Entry Task: Jan 7th Monday

Provide the Kc expression for this equation.

3 Ca2+(aq) + 2 PO4

3-(aq) ↔ Ca3(PO4)2 (s)

You have 5 minutes

Agenda:

• Discuss Kc Equilibrium constant ws• HW: – Ch. 15 sec. 4- Calculating Equilibrium constants

PLEASE MUDDLE THROUGH IT!!

1. What are the 3 conditions necessary for equilibrium?

• Must have a closed system• Must have a constant temperature• Ea must be low enough to allow a

reaction.

2. What is a forward reaction versus a reverse reaction?

• In a forward reaction, the reactants collide to produce products and it goes from left to right.

• In a reverse reaction, the products collide to produce reactants and it goes form right to left.

3. Why does the forward reaction rate decrease as equilibrium is approached?

• As the reaction goes to the right, the reaction concentration decreases and therefore, there are less reactant collisions causing the forward rate to decrease.

4. What are the characteristics of equilibrium or define equilibrium?

• Forward rate is equal to the reverse rate.• The concentration of reactants and products

are constant.(not equal)• Macroscopic properties are constant (color,

mass, density, pressure, concentrations).

5. As a reaction is approaching equilibrium describe how the following change. Explain

what causes each change.a. Reactant concentration• As the reaction goes to the right, the reaction

concentration decreases.b. Products concentration• As the reaction goes from left to right, the

concentration of the products increases.

5. As a reaction is approaching equilibrium describe how the following change. Explain

what causes each change.

c. Forward reaction rate. The reaction concentration decreases and therefore, there are less reactant collisions causing the forward rate to decreases.d. Reverse reaction rateThe concentrations of the products increases, therefore there are more product collisions causing the reverse reaction rate to increase.

6. What is equal at equilibrium?

The forward and reverse rates are equal.

7. What is constant at equilibrium?

The reactant and product concentrations and the macroscopic properties are constant.

For the following three reactions,*Write the Kc expression AND calculate the value of Kc.*State whether the equilibrium is product-favored, reactant-favored, or fairly even ([products] [reactants]).

At equilibrium: [N2] = 1.50 M[H 2] = 2.00 M[NH3]= 0.01 M

8. N2 (g) + 3 H2(g) 2 NH3(g)

3

2

][][

][

22

3

HN

NHKc

3

2

]00.2[]50.1[

]01.0[cK

Kc = 8.33 x 10-6Small Kc favors reactant

For the following three reactions,*Write the Kc expression AND calculate the value of Kc.*State whether the equilibrium is product-favored, reactant-favored, or fairly even ([products] [reactants]).

9. HF(aq) H+(aq) + F-(aq)At equilibrium:

[HF] = 0.55 M[H+] = 0.001 M[F-]= 0.001 M

][

][][

HF

FHKc

]55.0[

]001.0[]001.0[cK

Kc = 1.8 x 10-6Small Kc favors reactant

For the following three reactions,*Write the Kc expression AND calculate the value of Kc.*State whether the equilibrium is product-favored, reactant-favored, or fairly even ([products] [reactants]).

10. Fe3+(aq) + SCN-(aq) FeSCN2+(aq)

][]3[

]2[

SCNFe

FSCNKc

Kc = 1.8 Kc is very close to 1 so even

At equilibrium: [Fe3+] = 0.55 M[SCN-] = 0.001 M

[FeSCN2+]= 0.001 M

]001.0[]55.0[

]001.0[cK

Summarize:Fill in the blanks with product-favored,

reactant-favored, and approximately equal

Kc state of equilibrium

Kc >> 1

Kc << 1

Kc 1

Products favored

Reactant favored

~Equal~

11. Write equilibrium expressions for each of the following reactions:

a) CaCO3(s) CaO(s) + CO2(g)

b) Ni(s) + 4CO(g) Ni(CO)4(g)

c) 5CO(g) + I2O5(s) I2(g) + 5CO2(g)

][ 2COKc

4][

]4)([

CO

CONiKc

5

522

][

][][

CO

COIKc

11. Write equilibrium expressions for each of the following reactions:

d) Ca(HCO3)2(aq) CaCO3(s) + H2O(l) + CO2(g)

e) AgCl(s) Ag+(aq) + Cl-(aq)

])([

][

23

2

HCOCa

COKc

][][

ClAgKc

12. Write the equilibrium expression in terms of partial pressures (Kp) for each of the following reactions.

Rate the reactions in order of their increasing tendency to proceed toward completion (products):

___ ___ ___ ___

(a) 4NH3(g) + 3O2(g) 2N2(g) + 6H2O(g)

Kp = 1 x 10228

34

62

][][

][][

23

22

ONH

HONp

PP

PPK

12. Write the equilibrium expression in terms of partial pressures (Kp) for each of the following reactions.

Rate the reactions in order of their increasing tendency to proceed toward completion (products):

___ ___ ___ ___

(b) N2(g) + O2(g) 2NO(g) Kp = 5 x 10-31

][][

][

2

2

2

2

ON

NOp

PP

PK

12. Write the equilibrium expression in terms of partial pressures (Kp) for each of the following reactions.

Rate the reactions in order of their increasing tendency to proceed toward completion (products):

___ ___ ___ ___

(c) 2HF(g) H2(g) + F2(g) Kp = 1 x 10-13

2][

][][ 22

HF

FHp

P

PPK

12. Write the equilibrium expression in terms of partial pressures (Kp) for each of the following reactions.

Rate the reactions in order of their increasing tendency to proceed toward completion (products):

___ ___ ___ ___

(d) 2NOCl(g) 2NO(g) + Cl2(g) Kp = 4.7 x 10-4

2

2

][

][][ 2

NOCl

ClNOp

P

PPK

12. Write the equilibrium expression in terms of partial pressures (Kp) for each of the following reactions.

Rate the reactions in order of their increasing tendency to proceed toward completion (products):

___ ___ ___ ___

(d) 2NOCl(g) 2NO(g) + Cl2(g) Kp = 4.7 x 10-4

(c) 2HF(g) H2(g) + F2(g) Kp = 1 x 10-13

(b) N2(g) + O2(g) 2NO(g) Kp = 5 x 10-31

(a) 4NH3(g) + 3O2(g) 2N2(g) + 6H2O(g) Kp = 1 x 10228

Lowest value is reactant favored- least likely to complete reaction

B C D

Highest value is product favored- most likely to complete reaction

A

13. (a) Write the Kc expression for2 SO2(g) + O2(g) 2 SO3(g)Calculate the value of Kc:

At equilibrium: [SO2] = 1.50 M[O2] = 1.25 M[SO3]= 3.50 M

][][

][

22

23

2 OSO

SOKc

]25.1[]50.1[

]50.3[2

2

cK

Kc = 4.36

13. (b) If we reverse the equation, it is:2 SO3(g) 2 SO2(g) + O2(g)

Write the Kc expression for this equation and calculate the new value of Kc:At

equilibrium:[SO2] = 1.50 M[O2] = 1.25 M[SO3]= 3.50 M

23

22

2

][

][][

SO

OSOKc

Kc = 0.229

2

2

]50.3[

]25.1[]50.1[cK

13. (b) If we reverse the equation, it is:2 SO3(g) 2 SO2(g) + O2(g)

Write the Kc expression for this equation and calculate the new value of Kc:

How does the expression and the value of Kc in 7(b) compare with those in 7(a)?

Kc = 1/Kc

(c) If we now multiply all of the coefficients by ½:

SO3(g) SO2(g) + ½ O2(g)Write the Kc expression for this equation and calculate the new value of Kc:

At equilibrium:

[SO2] = 1.50 M[O2] = 1.25 M[SO3]= 3.50 M

][

][][

3

2/122

SO

OSOKc

Kc = 0.479

]50.3[

]25.1[]50.1[ 2/1

cK

(c) If we now multiply all of the coefficients by ½:

SO3(g) SO2(g) + ½ O2(g)Write the Kc expression for this equation and calculate the new value of Kc:

How do they compare with 7(b)?

Kc = 0.229 Kc = 0.479

Kc = square root of Kc

(d) What would happen to the Kc expression and its value if we doubled the coefficients?

Kc = (Kc)2

Summarize:

Equation Kc expression Valuedoubled reversed halved

squaredInverse or 1/KcSquare root