Worksheet Of Equilibrium 5kd5n

WORKSHEET 1ST

EQUILIBRIUM Drs. HARIYANTO Public Senior High School 3 Malang Jl.Sultan Agung Utara 7 Malang Phone (0341) 324768 e-mail : [email protected] blog : http://hariyantobhawikarsu.wordpress.com

ACTIVITY 1

REVERSIBLE REACTION

Based on the products of reaction can or not react to produce the ractants, chemical reaction is classified into reversible and irreversible. What is reversible reaction? Or what is irreversible reaction? Please do the following acticity! Apparatus and matterial

No

Apparatus / material

size

Quantity

1 2 3 4 5 6 7 8 9 10

Beaker glass Graduated cylinder Pipets Stirring rods Petri dish Copper(II)sulfate Aqueous Sodium iodide Aqueous Sodium sulfate Wash bottle Aquadest

50 mL 10 mL 1M 1M -

2 pieces 2 pieces 2 pieces 2 pieces 1 pieces 2 grams 20 mL 20 mL 1 50 mL

Procedure 1. Place 1 gram of PbSO4 into beaker glass. Pour 10 mL of aqueous NaI 1 M, and drop by pipet into beaker glass and stir this mixture. Observe and write all the changes 2. Decantate that solution from the beaker glass, and wash the precitipation with aquadest. 3. Move a half of precitipation to other beaker glass. Drop carefully the aqueous Na2SO4 1 M about 10 mL and stir this mixture. Observe the color changes of this precitipation and compare the colors in the both beaker glass and write your observation.

Observation Matterial

The changes ( ie :color)

PbSO4 + NaI

………………………………………………………………………………

Precitipation I + Na2SO4

………………………………………………………………………………

Questions The chemical equations of the experiment above are written in the following equations: PbI2 (s) + Na2SO4 (aq) I. PbSO4(aq) + 2 NaI (aq) PbSO4(aq) + 2 NaI (aq) II. PbI2 (s) + Na2SO4 (aq)

No 1

Questions

Answers

How is the relation between reaction I and II? ………………………………………..

2

Are the products of reaction I react to produce the reactants?

3

………………………………………..

That reaction is called the reversible reaction, so what is the reversible reaction? The opposite of reversible is irreversible, so what is irreversible reaction?

4

ACTIVITY 2

……………………………………….. ………………………………………..

THE ANALOGY OF DYNAMIC EQUILIBRIUM

Before we define the of equilibrium and dynamic equilibrium, please do the following activity.

Apparatus and matterial

No

Apparatus/matterial Size

Quantity

1 2 3

Test tube rack The test tube Pipe

-

4 5

The color liquid (KMnO4) Petri dish

-

1 pieces 2 pieces (identic) 2 pieces (different diameter) 10 mL 1 buah

-

Procedure

PIPE 1

PIPE 2 - TUBE A AND B ARE IDENTIC - DIAMETER PIPE 1 > PIPE 2

TUBE A

KMnO4

1. 2. 3.

TUBE B EMPTY

Fill the tube A 10 mL of aqueous KMnO4. Place the pipe 1 into the tube A Place the glass pipe 2 into the tube B that is no any solutions. By using the pipe 1 move the solution in the tube A to the tube B. And move the solution in tube B into tube A by using the pipe 2 until the volume of the solutions in the both tubes are constant.

Observation and the analogies. If the tube A is analogized as reactants, and the tube B is products. The volume of solution is analogized as concentration of substances, complete the following table!

No 1

Experiments The initial Volume : a. tube A is full of solution

Analogy in equilibrium The initial concentration : a. the concentration of reactants : …………………………………………………………………

b. tube B is no solutions

b. the concentration of

products : …………………………………………………………………

2

The Volume of solution when the solution is moved a. The solution in tube A moves to tube B, and in the same time the solution in tube B moves to tube B continuously.

Concentration of substances when reaction is processing a. The process of reactions i ………………………………………………………………… …………………………………………………………………

b. The Volume of solutios in tube A and tube B 1) The volume of solution in

b. The concentration of reactant and product : 1) the initial concentration of

tube A is decreasing

Reactants…………………………………

tube B is increasing.

Products ……………………………………

2) When the equilibrium is reached, the volume of solutions in he both tube are constant

2)When the equilibrium is reached, concentration of reactants and products are ………………………………………………………………… …………………………………………………………………

Questions No 1

2

3

4

Questions The changes of ractants to products (the forward reaction) or the products to reactants (the reverse reaction), a. Occurs in the same time or not? b. Occurs continuously or not? c. The rate of the forward reaction and the reverse reaction are equal or not? How is the changes of concentration of reactants or products when the equilibrium is reached? If the answers of questions no 1, 2, and 3 are the characterictic of the dynamic equilibrium, define the dynamic equilibrium, please! The equilibrium reaction symbolized by two directions ( ), please write the equilibrium equation between reactants (R) and Products (P)

Answers ……………………………………….. ……………………………………….. ………………………………………..

……………………………………….. ………………………………………..

………………………………………..

………………………………………..

CONCLUSION Based on the activity 1 and 2, please conclude the definition of reversible reaction, irreversible reaction, equilibrium, and dynamic equilibrium

EXERCISES

01. Give 2 examples the reversible reactions in daily life!

02. Give 2 examples the irreversible reactions in daily life!

03. Observe the following equation: R ↔ 2 P Explain ! the quantity of R and P when the equilibrium is reached?

04. Plot the graph of concentration of R, [R] and P, [P] versus time (t) [ ] (M)

t (s) 05. How are the values of the concentration of R and P when the equilibrium is reached?

06. Plot the graph the rate of the forward raction and the reverse reaction (v) versus time (t) v (M/s)

t (s) 07. How are the rates of the forward reaction and the reverse reaction when the equilibrium is reached?

WORKSHEET 2nd

THE EQUILIBRIUM LAW Drs. HARIYANTO Public Senior High School 3 Malang Jl.Sultan Agung Utara 7 Malang Phone (0341) 324768 e-mail : [email protected] blog : http://hariyantobhawikarsu.wordpress.com

PRE-KNOWLEDGES No 1

Questions Write 2 examples of equilibrium equations.

answers ………………………………………………………………………… …………………………………………………………………………

2

3

Mention 4 kinds of physical state of substance!

…………………………………………………………………………

Write the formula of concentration / molarity of solution.

………………………………………………………………………… …………………………………………………………………………

ACTIVITY 1

HOMOGENEOUS AND HETEROGENEOUS EQUILIBRIUM

Based on the physical state of substances that are involved in reaction, equilibrium is classified into homogeneous and heterogeneous equilibrium. What is the homogeneous equilibrium? What is the heterogeneous equilibrium? Observe and complete the following table!. No The equilibrium equations

1 2 3 4

H2(g) + I2(g) ↔ 2HI(g) Ag+ (aq) + 2NH3(aq) ↔ Ag(NH3)2 + (aq) CaCO3(s)<=>CaO (s) + CO2(g) PbI2(s) <=>Pb2+(aq)+2I-(aq)

The physical state of substance in reactions …………………. …………………. …………………. ………………….

homogeneous / heterogeneous equilibrium Homogeneous Homogeneous Heterogeneous Heterogeneous

CONCLUSION Based on the table above, conclude the definition of homogeneous and heterogeneous equilibrium.

ACTIVITY 2

EQULIBRIUM CONSTANT (KC)

What is the equilibrium constants? What is the quantitative relationship between the concentration of reactants and product at equilibrium? To define the Kc, complete the following table! 1). H2(g) + I2(g) ↔ 2HI(g) at 250C No

Concentration of spicies at equilibrium [H2]

[I2]

[HI]

A

B

C 2

[HI] [H2] [I2]

[HI] [H2] [I2]

[H2] [I2] [HI]

1

0,00421

0,00421

0,0294

…………………

…………………

…………………

2

0.0110

0,00420

0,00473

…………………

…………………

…………………

3

0,0110

0,0550

0,0173

…………………

…………………

…………………

4

0,0430

0,0620

0,3580

…………………

…………………

…………………

2). Ag+ (aq) + 2NH3(aq) ↔ Ag(NH3)2 + (aq) at 250C No

Concentration of spicies at equilibrium [Ag(NH3)2+ ]

A

B

C

[Ag(NH3)2+ ] [Ag+] [NH3]

[Ag(NH3)2+ ]2 [Ag+] [NH3]

[Ag+][NH3][Ag(NH3)2+ ]

[Ag+]

[NH3]

1

0,00100

0,00500

0,401

…………………

…………………

…………………

2

0.00100

0,00100

0,016

…………………

…………………

…………………

3

0,00200

0,00200

0,128

…………………

…………………

…………………

4

0,00200

0,00100

0,0322

…………………

…………………

…………………

QUESTIONS No

Questions

1

For each table above a. Which column that the value is constants? b. What is relation between the formula that has the constant value and the reaction?

2

If the formula that has the constants value is called equilibrium constant, write the formula of Kc of the following equation:

answers a. table 1……………………………….. table 2……………………………….. b. ………………………………………..

…………………………………………… 3

mA+nB↔pC+qD where m,n,p and q are coeffient Based the data of table 1 and 2, Please describe the definition of Kc

CONCLUSION

……………………………………………

EXERCISES Write the formula of the equilibrium of the following equations!: a. N2(g) + 3 H2(g) ↔ 2 NH3(g) b. 2SO2(g) + O2 (g) <=> 2 SO3 (g) c. Cu2+(aq) + 4 NH3(aq) <=> Cu (NH3)42+ d. N2O4(g) <=> 2 NO2(g) e. 2N2O4 (g) + O2 (g) <=>2 N2O5 (g) f. 2 HBr (g) <=> H2 (g) + Br2 (g) g. 4 NH3 (g) + 3 O2 (g) <=> 2 N2 (g) + 6 H2O (g) h. N2 (g) + O2 (g) <=> 2 NO (g)

Solution

ACTIVITY 3

THE VALUE OF THE EQUILIBRIUM CONSTANT

What are the factors that influence the value of Kc? Observe the following tables. Table 1 : The relation of Kc and temperature temperature 800K 1000K

Reactions 2 SO2(g) + O2 (g) <=> 2 SO3 (g) 2 SO2(g) + O2 (g) <=> 2 SO3 (g)

Kc 59000 256

Table 2 : The relation between Kc and the balanced reaction, if the reaction is inverted Temperature 1000K 1000K Table 3

Reaktions 2 SO2(g) + O2 (g) <=> 2 SO3 (g) 2 SO3(g) <=> 2 SO2 (g) + O2 (g)

Kc 256 1/256

: The relation between Kc and the balanced reaction, if the coefficient is multiplied by factor n

Temperature 1000K 1000K

Reaction 2 SO2(g) + O2 (g) <=> 2 SO3 (g) SO2(g) + ½ O2 (g) <=> SO3 (g)

Kc 256 16

Table 4

: determining of Kc of the total reaction some equilibrium reactions Temperature Reaction 1000K A(g) + B (g) <=> C (g) 1000K C(g) + D (g) <=> E (g) + 1000K A(g) + B (g)+ D (g) <=> E (g)

from addition of Kc A B A.B

QUESTIONS No 1

Questions Which factor that influence the value of Kc a. table 1 b. table 2 c. table 3 d. table 4

Answers table 1……………………………….. table 2……………………………….. table 3……………………………….. Table 4………………………………..

2

Constant or change the value of Kc at the same temperature?

……………………………………….. ………………………………………..

3

How is the value of Kc if the reaction is inverted?

……………………………………….. ………………………………………..

4

How is the value of Kc if the coefficient

……………………………………….. ………………………………………..

numbers are multiplied by factor n? 5

What is the value of Kc of the new equilibrium if the total of reaction is obtained from addition of the equilibrium reaction 1 and 2?

……………………………………….. ………………………………………..

CONCLUSION

EXERCISES

01. Given that the equilibrium constant, Kc of the following reaction : N2O4(g) <=> 2 NO2(g) at 1000 K is 16 Determine the value of Kc at 1000 K of the following reactions! a. 2NO2(g) <=> N2O4(g) b. 2N2O4(g) <=> 4 NO2(g) c. NO2(g) <=> ½ N2O4(g) Answer :

: A(g) + B2 (g) <=> AB2 (g) Kc = 2 2AD2(g) + 2B2(g) <=> 2AB2(g)+4D(g) Kc = 9 Determine the value of Kc at the same temperature of the following reactions: A(g) + 2D (g) <=> AD2(g)

02. Given

Answer :

ACTIVITY 4

THE HETEREGENEOUS EQUILIBRIUM CONSTANT

At the previous activities, the physical state of substances that involved in equilibrium are homogeneous. How to write the formula of Kc of the heterogeneous equilibrium?

Observe and complete the following table! No

The equilibrium equations

The formula of Kc

1

CaCO3(s)<=>CaO (s) + CO2(g)

Kc = [CO2]

The physical state of substances ……………………

2

PbI2(s) <=>Pb2+(aq)+2I-(aq)

Kc=[Pb2+][I-]2

……………………

………………

3

H2O(l)<=> OH-(aq) + H+(aq)

Kc=[H+][OH-]

……………………

………………

4

2NH3(g)+CO2(g) <=> CO(NH2)2(s) + H2O(l)

Kc=

……………………

………………

1 [NH3]2[CO2]

The physical state in Kc ………………

CONCLUSION Based on the activities above , conclude the formula of Kc based on the physical state of substances that are involved in heterogeneous equilibrium.

EXERCISES 01. Write the formula of Kc of the following equations a. C(s) + H2O(g) CO (g) + H2 (g) b. H2S (g) + I2 (s) ↔ 2HI (g) + S (s) c. AgCl (s) ↔ Ag+ (aq) + Cl- (aq) d. 2 NaHCO3 (s) ↔ Na2CO3 (s) + CO2 (g) + H2O (g) e. 2 BaO (s) <=> 2 BaO (s) + O2 (g) f. Ag2CrO4 (s) <=> 2 Ag+ (aq) + CrO42- (aq) Answers

WORKSHEET 3rd

DETERMINING THE VALUE OF EQUILIBRIUM CONSTANT Drs. HARIYANTO Public Senior High School 3 Malang Jl.Sultan Agung Utara 7 Malang Phone (0341) 324768 e-mail : [email protected] blog : http://hariyantobhawikarsu.wordpress.com

PRE-KNOWLEDGES No 1

2

Questions

Answers

Write the formula of Kc of the following equation : pA (g)+ qB(g)<=> mC(g) + n D(g) where : p,q,m and n are the coefficients of reaction

………………………………………………………………

Calculate the molarity of solution when 0.5 moles of SO2 gas is dissolved in 200 mL of solution

………………………………………………………………

ACTIVITY 1

CALCULATING THE VALUE OF EQUILIBRIUM CONTANT KC

How to calculate the Kc when the molarity of reactants and products at the equilibrium are given? How to calculate the Kc when the number of moles of reactants and products at the equilibrium are given ? How to calculate the Kc when the number of moles of reactants and products at the equilibrium are not given ?

Come on we study while you complete the following table! No 1

Problems Equilibrium is established in the reaction : 2 SO3(g) <=> 2 SO2 (g) + O2 (g) If equilibrium concentration are 0.04M of SO3 gas, 0.02M of SO2 gas, and 0.01M of O2 gas, What is the value of Kc?

Solutions [……] [……] Kc= [……] ……x…… =

= …………….. ………

2

In 2 liters of volum there is quilibrium position of reaction : N2O4(g) <=> 2 NO2 (g) If the equilibrium is reached there are 0.04 moles of N2O4 gas, 0.02 moles of NO2 gas, calculate the value of Kc at that temperature!

[……]2 Kc= [……]

2

⎡0.04moles ⎤ 2liters ⎥⎦ ⎢⎣ = Kc = 0.02moles 2liters …………. 3

4

In 2 liters of volume there is equilibrium position of reaction: C (s) + H2O (g) <=> H2 (g) + CO (g) If the amount at equilibrium the amount were 0.1 moles of C; 0.1 moles of H2O; 0.2 moles of H2, and 0.2 moles of CO, What is the value of Kc? The initial amount of the reactants present in 2 liters of mixture were 2 moles of A and 3 moles of B according to the following equation : A (g) + 2 B (g) <=> AB2(g) At equilibrium, the amount were 1,2 moles of A. Calculate the equilibrium constant,Kc.

[……] [……] Kc= [……]

=

(....mol ....L )(.....mol ....L ) = ......... ......mol

.....L

A (g) + 2 B (g) <=> AB2(g) Initial : 2 moles 3 mol es 0 Reaction : …moles ….moles_ …..moles + Equilibrium: 1,2 moles….moles ….moles [……] Kc=

[……]2[…..]

.....mol

=

5

The equilibrium constant reaction A (g) + B (g) <=> C (g) is 1/8. Determain the initial amount of A that react with 3 moles of B in 1 L of volume, if at equilibrium the amount of C is 1 mole.

...... 2L = = ...... ...... ⎛⎜1,2mol ⎞⎟ x ....mol 2L ⎠ 2L ⎝ 2

A (g) + B (g) <=> C(g) Initial : x moles 3 moles 0 Reaction : …moles ….moles __ …..moles + Equilibrium: … moles….moles 1 mole [……] Kc = [……].[…..]

1mol 1/8 =

....mol

1L

1L .... x mol

1L

x =……..mol So the initial amount of A is……….moles

CONCLUSION Based on the activities above, conclude the steps of determaining the value of Kc

EXERCISES 01. Equilibrium is established in the reaction N2 (g) + 3 H2(g) <=>2 NH3(g). At equilibrium the amount are 0.1 M of N2 , 0.2 M of H2, and 0.2 M og NH3. Calculate the value of Kc..

02. In 2 L of volume at 1270C the amount at equilibrium position are 4 moles of A; 2 moles of B2 and 1 mole of AB according to the reaction : 2 A (g) + B2 (g) <=> 2 AB (g). Calculate the value of Kc at 1270C.

03. In 2 liters of volume present are the initial amount 2 moles CO(g) and 2 moles of H2O g) according to the following equation : CO (g) + H2O (g) <=> CO2 (g) + H2 (g). If the amount at equilibrium is 1,5 mol es of H2, calculate the value of Kc !

04. The equilibrium constant of reaction: 2 HI (g) <=> H2 (g) + I2 (g) is 0,64. If the initial amount of HI is 2 moles in 2 L of volume, Calculate the amount of hydrogen at equilibrium

05. The equilibrium constant of equation : 2 HCl (g) <=> H2 (g) + Cl2 (g) is 0,25. At equilibrium, the amount of chlorine is 1 mol, determain the initial amount of HCl!

06. Equilibrium is established in the reaction : S (s) + O2 (g) <=>SO2 (g). In 2 L of volume the amount at equilibrium position are 0.4 moles of S; 0.4 moles of O2, and 0.8 moles of SO2, calculate the value of Kc!

07. In 1 L of volume the initial amount are 0.1 moles of C, and 0.1 moles of O2 acording to the equation : 2 C (s) + O2 (g) <=> 2 CO (g). At equilibrium the amount 0.08 moles of CO, calculate the value of Kc!

08. The equilibrium constant of the reaction : AgI (s) <=> Ag+ (aq) + I-(aq) is 10-16. Jif the initial amount of AgI is 0.01 mol AgI determain the concentration of Ag+ ion at equilibrium!

Worksheet 4th of equilibrium

EQUILIBRIUM CONSTANT IN OF PARTIAL PRESSURES ( KP) Drs. HARIYANTO Public Senior High School 3 Malang Jl.Sultan Agung Utara 7 Malang Phone (0341) 324768 e-mail : [email protected] blog : http://hariyantobhawikarsu.wordpress.com

PRE-KNOWLEDGES No 1

QUESTIONS

ANSWERS

What is partial pressures of gases in mixture? ……………………………………… ………………………………………

2

3

From the general equation of ideal gas : p.V = n.R.T where : p = pressure of gas(atm) V = volume of gas (L) n = number of mole of gas (mole) R = 0.082 L.atm/K.mol T = temperature (K) What is the quantitative relation between pressure and number of mole? In a flask presents 1 mole of A gas; 3 mole B gas, and 2 moles C gas,. Given that the pressure of mixture is 12 atm, calculate : a. The partial pressure of A b. The partial pressure o f B c. The partial pressure of C

ACTIVITY 1

……………………………………… ……………………………………… ………………………………………

……………………………………… ……………………………………… ……………………………………… ……………………………………… ……………………………………… ………………………………………

THE QUANTITATIVE RELATIONSHIP BETWEEN Kp AND THE PARTIAL PRESSURES OF GASES IN REACTION

What is the equilibrium constant of pressure? What is the quantitative relationship between Kp and the partial pressures of reactant gases, product gases? Complete the following table that obtained from the equilibrium equation :

. H2(g) + I2(g) ↔ 2HI(g) at 270C No

The partial pressure at equlibrium pH2 pI2 pHI

A pHI pH2. pI2

B (pHI)2 pH2. pI2

C pH2. pI2. pHI

1

0.2

0.2

0.2

…………………

…………………

…………………

2

0.3

0.3

0.3

…………………

…………………

…………………

3

0.4

0.4

0.4

…………………

…………………

…………………

4

0.5

0.5

0.5

…………………

…………………

…………………

QUESTIONS No 1

2

3

Questions From each table above a. Which column that the value is constants? b. What is relation between the formula that has the constant value and the reaction?

Answers a………..……………………………….. b. ………………………………………..

If the formula that has the constants value is called equilibrium constant, write the formula of Kp of the following equation: mA+nB↔pC+qD where m,n,p and q are coeffient Based the data of table above, Please describe the definition of Kp

……………………………………………

CONCLUSION

EXERCISES 01. Write the formula of Kp, according to the following equation: b. N2(g) + 3 H2(g) ↔ 2 NH3(g) b. 2 SO2(g) + O2 (g) <=> 2 SO3 (g) c. Cu2+(aq) + 4 NH3(aq) <=> Cu (NH3)42+ d. N2O4(g) <=> 2 NO2(g) e. 2N2O4 (g) + O2 (g) <=>2 N2O5 (g) f. 2 HBr (g) <=> H2 (g) + Br2 (g) g. 4 NH3 (g) + 3 O2 (g) <=> 2 N2 (g) + 6 H2O (g) h. N2 (g) + O2 (g) <=> 2 NO (g)

02. Given that the equilibrium equation : N2 (g) + 3 H2 (g) <=>2 NH3 (g) . If at equilibrium position, the partial pressures of gases N2, H2 dan NH3 are 4 atm, 2 atm dan 5 atm, determine the value of Kp of that reaction.

03. The pressure of gas mixture in flask is 2.0 atm. The equilibrium equation : A(g) + 2B(g) <=>2 C(g). At equilibrium position presents 0,2 moles of A gas, 0,1 molesof B gas,and 0,2 moles of C gas, Determain the value of Kp.

04. The value of Kp equilibrium equation: 2C(s) + O2(g) <=>2CO(g) is 9. If the partialpressure of CO2 gas = 6 atm, Calculate the partial pressure of O2 gas.

05. The value of Kp of equilibrium equation : 2 X(g) <=> 3 Y(g) is 1/8. At equilibrium the partial pressure of X gas = 8 atm, determine the partial pressure of Y gas.

06. Given that 6 moles of SO3 gas is decomposed according to the equation : 2SO3(g) <=>2SO2(g) + O2(g). At equilibrium presents 1 mole of O2 gas, and the total of pressure is 14 atm, determine the value of Kp.

07. The pressure of equilibrium equation : CaCO3(s) <=> CaO(s) + CO2(g) is 0.75 atm, determine the the value of Kp.

ACTIVITY2 2

THE QUANTITATIVE RELATIONSHIP BETWEEN Kp AND Kc

What is the quantitative relationship between Kp and Kc? To express the quantitative relationship between Kp and Kc,Please do the following activity.

The equilibrium equation : m A (g) + n B (g) <=> p C (g) + q D (g)

No 1

Questions

Answers

Write the formula of Kc and Kp of the equation above. Kc = ……………………. (1) Kp = ………………………(2)

2

3

From : P.V = n.R.T and M = n/V, Write the quantitative relationship between P and M or [ ] Write the foemula Kpfrom the equation (2), and substitute P from equation (3). In equation (4), you get the quantitative relationship between Kp and M or [ ]

P.V = n.R.T P

= n/V. R. T

P

= ………….

(3)

Kp = ……………………… Kp =………………………..(4)

4

Substitute [ ] in equation (1) from the equation (4). Kp = ……………………… Kp = Kc. …………………. (5)

5

From equation(5), if (p+q) - (m+n) are written as ∆n, so you get the quantitative relationship between Kp and Kc

Kp =………………………..(6)

CONCLUSION

EXERCISES 01. By calculating ∆n, Write the quantitative relation between Kp and Kc according to equation follows: a. 2 HCl (g) <=> H2(g) + Cl2 (g) b. 2 O3 (g) <=> 3 O2 (g) c. SnO2 (s) + 2 H2 (g) <=> Sn (s) + 2 H2O (g) d. 2 BaO2 (s) <=> 2 BaO (s) + O2(g)

02. The value of Kp of the equilibrium equation : H2(g) + I2 (g) <=> 2 HI (g) at 4460C is 3,14x102. Calculate the value of Kc at the same temperature.

03. The value of Kc of the equilibrium :N2 (g) + 3 H2(g) <=> 2 NH3(g) at 5270C is 4,5 x 10-2. If R= 0,082 L.atm/K.mol, Determine the value of Kp.

WORKSHEET 5th

DISSOCIATION AND DISSOCIATION DEGREE Drs. HARIYANTO Public Senior High School 3 Malang Jl.Sultan Agung Utara 7 Malang Phone (0341) 324768 e-mail : [email protected] blog : http://hariyantobhawikarsu.wordpress.com

ACTIVITY 1

DISSOCIATION

What is dissociation? Is dissociation a part of equilibrium equation? After you complete the following table, I hope you can define the dissociation. No

Equilibrium

1

2SO3(g) ↔ 2SO2(g) + O2(g)

Kind of Mr of reactants reactant 1 SO3= 80

2

NH4Cl(s) ↔ NH3(g) + Cl2(g)

1

………..

3

PCl5(g) ↔ PCl3(g) + Cl2(g)

1

………..

5

2BaO2(s) ↔ 2BaO (s) + O2(g)

1

………..

6

H2(g) + I2 (g) ↔ 2 HI(g)

……..

………..

7

2H2O(l) ↔ 2H2(g) + O2(g)

……..

………..

8

2CaO(s) + O2(g) ↔2CaO2 (s)

……..

………..

Mr of ech product SO2 = 60 O2 = 32 …………. …………. …………. …………. …………. …………. …………. …………. …………. …………. …………. ………….

Dissociation or not Dissociation Dissociation Dissociation Dissociation Not Dissociation Not Dissociation Not Dissociation

QUESTIONS Based on the the data of table above, answer the following questions! No Questions Answers 1 How many kind of substances are the reactants in dissociation? ……………………………………………….. 2

Observe and compare the relative molecule mass (Mr) of reactant and the relative molecule mass of each product in dissociation process? Please choose the correct! Mr of reactant is ( < than / > than / = with ) Mr of each product

Mr of reactant is …..........Mr of each product

3

4

If the substance that has the smaller Mr is the simpler substance, which the simpler substance is in dissociation process? Reactant or product? Based on the answers of the questions no 1, 2, and 3. Please define the dissociation process!

………………………………………………..

………………………………………………..

CONCLUSION Based on the answer of no 4 above, conclude the definition of dissociation!

EXERCISE 1 01. Which are the dissociation process according to the following equations?: a. 2H2O2(l) ↔ 2H2O(l) + O2(g) b. 2NH3(g) ↔ N2(g) + 3 H2(g) c. I2(g) ↔ 2 I(g) d. SO2(g) + ½ O2(g) ↔ SO3(g) e. 4NO2(g) + O2(g) ↔ 2N2O5(g) f. N2O4(g) ↔ 2NO2(g) g. H2O(l) ↔H2O(g) h. CH3COO-(aq) + H2O(l) ↔ CH3COOH(aq) + OH-(aq) i. CO(g) + H2O(g) ↔ CO2(g) + H2(g) j. 2HI(g) ↔ H2(g) + I2(g)

Answers to the exercise 1

ACTIVITY 2

2

DISSOCIATION DEGREE (α)

What is the dissociation degree? What is the relation between dissociation degree and the initial amount and the decomposed amount of substance? Please observe and fill the blank by calculating the initial amount or the decomposed amount or the amount at equilibrium!

No

Equations

The initial amount of reactants

The decomposed amount of reactants

The amount of reactants at equilibrium

Dissociation degree (α)

1 2 3 4 5 6

2NH3(g) ↔N2(g) + 3H2(g) 2HI(g) ↔ H2(g) + I2(g) N2O4(g) ↔ 2 NO2(g) PCl5(g) ↔ PCl3((g) + Cl2(g) 2BaO2(s) ↔ 2BaO (s) + O2(g) NH4Cl(s) ↔ NH3(g) + Cl2(g)

0.5 moles 1.2 moles 0.8 moles 0.6 moles 0.8 moles 1.0 moles

0.3 moles 0.9 moles ….moles ….moles ….moles ….moles

0.2 moles 0.3 moles 0.6 moles 0,36 moles 0.2 moles 0.3 moles

0.60 0.75 0.25 0,40 ….. …..

7 8

2 O3 (g) <=> 3 O2 (g) 2NOg) ↔N2 (g) + O2 (g)

2.0 moles 4.0 moles

….moles ….moles

….moles ….moles

0.36 0.30

CONCLUSION Based on the table above, if the initial amount of reactant is symbolized by symbol decomposed amount of reactant is

“ a“ and the

“x”, Write the formula of dissociation degree!

EXERCISE 2 01. Given that the equilibrium : I2(g) ↔ 2 I(g). If the initial amount of I2 is 0.40 moles and at equilibrium presents 0.15 moles of I, determine the dissociation degree of I2.

02. At equilibrium of : 2HI(g) ↔ H2(g) + I2(g) presents are 0.80 moles of HI(g); 0.20 moles of H2(g) and 0.20 moles of I2(g). Determine the value of dissociation degree of HI.

03. Given that the equilibrium : 2HI(g) ↔ H2(g) + I2(g), The initial amount HI is 0.75 mol and 20% of HI is dissociated. a. determine the amount of each substance at equilibrium b. the value of Kc

04. At 270C there is the equilibrium equation : PCl5(g) ↔ PCl3((g) + Cl2(g). The initial amount of PCl5 is 0.25 moles, and at equilibrium the amount is 0.15 moles Cl2. a. calculate the dissociation degree of PCl5 b. determine the amount of each substance at equilibrium c. determine the value of Kc d. determine the value of Kp

05. Given that the equilibrium : N2O4(g) ↔ 2 NO2(g). If the comparasion of the number of mole of N2O4 and NO2 at equilibrium position is 1 : 2, determine the value of dissociation degree of N2O4.

06. At 2.0 liters of volume there is the equilibrium : 2SO3(g) ↔ 2SO2(g) + O2(g). The initial amount of SO3 is 80 grams. If the proportional the number of mole of SO3 and O2 at equilibrium is 4 : 3, calculate: a. dissociation degree of SO3 b. the value of Kc at the condition

07. Equilibrium is established in the reaction : 2SO3(g) ↔ 2SO2(g) + O2(g) at 27oC; 8.2 atm. The initial of SO3 is 0.20 moles. At equilibrium the total volume of gases is 0.84 liters, determine the value of dissociation degree of SO3.