x 0.00157 = 0.0255 M

Check:

]][I[H[HI]K

22

2

c

54.30.00257)(0.00466)(

(0.0255)K2

c

30

There is a simple way to tell which way the reaction will proceed to equilibrium.

31

There is a simple way to tell which way the reaction will proceed to equilibrium. Consider

H2(g) + I2(g) 2 HI(g)

and suppose all three gases are present at the start of the reaction.

32


H2(g) + I2(g) 2 HI(g)


The reaction quotient is where the initial concentrations are employed.

]][I[H[HI]Q

22

2c

33


H2(g) + I2(g) 2 HI(g)


The reaction quotient is where the initial concentrations are employed. If Qc > Kc there must be too much HI in the mixture

for equilibrium to exist. Consequently, the reaction will proceed from right to left, until Qc becomes equal to Kc.

]][I[H[HI]Q

22

2c

34


H2(g) + I2(g) 2 HI(g)


The reaction quotient is where the initial concentrations are employed. If Qc > Kc there must be too much HI in the mixture

for equilibrium to exist. Consequently, the reaction will proceed from right to left, until Qc becomes equal to Kc.

If Qc < Kc, the reaction will proceed left to right.

]][I[H[HI]Q

22

2c

35

In many equilibrium problems, the equilibrium constant is very small. This allows a simplification of the mathematics, so that the general quadratic equation does not have to be solved.

36

In many equilibrium problems, the equilibrium constant is very small. This allows a simplification of the mathematics, so that the general quadratic equation does not have to be solved.

Example 3: 1.00 moles of N2 is placed in a 1.00 liter vessel along with 1.10 moles of O2 at 25 oC. Calculate the amount of NO formed at 25 oC. Kc = 4.8 x 10-31 at 25 oC for the reaction

N2(g) + O2(g) 2 NO(g).

37


N2 O2 NO

38


N2 O2 NO

M1.00l1.00

mol1.00 M0

M1.10l1.00

mol1.10

39


N2 O2 NO

M1.00l1.00

mol1.00

y- y- y2

M0M1.10l1.00

mol1.10

40


N2 O2 NO

1.00 – y 1.10 – y 2y

M1.00l1.00

mol1.00

y- y- y2

M0M1.10l1.00

mol1.10

41


N2 O2 NO

1.00 – y 1.10 – y 2y

M1.00l1.00

mol1.00

y- y- y2

M0

]][O[N[NO]K

22

2c

M1.10l1.00

mol1.10

42

Now

y)-y)(1.10-(1.00y)(2 2

]][O[N[NO]K

22

2c

43

Now

At this point, always stop and look for any possible simplifications.

y)-y)(1.10-(1.00y)(2 2

]][O[N[NO]K

22

2c

44

Now

At this point, always stop and look for any possible simplifications. Notice that we do not have a perfect square on the right-hand side.

y)-y)(1.10-(1.00y)(2 2

]][O[N[NO]K

22

2c

45

Now

At this point, always stop and look for any possible simplifications. Notice that we do not have a perfect square on the right-hand side. In this case there is a mathematical simplification on the right-hand side of the equation –

y)-y)(1.10-(1.00y)(2 2

]][O[N[NO]K

22

2c

46

Now

At this point, always stop and look for any possible simplifications. Notice that we do not have a perfect square on the right-hand side. In this case there is a mathematical simplification on the right-hand side of the equation – what is it?

y)-y)(1.10-(1.00y)(2 2

]][O[N[NO]K

22

2c

47

Now

At this point, always stop and look for any possible simplifications. Notice that we do not have a perfect square on the right-hand side. In this case there is a mathematical simplification on the right-hand side of the equation – what is it?

Notice that Kc is very small, which means very little product will be formed.

y)-y)(1.10-(1.00y)(2 2

]][O[N[NO]K

22

2c

48

Since y is very small, we try the approximations:

1.00 – y 1.00 1.10 – y 1.10

49


1.00 – y 1.00 1.10 – y 1.10 We will check momentarily the quality of these

approximations. The equilibrium constant expression simplifies to

50




Kc

))(1.10(1.00y4

2

51




Kc

))(1.10(1.00y4

2

31c

2 5.3x10K0)(1.00)(1.14y

52




Kc

So y = 3.6 x 10-16 M.

))(1.10(1.00y4

2

31c

2 5.3x10K0)(1.00)(1.14y

53

Since y = 3.6 x 10-16 M the approximations

1.00 – y 1.00 1.10 – y 1.10 are justified. At equilibrium: [NO] = 2 y = 7.2 x 10-16 M

54

When can approximations be made to simplify equilibrium constant expressions?

55


The only places where approximations can be made are (where A and B are concentrations):

56



(A – y) A where A >> y

57



(A – y) A where A >> yand (B + y) B where B >> y

58



(A – y) A where A >> yand (B + y) B where B >> y

Never in places like y (A – y) or

yy) - (A

59

Le Châtelier’s Principle

60

Le Châtelier’s Principle We will examine the effect of the following on the

equilibrium constant.