Chapter 17 Section 1: When has a chemical reaction reached

advertisement

Chapter 17

Section 1:

1.

When has a chemical reaction reached equilibrium?

2.

What is a reversible reaction?

3.

Which direction is a chemical reaction going in when equilibrium is reached?

4.

Are the concentrations of reactants and products equal at equilibrium? Explain.

5.

What is the equilibrium constant expression?

6.

What is the equilibrium constant, K eq

?

7.

Describe the difference between a homogeneous equilibrium and a heterogeneous equilibrium.

8.

How is a homogeneous equilibrium expressed?

Practice problem: Write the equilibrium expression for the following reaction:

N

2

O

4

(g) 2NO

2

(g)

9.

How is a heterogeneous equilibrium expressed?

Practice problem: Write the equilibrium expression for the following reaction:

C

10

H

8

(s) C

10

H

8

(g)

10.

What does the data in Table 17.1 demonstrate about equilibrium constants (K eq

)?

11.

What does a large equilibrium constant tell you about the concentration of products and reactants at equilibrium? Small equilibrium constant?

12.

What are the three characteristics of all chemical reactions that reach equilibrium?

Complete practice problems:

#1b-e on p. 601

#3b-e on p. 603

#5,6,7 on p. 605

Section 2

13.

What is Le Chatelier’s Principle?

14.

How is O

2

circulation in the body an example of Le Chatelier’s Principle?

15.

List the five factors that influence equilibrium and describe their affect. a.

________________________________ : b.

________________________________: c.

_________________________________:

d.

______________________________: e.

_______________________________:

Section 3

Problem: The equilibrium constant for the reaction: 2H

2

S 2H

2

(g) + S

2

(g) is 2.27 X 10 -3 at

1405K. What is the concentration of H

2

(g) if [S

2

]=0.540 mol/L and [H

2

S] = 0.184 mol/L?

K eq

= [H

2

] 2 [S

2

] = 2.27 X 10 -3 = [H

2

] 2 (0.540)

[H

2

S] 2 (0.184) 2

[H

2

] = (2.27 X 10 -3 )(0.184) 2 = 0.012M

0.540 p. 613 practice problem #18: [CH

3

OH] = 10.5

[CO] [H

2

] 2 a.

10.5 = 1.32 , [CO] = 1.32 = 0.144M

[CO](0.933) 2 10.5 (0.033) 2 b.

10.5 = 0.325 , [H

2

] = 0.325 = 0.169M

1.09[H

2

] 2 (10.5)(1.09) c.

10.5 = [CH

3

OH] , [CH

3

OH] = (10.5)(3.85)(0.0661) 2 = 0.177M

(0.0661) 2 (3.85)

Ionic substances dissociate when they dissolve. Example: NaCl(aq) Na + + Cl Table salt is highly soluble in water so this process proceeds to the right and there is virtually no reverse process.

When substances do not readily dissolve, there is a reverse process. One example is barium sulfate,

BaSO

4

. Barium ions, Ba 2+ , are toxic to humans yet barium sulfate solution is safely ingested during X ray procedures involving the digestive tract. Explain this. (p. 614-615)

16.

How is a product solubility constant expressed?

17.

What the K sp

value tell us about a substance?

Example problem: What is the solubility of AgI in mol/L at 298K?

In water, AgI(s) Ag + (aq) + I (aq).

K sp

= [Ag + ][I ] = 8.5 X 10 -17 at 298K

According to the chemical equation, one mole each of Ag + and I + are formed when one mole of AgI

dissolves. So we can substitute x for both Ag + and I :

(x)(x) = 8.5 X 10 -17 , x = 9.2 X 10 -9 mol/L. This is the solubility.

18.

Calculate the solubility for the following ionic compounds at 298K: a.

PbCrO

4

Pb 2+ + CrO

4

2-

K sp

= 2.3 X 10 -13

K sp

= [Pb 2+ ][CrO

4

2] = X 2 = 2.3 X10 -13 , x = 2.3 X 10 -13 = 4.8 X 10 -7 M b.

AgCl Ag +1 + Cl -1

K sp

= 1.8 X 10 -10

K sp

= [Ag +1 ][Cl -1 ] = x 2 = 1.8 X 10 -10 = 1.3 X 10 -5 M c.

CaCO

3

Ca 2+ + CO

3

2-

K sp

= 3.4 X 10 -9

K sp

= [Ca 2+ ][CO

3

2] = x 2 = 3.4 X 10 -9 = 5.8 X 10 -5 M d.

CaF

2

Ca 2+ + 2 F , if x = [Ca 2+ ] then 2 x = [F ] because 2 mol of F are produced for every 1 mol of Ca 2+

K sp

= 3.5 X 10 -11

K sp

= [Ca 2+ ][F ] 2 = (x)( 2 x) 2 = 4x 3 = 3.5 X 10 -11

[Ca 2+ ] = x = 3 3.5 X 10 -11 = 2 X 10 -4

4

[F ] = 2 x = 2 (2 X 10 -4 ) = 4 X 10 -4 e.

Ag

2

CrO

4

2 Ag + + CrO

4

-2 , if x = [CrO

4

2] , then 2 x = [Ag + ]

K sp

= 1.1 X 10 -12

K sp

= [Ag + ] 2 [CrO

4

2] = ( 2 x) 2 x = 4x 3 = 1.1 X 10 -12 x = 3 1.1 X 10 -12

4

[CrO

4

2] = x = 6.5 X 10 -5 M

[Ag + ] = 2 x = (2)(6.5 X 10 -5 ) = 1.3 X 10 -4 M

19.

What is a common ion? The common ion effect?

20.

How would adding K

2

CrO

4

to a solution of PbCrO

4

affect the solubility of PbCrO

4

? (hint: use the common ion effect to answer this)

21.

Would adding sodium sulfate, Na

2

SO

4

, to the BaSO

4

that is ingested to X-ray the intestine make the BaSO

4

safer to ingest? Explain.

Download