thermodynamics energy transformations. n 2 o 4 2 no 2 initially forward reaction rapid – as some...

Chemical Equilibrium

Thermodynamics• Energy transformations

N2O4 2 NO2 • Initially forward reaction rapid– As some reacts [N2O4] so rate forward

• Initially Reverse reaction slow– No products

• As NO2 forms– Reverse rate

• Eventually rateforward = ratereverse

– Equilibrium

5

Reaction Reversibility Closed system• Equilibrium can be reached

from either direction• Independent of whether it

starts with “reactants” or “products”

• Always have the same composition at equilibrium under same conditions

N2O4 2 NO2

6

• For given overall system composition• Always reach same equilibrium concentrations• Whether equilibrium is approached from forward or

reverse direction

N2O4 2 NO2

Reactants ProductsEquilibrium

If equilibrium is established, concentrations don’t change any more.

Each set of equilibrium concentrations

Equilibrium position

Check it yourself!

TRnVp TRcTRV

np

Relation between Kp and Kc

31

2

22

3

HN

NH

c cc

cK TRcp ii

ii cTR

p

31

2

22

3

RT

p

RT

p

RT

p

cHN

NH

K

231231

211

22

3

RT

KRTpp

pK p

HN

NH

c

pc KRTK 2

p

n

c KRTK

ctantsreagproductsg nnn

17

Learning CheckConsider the reaction: 2NO2(g) N2O4(g)

If Kp = 0.480 for the reaction at 25°C, what is value of Kc at same temperature?

•n = nproducts – nreactants = 1 – 2 = –1

Δncp (RT)KK

1np

c)2980821.0(

480.0

(RT)

KK

K

Kc = 11.7

Interpreting Kp or Kc

Means product rich mixture

2SO2(g) + O2(g) 2SO3(g) Kc = 7.0 1025 at 25 ° C

Means reactant rich

mixture H2(g) + Br2(g) 2HBr(g)

Kc = 1.4 10–21 at 25 °C

Reaction goes only ~ halfway

• more than phase involved

Concentrations of pure liquids and solids do not change with changing the amounts of these substances.

][][

][' 2

3 COKCaO

CaCOK c

Pure solids and liquids do not appear in the expression for equilibrium constant.

Reaction Quotient Q

mequilibriu

c HN

NHK

3

22

2

3

][][

][

timeanyHN

NHQ

3

22

2

3

][][

][

• If Q=K → equilibrium• Q ≠ K → not at equilibrium

tries to reach equilibrium

mequilibriu

c HN

NHK

3

22

2

3

][][

][

timeanyHN

NHQ

3

22

2

3

][][

][

For the synthesis of ammonia at 500oC, the equilibrium constant is 6.0×10-2. Predict the direction in which the system will shift to reach equilibrium in each of the following cases:

Q >> KQ will decrease with time

More reactants, less productsSystem shifts to the right

Reversed reaction dominatesThe amount of N2 and H2 will increase

The amount of NH3 will decrease

Your turn!

N2O4 was placed in a flask and allowed to reach equilibrium at a temperature where Kp 0.133. At equilibrium, the pressure of N2O4 was found to be 2.71 atm. Calculate the equilibrium pressure of NO2(g). N2O4 2 NO2

At a certain temperature a 1.00-L flask initially contained 0.298 mol PCl3(g) and 8.70×10-3 mol PCl5(g). After the

system had reached equilibrium, 2.00×10-3 mol Cl2(g) was

found in the flask.

no

change

neq

ceq6.70×10-3 0.300 2.00×10-3

Carbon monoxide reacts with steam to produce carbon dioxide and hydrogen. At 700 K the equilibrium constant is 5.10. Calculate the equilibrium concentrations of all species if 1.000 mol of each component is mixed in a 1.000-L flask.

co

chang

e

ceq

)(2)(2)(2)( gggg HCOOHCO

Q < K Shift to the right

x=0.387 mol/L

Assume that the reaction for the formation of gaseous hydrogen fluoride from hydrogen and fluorine has an equilibrium constant of 1.15×10-2 at a certain temperature. In a particular experiment, 3.000 mol of each component was added to a 1.500-L flask. Calculate the equilibrium concentrations of all species.

)()(2)(2 ggg HFHF

co

chang

e

ceq

Q < K Shift to the right

x=1.528 mol/L

3.000 mol H2 and 6.000 mol F2 are mixed in a 3.000-L flask. Assume that the equilibrium constant for the synthesis reaction at this temperature is 1.15×10-2. Calculate the equilibrium concentration of each component.

Quadratic equation

a

cabbx

cxbxa

2

4

0

2

2

Assume that gaseous hydrogen iodide is synthesized from hydrogen gas and iodine vapor at a temperature where the equilibrium constant is 1.00×10-2. Suppose HI at 5.000×10-1 atm, H2 at 1.000×10-2 atm, and I2 at 5.000×10-3 atm are mixed in a 5.000-Lflask. Calculate the equilibrium pressures of all species. Q > K

1.0 mol NOCl is placed in a 2.0-L flask, what are the equilibrium concentrations?

NOCl NO Cl2

co

chang

e

ceq

x=0.010 mol/L

How valid is this approximation?

LmolNOCl 48.0][

%5%

%4%10050.0

48.050.0%

%100][

][][%

Difference

Difference

NOCl

NOClNOClDifference

assumed

resultassumed

Approximation justified!!

2( ) 2( ) ( )2g g gH I HI

If the conc. of a substance is increased, the equilibrium will shift in a way that will decrease the conc. of the substance that was added.

I. Concentration Changes

2( ) 2( ) ( )2g g gH I HI

H2

I2

HI

Decrease HI - forward

Decrease I2 - backward

Increase H2 - forward

Increase HI - backward

Where will the reaction shift?

Note:

adding/removing a solid/liquid to an equilibrium system will not cause any shift in the position of equilibrium because this doesn’t change their concentration.

addition of an inert gas such as He, Ar, Kr, etc. at constant volume, pressure and temperature does not affect the equilibrium because it doesn’t change the concentrations or partial pressures.

3( ) ( ) 2( )s s gCaCO CaO CO

Increase CO2 :

Increase CaO

Decrease CO2

Adding Kr

- backward

- No effect

- forward

- No effect

An increase in pressure (decrease in volume) shifts the position of the equilibrium in such a way as to decrease the pressure by decreasing the number of moles of gaseous component.

When the volume is increased (pressure decreased), a net reaction occurs in the direction that produces more moles of gaseous component (increase in system pressure).

The effect of pressure (volume) on an equilibrium system.

2( ) 2( ) 3( )3 2g g gN H NH

Decrease in pressure :

Decrease in volume:

backward

forward

TRnpV

Sample Problem 17.12

SOLUTION:

Predicting the Effect of a Change in Volume (Pressure) on the Equilibrium Position

PROBLEM: For the following reactions, predict the direction of the reaction if the pressure is increased:

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

(b) S(s) + 3F2(g) SF6(g)

(c) Cl2(g) + I2(g) 2ICl(g)

(a) CO2 is the only gas present. The equilibrium will shift to the direction with less moles of gas. Answer: backward

(b) There are more moles of gaseous reactants than products.

Answer: forward

(c) There are an equal number of moles of gases on both sides of the reaction. Answer: no effect

Effect of Temperature Changes• T shifts reaction in direction that absorbs heat

(enhances the endothermic change)• T shifts reaction in direction that produces heat

(enhances the exothermic change)

• Changes in T change the value K – K depends on T– T of exothermic reaction makes K smaller• More heat (product) forces equilibrium to reactants

– T of endothermic reaction makes K larger• More heat (reactant) forces equilibrium to products

Effect of Change in Temperature

Cu(H2O)42+(aq) + 4Cl–(aq) CuCl4

2–(aq) + 4H2O blue yellow– Reaction endothermic– Adding heat shifts equilibrium toward products– Cooling shifts equilibrium toward reactants

Ice water

Boiling water

Your Turn!The equilibrium between aqueous cobalt ion and the chlorine ion is shown:

[Co(H2O)6]2+(aq) + 4Cl–(aq) [Co(Cl)4]2–(aq) + 6H2O(ℓ)

pink blue

It is noted that heating a pink sample causes it to turn violet.The reaction is:A. endothermicB. exothermicC. cannot tell from the given information

LE CHATELIER’S PRINCIPLE

The rates of both the forward and reverse reactions are increased by the same amount. NO SHIFTadd a catalyst

more heat/ energy needs to be produced to make up for the loss

towards heat/ energyexothermic reaction is favored

decrease temperature of system

extra heat/ energy must be used up

away from heat/ energyexothermic reaction is favored

increase temperature of system

for gas: pressure decrease = volume increase

towards more moles of gasdecrease pressure of system

for gas: pressure increase = volume decrease

towards fewer moles of gasincrease pressure of system

need to produce more of substance to make up for what was removed

towards substancedecrease concentration of a substance

extra concentration needs to be used up

away from substanceincrease concentration of a substance

WHY?SHIFTSTRESS

N2(g) + 3H2(g) 2NH3(g) + 22.0 kcal

10. Decrease press.

9. Increase press.

8. Decrease temp

7. Increase temp

6. Remove NH3

5. Remove H2

4. Remove N2

3. Add NH3

2. Add H2

1. Add N2

[NH3][H2][N2]ShiftStress

CO2 is added

CO is added

O2 is added

CO is removed

O2 is added

pressure is increased

pressure is decreased

temp. is increased

temp. is decreased

catalyst is added

2CO2(s) + heat 2CO(s) + O2(g)stress

Given the reaction:N2(g) + O2(g) + 182.6 kJ 2NO(g)

Which change would cause an immediate increase in the rate of the forward reaction?

(1) increasing the concentration of NO(g)

(2) increasing the concentration of N2(g)

(3) decreasing the reaction temperature(4) decreasing the reaction pressure

DO NOW

2POCl3(g) + energy 2PCl3(g) + O2(g)

Which changes occur when O2(g) is added to the system?

1) The equilibrium shifts right and [PCl3] increases.2) The equilibrium shifts right and [PCl3] decreases.3) The equilibrium shifts left and [PCl3] increases.4) The equilibrium shifts left and [PCl3] decreases

DO NOW

2SO2(g) + O2(g) 2SO3(g) + heat

For each of the following stresses, state in which direction the equilibrium will shift and what will happen to the concentration of SO2. Explain.

1) concentration of O2 is decreased

2) pressure is decreased3) temperature is decreased4) a catalyst is added

QUIZ