UNIT 2: KINETICS AND EQUILIBRIUMcarussell.weebly.com/uploads/3/8/8/3/38837901/ke_notes.pdf · a...

CHEMISTRY 3202

UNIT 2: KINETICS AND EQUILIBRIUM

OVERVIEW OF UNIT

● Chemical reactions occur at different rates under different conditions. We group all the concepts related to the rates of chemical reactions under the heading kinetics.

● Some chemical reactions are reversible, that is, reactants become products while products become reactants again. In some systems, the opposing reactions occur at the same time. When the opposing rates of change are equal, a chemical system is said to be in equilibrium.

OVERVIEW OF UNIT

We will:● explain and predict how changes in reaction

conditions affect chemical systems● See how conditions such as temperature and

pressure can be manipulated to favor the formation of products

● Identify the major factors that affect reaction rates

OVERVIEW OF UNIT

● Explore examples of systems at chemical equilibrium and make predictions about how equilibria are affected by changes in reaction conditions

● Explore quantitatively chemical equilibrium

Throughout the unit, we will encounter examples of reactions and equilibria in nature, living systems, and commercial/industrial settings.

DID YOU KNOW

An example of this in real life would be:● Chemical engineers design equipment and

chemical processes that maximize product formation in a cost efficient manner

Part 1: Reaction Kinetics (Chp. 12)

● Reaction Kinetics is the study of the rate of a chemical reaction

Qualitative: ● Reactions may be described as being FAST

or SLOW ● Fast – burning, explosions, precipitation ● Slow – rusting, fermentation

Reaction Kinetics

Quantitative:

The rate of a reaction measures how fast products are formed or how fast reactants are consumed

RATE = CHANGE IN QUANTITY

CHANGE IN TIME

POSSIBLE UNITS??

Measuring Reaction Rate

● The method used to determine reaction rate will depend on the reaction being studied. (p. 466)

Methods:

1. Monitor pH if there is an acid or base in the equation

2. Record gas volume or changes in pressure if there is a gas in the reaction

Measuring Reaction Rate

Methods: (cont’d)

3. record changes in mass if solids are present

4. monitor absorption of light if there is a color change

5. changes in electrical conductivity indicate changes in ion concentration

MC: What could we use to measure the rate of this reaction?

Cu(s) + 2 AgNO3(aq) 2 Ag(s) + Cu(NO3)2(aq)

a) pressure c) gas volume

b) pH d) mass

ANSWER: d) because a solid is present

MC: What could we use to measure the rate of this reaction?

SO3(g) + H2O(l) H2SO4(aq)

a) pressure c) gas volume

b) pH d) mass

ANSWERS:

a) & c) because a gas is present

b) because an acid is being produced

ANSWER:

c) because the pH starts off as fairly acidic (low pH) and the ph gradually becomes less acidic/more basic (pH increases). c) is the only equation that starts off with an acid and as the equation progresses, the acid is slowly used up. Thus the pH increases.

What determines RATE??

● All chemical reactions are bond breaking/bond forming events

● The rate of a reaction depends on how quickly bonds are broken and how rapidly new bonds form.

● Kinetic Molecular Theory (KMT) and Collision Theory are used to explain reaction rates.

Kinetic Molecular Theory (KMT)

● Matter is made of particles (atoms, ions, or molecules) in continuous motion

● An increase in temperature: increases the speed of particles reduces the forces of attraction between

particles

Kinetic Molecular Theory (KMT)

● KMT is supported by:

Diffusion – particles of a gas spread to fill their container (‘perfume in a room’) solids dissolve uniformly in liquids over

time.

Pressure – a balloon remains inflated because gas particles are continuously hitting the sides of the balloon

MC: Which observation best supports the Kinetic Molecular Theory?

(A) Acetic acid odour is detected from across the room.

(B) Liquid water freezes at 0°C under standard conditions.

(C) Nitrogen dioxide gas is dark brown in colour.

(D) When burned, butane produces more heat per mole than propane.

ANSWER:

(A) because is the smell is detected from across the room that means that the gas particles of the acetic acid vapor spread to fill out their container (the room). Thus we smell it from across the room.

Collision Theory

● states that reacting particles must collide with one another in order for a reaction to occur

However:● Particles must collide with proper orientation● The collisions must have enough intensity to

break old bonds and allow new bonds to form

Collision Theory

● to increase reaction rate you must increase the number of successful collisions between reactant molecules

● http://phet.colorado.edu/en/simulation/reactions-and-rates

Reaction Rates Simulation

Correct Orientation

Why is the correct orientation correct?● Essentially atoms must come in such a way

such that positive charge comes together with negative charge (opposites attract) and in such a way where there is enough room for it to happen.

Correct Orientation

Sufficient Collision Intensity

● Basically, unless particles collide with each other with enough intensity, then bonds will not break and thus reform. Therefore no reaction will occur, even if the orientation of the collision was correct.

Sufficient Collision Intensity

Paper, plastic and many other materials burn in air (oxygen gas). Oxygen molecules in the air are in a state of constant motion (at 20°C they move at about 1650 km/h) and that they are likely colliding with your notebook right now.

● So why isn't your notebook on fire (i.e. reacting with oxygen in the air)?

Sufficient Collision Intensity

● Basically, the collisions between the carbon rich cellulose fibers in your notebook and oxygen gas in the air are not intense enough to rearrange the chemical bonds in these substances.

MC: Which must occur for a chemical reaction to take place?

(A) addition of a catalyst

(B) addition of energy

(C) collisions between reacting particles

(D) formation of a reaction intermediate

ANSWER:

C) because this is what the Collision Theory states and if there are no collisions, then there are no bond breaking and bond reforming moments. So no reaction would happen

Factors Affecting Reaction Rate

1. Concentration (only for (aq) or (g) reactants)

– an increase in the concentration of a reactant usually increases the rate of a chemical reaction

- the rate increases because there are:

- more particles resulting in

- more collisions between particles &

- more successful collisions.

Factors affecting reaction rates

2. Temperature

- an increase in the temperature increases the rate of a chemical reaction

- the higher temperature results in:

- faster moving particles

- more collisions between particles

- more intense collisions

NOTE: A temperature increase of 10 ºC usually causes reaction rate to double.

More successfulcollisions

FasterRate

Factors Affecting Reaction Rate

3. Nature of Reactants – compounds with fewer bonds to break will react more rapidly than compounds with many bonds

eg. propane (C3H8) burns faster than candlewax (C25H52) because it has fewer bonds

Factors affecting reaction ratesPropane Candlewax

Factors Affecting Reaction Rate

3. Nature of Reactants

- compounds with weak bonds react more rapidly than compounds with strong bonds

– ions will react more rapidly than atoms and molecules

Factors Affecting Reaction Rate

4. Surface Area

- crushing a solid to produce a powder, or changing a substance to the gas phase, exposes more particles for collision

-if more particles are available for collision there will be:

- more collisions

- more successful collisions

FASTER RATE

Factors affecting reaction rates

For example:

If we did an experiment that involved dropping an alka seltzer tablet, pieces of an alka seltzer tablet and a crushed alka seltzer tablet into their own glass of water at room temperature, we should find that the crushed tablet would react noticeably faster than the other two trials. The reason is because there was more surface area to react with the water when the tablet was crushed.

Which reaction occurs faster at room temperature?

(A) 2H2(g) + O

2(g) → 2H

2O(liq)

(B) 2Ag+(aq) + CrO4

2-(aq) → Ag2CrO

4(aq)

(c) Pb(s) + 2HCl(aq) → PbCl2(s) + H

2(g)

(D) CH4(g) + 2O

2(g) → CO

2(g) + 2H

2O(g)

ANSWER:

(B) because ions are generally the quickest types of species to react since they don't have bonds to break. They just come together to form compounds.

MC: Which factor explains why coal dust is

explosive?

(A) concentration

(B) pressure

(C) surface area

(D) temperature

ANSWER:

(c) because coal dust has far more surface area susceptible to a reaction that would lead to spontaneous combustion than chunks of coal

http://en.wikipedia.org/wiki/Coal_dust

PRACTICE

p. 484; #’s 1 & 2

p. 486; #’s 1,2, 4, 6, & 7

KINETICS & EQUILIBRIUM SHEET #1

Factors affecting reaction rates

5. Catalysts

- a catalyst increases the reaction rate by providing a different reaction pathway or mechanism with a lower activation energy

- a catalyst IS NOT consumed by a chemical reaction.

Potential Energy Diagrams (p. 473)

● PE diagrams show changes in potential energy (stored chemical energy) during chemical reactions

● Exothermic reactions release more energy than they absorb (eg. burning)

● Endothermic reactions absorb more energy than they release (eg. photosynthesis)

Potential Energy Diagrams

● ΔH represents the heat of reaction or enthalpy of reaction

● ΔH is the difference between the PE of the reactants and the PE of the products

● The minimum energy needed for a chemical reaction to occur is the activation energy

Potential Energy Diagrams

● the activated complex for a reaction is a temporary, unstable, intermediate species that quickly decomposes to products

eg. H2 + I2 → H2I2 → 2 HI

ACTIVATEDCOMPLEX

Endothermic PE Diagram

Exothermic PE Diagram

ΔH & equations ● The energy term may be included in a chemical

equation● eg. CO(g) + 2 H2(g) → CH3OH(g) + 65 kJ

OR written as ΔH to the right of the equation.

eg. CO(g) + 2 H2(g) → CH3OH(g) ΔH = - 65 kJ

Energy is Produced/Released

EXOTHERMIC

ΔH & equations

Another eg.

eg. N2(g) + O2(g) + 90 kJ → 2 NO(g)

OR N2(g) + O2(g) → 2 NO(g) ΔH = + 90 kJ

Energy is REQUIRED ENDOTHERMIC

Formula: (OPTIONAL)

Eaforward - Eareverse = ΔH

This formula is NOT necessary if you prefer using the PE diagram.

Ea (forward Ea (reverse)

ΔH Endo or Exo

25 -30

50 20

150 250

65 28

Sketch a PE diagram for each reaction

06:52 PM

Reaction Progress

PE

∆HEafwd

CO2 + H2O

C6H12O6 + O2

Photosynthesis

Earev

06:52 PM

Reaction Progress

PE

∆H

Eafwd

CO2 + H2O

C6H12O6 + O2

Respiration

Earev

06:52 PM

p. 474

∆H

06:52 PM

∆H

06:52 PM

∆H

p. 475

06:52 PM

Reaction Progress

PE

no catalyst

catalyzed

Effect of a catalyst Endothermic

06:52 PM

Reaction Progress

PE

no catalystEXOTHERMIC

06:52 PM

● Sample problem: p. 475● Questions:

p. 476; #’s 2, 4

p. 484; #’s 1 – 4

06:52 PM

Reaction Mechanisms (pp. 477 – 485)

reaction mechanism – the steps that occur in a chemical reactionelementary reaction - each step in a reaction mechanismreaction intermediate – a molecule, atom or ion formed in one step and consumed in a later step NOTE: reaction intermediates are NOT included in the overall equation

06:52 PM

Reaction Mechanisms

eg. #1

Step #1 NO(g) + O2(g) NO3(g)

Step #2 NO3(g) + NO(g) 2 NO2(g)

Overall

Equation:2 NO(g) + O2(g) 2 NO2(g)

06:52 PM

HBr + O2 → HOOBr fast

HOOBr + HBr → 2 HOBr slow

2 HOBr + 2 HBr → 2 H2O + 2 Br2 fast

p. 478 #’s 5 – 8

06:52 PM

Reaction Mechanisms

rate-determining step (RDS)

- the slowest step in a reaction mechanism

- to increase the rate of a reaction you must speed up the RDS

- increasing the concentration of a reactant will increase the rate ONLY IF the reactant is in the RDS

06:52 PM

Reaction Mechanisms

PE diagrams

- every step in a reaction mechanism has an activation energy which can be drawn on a PE diagram

06:52 PM

Reaction Progress

PE

Reaction Mechanisms

3-step mechanism

#1

#2

#3RDS ??

69

06:52 PM

Reaction Mechanisms

eg:

Step #1 H2CO2 + H+ H2CO2H+ fast

Step#2 H2CO2H+ HCO+ + H2O slow

Step #3 HCO+ CO + H+ fast

06:52 PM

Reaction Mechanisms

eg:

Overall H2CO2 H2O + CO

Omit H+ - catalyst

Omit H2CO2H+ & HCO+ - reaction intermediates

06:52 PM

Reaction Mechanisms

Reaction Progress

PE

H2CO2 + H+CO + H+

p. 829 # 128

Step 1 H2(g) + NO(g) → H2O(g) + N(g)

Step 2

Step 3 H2(g) + O(g) → H2O(g)

2H2(g) + 2NO(g) → N2(g) + 2H2O(g)

Part 2: Chemical EquilibriumEquilibriumA balancing Act!

Text Ch 13: p 488 - 541

Part 2: Chemical Equilibrium

● All reactions we have done have shown reactants being converted 100% to products

● Many reactions are reversible with some products being converted back to reactants

Part 2: Chemical Equilibrium

● Dynamic equilibrium occurs when 2 opposing processes occur at the same rate

● A chemical equilibrium occurs when two opposing chemical reactions occur at equal rates.

Types of Equilibria

1. Phase EquilibriaAn equilibrium may be established between different phases of a compound in a sealed container

eg. H2O(l) in a sealed container

Types of Equilibria

Initially: H2O(l) changes to H2O(g)

H2O(l) → H2O(g)

Gradually: H2O(g) changes to H2O(l)

H2O(l) ← H2O(g)

Types of Equilibria

Using equilibrium notation:

H2O(l) ⇌H2O(g)

Types of Equilibria

2. Solubility Equilibria occur in saturated solutions when NaCl(s) is placed in water, the initial

rate of dissolving is fast

NaCl(s) NaCl(aq)

Types of Equilibria

as more solid dissolves, the rate of dissolving slows and recrystallization begins.

eg. NaCl(s) NaCl(aq)

Types of Equilibria

when the solution is saturated there are NO VISIBLE CHANGES

At equilibrium, the RATE of dissolving and the RATE of recrystallization are EQUAL.

eg. NaCl(s) NaCl(aq)

equilibrium

Types of Equilibria

3. Chemical Equilibrium Chemical reactions that are reversible usually

result in chemical equilibrium

eg. NO2 gas changing to N2O4

Types of Equilibria

Initially the forward rate is high

eg. 2 NO2(g) N2O4(g)

as more product forms, the reverse reaction begins and increases in rate.

eg. 2 NO2(g) N2O4(g)

Types of Equilibria

eventually the forward rate slows and the reverse rate increases such that the FORWARD AND REVERSE RATES ARE EQUAL

eg. 2 NO2(g) N2O4(g)

http://www.chm.davidson.edu/ronutt/che115/EquKin/EquKin.htm

Conditions for Equilibrium (p. 492)

1. Macroscopic properties are constant

ie. NO OBSERVABLE CHANGE

2. Forward and reverse rates must be equal

3. A CLOSED SYSTEM is required for equilibrium

Conditions for Equilibrium

4. Equilibrium may occur from either directioneg. 2 NO2(g) N2O4(g)

OR N2O4(g) 2 NO2(g)

Kinetics & Equilibrium #4

Shifts in Equilibrium

Equilibrium occurs when the forward rate equals the reverse rate.

Changes in concentration, temperature and pressure/volume can cause the forward or reverse reaction rate to change.

Shifts in Equilibrium

● Eventually, a new equilibrium will be established with different reactant and product concentrations

● Le Châtelier’s Principle (LCP) is used to predict changes in concentrations when a stress (change in C, T, P/V) is applied to a system at equilibrium.

Le Châtelier’s Principle (p.520)

When a stress is applied to a system at equilibrium, the system will adjust or shift to relieve the stress.

– A change(stress) is applied to a system at equilibrium

– Forward or reverse rate will change– New equilibrium established

Le Châtelier’s Principle

1. Changes in concentration

An increase in concentration on one side of an equation favors or drives the reaction to the opposite side.

eg. What will happen if CO is added to this system at equilibrium?

CO(g) + 2 H2(g) --> CH3OH(g)

LCP- the system will ‘shift’ to the right to use the

CO and produce more CH3OH

- some H2 will be used

[CH3OH] will increase

[H2] will decrease

[CO] ‘spikes’ and then drops to a value higher than it was before the change

?? possible graph ??

forward rate increases;reverse rate catches up

mol/L

time

CO

CH3OH

H2

Le Châtelier’s Principle

An equilibrium shifts away from a substance that increases in

concentration

OR

toward a substance that decreases in concentration.

Le Châtelier’s Principle

decrease [CO] → equilibrium shifts to the left

increase [H2] → equilibrium shifts to the right

decrease [CH3OH] → equilibrium shifts to the right

CO(g) + 2 H2(g) → CH3OH(g)

What happens if we:

Le Châtelier’s Principle

IMPORTANT NOTE:

- adding a solid does NOT change molar concentration

- changing the amount of a solid in an equilibrium will NOT cause a shift

Le Châtelier’s Principle

eg.

CaCO3(s) → CaO(s) + CO2(g)

add CaCO3(s) → NO SHIFT

add CO2(g) → Shifts to the left

What happens if we:

Le Châtelier’s Principle 2. Temperature

- Raising the temperature of an exothermic equilibrium favors the formation of reactants.

eg. CO(g) + 2 H2(g) →CH3OH(g) + 65 kJ

What happens if we increase the temperature in this equilibrium?

Le Châtelier’s Principle- Raising the temperature of an endothermic equilibrium favors formation of products.

eg. CaCO3(s) + heat → CaO(s) + CO2(g)

What happens if we increase temperature in this equilibrium?

Le Châtelier’s Principleeg. How would an increase in temperature

affect these equilibria?

N2(g) + 3 H2(g) --> 2 NH3(g) + heat

2 SO3(g) → 2 SO2(g) + O2(g) ΔH = +197 kJ

2 SO3(g) + 197 kJ → 2 SO2(g) + O2(g)

**A change in temperature changes Keq**

Le Châtelier’s PrincipleOR

- Raising the temperature shifts the equilibrium away from the energy term.

- Decreasing the temperature shifts the equilibrium toward the energy term.

CO(g) + 2 H2(g) → CH3OH(g) + 65 kJ

CaCO3(s) + heat → CaO(s) + CO2(g)

Le Châtelier’s Principle

3. Pressure/Volume An increase in pressure of a system at

equilibrium has the same effect as a decrease in the volume of the system.

(inverse relationship)

Le Châtelier’s Principle

Increasing the pressure of a system at equilibrium by reducing volume causes the equilibrium to shift in the direction that reduces pressure

ie. shift to the side with fewer molecules of GAS!!

Note: When the equilibrium is changed in this manner, only GAS atoms/molecules are

considered.

Le Châtelier’s Principle

eg. How would an increase in pressure - caused by a decrease in volume - affect these equilibria?

N2(g) + 3 H2(g) --> 2 NH3(g) (Shift right)

4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g) (Shift left)

Le Châtelier’s Principle

eg. Why would a change in pressure NOT affect the following equilibria?

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

● (equal # of gas atoms/molecules on both sides)

2 Ag(s) + Zn2+(aq) 2 Ag+

(aq) + Zn(s)

●No gas species present

Le Châtelier’s Principle

4. Catalyst- Does NOT cause a shift in equilibrium- Increases BOTH rates equally so

equilibrium is reached faster

5. Surface Area - same as a catalyst

Le Châtelier’s Principle

p. 533 #’s 1 – 3

Answers on p. 537

WORKSHEET #5

Equilibrium Constant (Keq)

For any system at equilibrium, there is a mathematical relationship between reactant and product concentrations

Equilibrium Constant (K)

See p. 495 eg. N2O4(g) --> 2 NO2(g)

Equilibrium Constant (K)

For the general equilibrium below:

aP + bQ → cR + dS

Keq = [R]c [S]d

[P]a [Q]b

Equilibrium Constant (K)

- a, b, c, & d are coefficients used to balance the equation

- P, Q, R, & S are the reactants and products- Kc is sometimes used instead of Keq when

units are molar concentration

Equilibrium Constant (K)

eg. Write the expression for Keq for:

2 SO2(g) + O2(g) → 2 SO3(g)

2 HCl(g) --> H2(g) + Cl2(g)

][O][SO

][SOK

22

2

23

eq

22

eq [HCl]

]H[][ClK 2

Equilibrium Constant (K)

NOTE!!

Solids or liquids ARE NOT included in the Keq or Kc expression because their concentration is constant

Equilibrium Constant (K)

eg. Write the expression for Kc for:

2 NaCl(s) + H2SO4(aq) → 2 HCl(g) + Na2SO4(aq)

CaCO3(s) → CaO(s) + CO2(g) ]SO[H

]SONa[[HCl]K

42

2

eq42

]CO[Keq 2

Interpreting K

Large vs small K values?

● K much larger than 1 - products favored● [products] > [reactants]

● K much smaller than 1 - reactants favored● [reactants] > [products]

Changing K (p. 497)

● For a given system at equilibrium, the value of the equilibrium constant depends only on temperature

ie. For any equilibrium, the only way to change the actual value of K is to change the temperature

Calculations with K

Types of K calculations:

1. Given equilibrium concentrations, find K

2. Find a missing concentration given K and other concentrations.

3. Given initial concentrations and equilibrium data, find K (ICE tables)

1.Given equilibrium concentrations, find K

eg. Calculate Kc for this equilibrium using the equilibrium concentrations given:

H2(g) + I2(g) → 2 HI(g)

[H2] = 0.22 mol/L[I2] = 0.30 mol/L[HI] = 1.56 mol/L

Ex: Given the equilibrium concentrations;

[CO(g)] = 0.105 mol/L,

[H2(g)] = 0.250 mol/L

[CH3OH(g)] = 0.00261 mol/L,

What is the value of Keq for the equilibrium below?

CO(g) + 2 H2(g) → CH3OH(g)

(A) 0.0994

(B) 0.398

(C) 2.51

(D) 10.0

Be Careful!!The other answers are possible if you mess up

the calculation

Ex: Given the equilibrium concentrations below, what is the value of Keq for

N2(g) + O2(g) → 2 NO(g) ?

[N2(g)] = 0.10 mol/L

[O2(g)] = 0.20 mol/L

[NO(g)] = 0.0030 mol/L

(A) 2.2×10−4

(B) 4.5×10−4

(C) 1.5×10−1

(D) 3.0×10−1

Ex: Find Keq for the equilibrium below:

4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g)

[NH3(g) ] = 0.100 mol/L

[O2(g) ] = 0.200 mol/L

[ NO(g) ] = 0.300 mol/L

[ H2O(g) ] = 0.250 mol/L

2. Find a missing concentration . . .

eg. Find the [HI] in the equilibrium below if Kc = 36.9, [H2] = 0.125 mol/L and [I2] = 2.56 mol/L.

H2(g) + I2(g) 2 HI→ (g)

3. ICE tables

● Keq may be calculated given initial concentrations and at least one equilibrium concentration

● Using the Initial concentration and the Change in concentration we can find the missing Equilibrium concentrations and calculate Keq

ICE tables

eg. 4.30 mol of NH3 was placed in a 1.00 L closed container to establish this equilibrium:

2 NH3(g) →N2(g) + 3 H2(g)

Calculate Kc if the equilibrium concentration of H2(g) = 0.500 mol/L

ICE tables

2 [NH3] → [N2] + 3 [H2]

I

C

E

4.30

0.500

By how much does the [H2] change?

0 0

?? ?

? ?

ICE tables

2 [NH3] --> [N2] 3 [H2]

I

C

E

4.30 mol/L 0 0

-2x +x +3x

4.30 - 2x x 3x

0.500 mol/L

[H2] = 0.500 mol/L

3x = 0.500 mol/L x = 0.167 mol/L

[N2] = x

= 0.167 mol/L

[NH3] = 4.30 - 2x

= 3.97 mol/L

K = (0.500)3 x (0.167)

(3.97)2

= 0.00132

ICE tables

eg. 2.00 mol of N2, 4.00 mol of H2 and 3.00 mol of NH3 were allowed to come to equilibrium in a 1.00 L container

2 NH3(g) → N2(g) + 3 H2(g)

Calculate Kc if the equilibrium concentration of NH3(g) = 3.50 mol/L

ICE tables

2 [NH3] --> [N2] 3 [H2]

I

C

E

3.00 2.00 4.00

+2x -3x

3.00 + 2x 2.00 - x 4.00 - 3x

-x

= 3.50

[H2] = 4.00 mol/L – 3x

= 3.25 mol/L

[N2] = 2.00 mol/L - x

= 1.75 mol/L

[NH3] = 3.50 mol/L

3.00 + 2x = 3.50 mol/L

x = 0.25 mol/L

K = (3.25)3 x (1.75)

(3.50)2

= 4.90

ICE tableseg. The oxidation of ammonia occurs

according to the following expression:

4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g)

0.800 mol of each chemical were placed in a 1.00 L container and there was 0.450 mol of NH3 at equilibrium. Calculate the equilibrium concentrations and Kc

4 NH3(g) 5 O2(g) 4 NO(g) 6 H2O(g)

0.800

-4x

0.800 - 4x

0.450 mol/L

0.800

-5x

0.800 - 5x

0.800

+4x

0.800 + 4x

0.800

+6x

0.800+ 6x

[NH3] = 0.450 mol/L

0.800 – 4x = 0.450 mol/L

x = 0.0875 mol/L

[O2] = 0.3625 mol/L

[NO] = 1.15 mol/L

[H2O] = 1.325 mol/L

K = 37000

ICE tables

eg. 2.30 mol of NH3 was placed in a 2.00 L closed container to establish this equilibrium:

2 NH3(g) → N2(g) + 3 H2(g)

Calculate Kc if 25 % of the NH3(g) reacts.

2 [NH3] --> [N2] 3 [H2]

I

C

E

1.15 0 0

-2x +x +3x

1.15 - 2x x 3x

= (0.75 x 1.15) = 0.8625 mol/L

mol/L1.15L2.00

mol2.30

V

nC

[NH3] = 0.8625 → 75% of 1.15

1.15 - 2x = 0.8625 mol/L

x = 0.1438 mol/L

[N2] = x

= 0.1438 mol/L

[H2] = 3x

= 0.4314 mol/L

K = 0.0155

ICE tables

eg. 0.800 mol of NH3 was placed in a 1.00 L closed container to establish this equilibrium:

2 NH3(g) → N2(g) + 3 H2(g)

Calculate Kc if 70% of the NH3 reacts.

→ Kc = 2.88