Unit 9- Kinetics and Equilibrium

• Activated complex (transition state)

• Activation energy• Catalyst• Enthalpy• Entropy• Equilibrium• Le Chatelier’s

principle

• Potential energy diagram

• Rate• Stress

Kinetics

– Chemistry concerned with the rates of reaction• Many factors affect how quickly a reaction will take place• In order for a reaction to take place- particles must collide

– Collision theory- in order for a Rx to occur, particles must collide with enough E and the right orientation• (Think about shooting pool)

Factors that affect Reaction Rates(change # of effective collisions)

• 1- nature of reactants- (ionic vs. covalent)– Covalent- slower- due to greater number of bonds

that need to be broken• More bonds= more E when collide to break • The fewer electrons that need to be rearranged, the faster the

reaction is

• 2- concentration-– More reactants= more collisions= faster Rx• Ex: burn paper in 20% O2 or 100% O2

• 3- surface area-– More SA = more chance of collision= increased

rate of Rx

• 4- pressure- (in gases)– Increase P= increase collisions= increased rate of

Rx

• 5- temperature-– Increase T= increase collisions (b/c particles have

high KE) = increased rate of Rx

• 6- catalyst-– Catalyst- substance that provides a different or

easier pathway for the Rx– Increases rate

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Reaction Coordinate

Activation Energy with CatalystActivation Energy with Catalyst

Activation Energy without Catalyst

Activation Energy without Catalyst

Potential E diagrams• Shows ΔPE during a chemical Rx• Axes- – Vertical- ΔPE – Horizontal- reaction coordinate- progress of Rx

• For Rx to occur:– Must have effective collisions– As particles move closer; KE is converted to PE

• Activation E- E required to form activated complex

• Activated complex- temporary, intermediate that will break apart to form new products or reform reactants

• Heat of Rx- ΔH- difference between PE of the reactants and products– Endothermic- ΔH is + (heat is absorbed)– Exothermic- ΔH is – (heat is released)– Table I in Reference Tables

Spontaneity in chemical reactions

• Spontaneous reaction- can proceed without applying external forces– Ex: H2O(s) H2O(l) • Spontaneous at 25OC and 1 atm• Not spontaneous at -25OC and 1 atm

– Ex: H2O(l) H2O(s) • Spontaneous at -25OC and 1 atm• Not spontaneous at 25OC and 1 atm

• Entropy and enthalpy determine spontaneity

Enthalpy

• Tendency to change to a state of lower E (decrease in PE)

• Exothermic rx’s because E in reactants is released (products have less E)

• Negative ΔH (exothermic) is favored

Entropy• Measure of disorder/randomness• Nature’s tendency- towards more disorder• Greater disorder = higher entropy• Ex of entropy change:– S L G (phase change)

– Compound free elements (chemical change)• Generally the side of the eqn with more

molecules has greater entropy

Equilibrium• When forward and reverse rx’s occur at the

same rate• Reactions that can go both ways are

represented by ↔ (or 2 arrows each pointing a different way)

• Can only occur if reactants and products can’t leave the system

Equilibrium con’t• Dynamic process• *at equilibrium- the quantities of reactants

and products aren’t equal; it’s the rate of the 2 rx’s that are equal

• 2 types- physical and chemical

Physical Equilibrium• Phase- – exists btwn solid and liquid• At melting/freezing pt

– In closed system- exists between liquid and gas• At evaporation/condensation pt

• Solution-– In a saturated solution- rate of

crystallization/dissolving is equal– In a closed system btwn gas and dissolved gas in a

liquid• * ΔT (increase) = more solid dissolved in soln;

less gas dissolved in a liquid

• Consider the two sodabottles to the right:o CO2 molecules are moving

randomly in all directions.o Some CO2 molecules are

moving out of the sodainto the space above.

o Other CO2 molecules arereturning to the soda fromthe space above.

• Which bottle goes flat faster, A or B?

Lessons in a soda bottle

a B

• At first, CO2 is only moving out of the soda into the air space above the liquid.o This is because, initially, all the CO2 is in the soda.

There is no CO2 in the air space above.• As the amount of CO2 in the space above the

soda increases, so does the likelihood (probability) that CO2 molecules will return to the soda.

• This means the speed with which CO2 returns to the soda increases as the amount of CO2 in the air space above the soda increases.

An Explanation (Part 1)

• As the amount of CO2 in the space above the soda increases, the amount of CO2 dissolved in the soda decreases.

• As the amount of CO2 dissolved in the soda decreases, so does the likelihood that CO2 molecules will leave the soda.

• This means the speed with which CO2 leaves the soda decreases as the amount of CO2 in dissolved in the soda decreases.

An Explanation (Part 2)

• Eventually, the speed at which the CO2 leaves the soda becomes equal to the speed at which the CO2 returns to the soda, and both processes proceed at the same rate.

• When two opposing processes occur at the same rate, the system is said tobe in equilibrium.

• Since air the space in bottle B is smaller, the CO2 molecules in the space are more crowded (concentrated) and equilibrium is reached sooner.

• As a result, bottle A goes flat faster.

An Explanation (Part 3)

Sodas discuss equilibrium.

Chemical Equilibrium

• Equilibrium occurs when forward and reverse reactions occur at equal rates.

A + B C + D• At equilibrium, macroscopic or observable

changes no longer occur (color, temperature, pressure, etc.)

Temperature affects equilibrium• Examine the reaction coordinate below.

o The activation energy of thereverse reaction is higher than that of the forward reaction.

o As a result, temperature changes effect the rates of the forward and reverse reactions differently, changing equilibrium.

o Increasing the temperature favors endothermic reactions because they need more energy for effective collisions than exothermic reactions do.

o Decreasing the temperature favors exothermic reactions.

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Reaction Coordinate

Le Châtelier’s Principle

• There are many factors that affect (stress) the equilibrium of a system

• Le Châtelier’s principle explains how a system at equilibrium responds/relieves these stresses

Collision Theory Analysis

• The reaction SO2(g) + NO2(g) = SO3(g) + NO(g)

is at equilibrium. What effect will addition of SO3(g)have? Why?o Addition of SO3(g) will cause more SO2(g) and NO2(g)

to form, while the amount of NO(g) decreases.o This is because addition of SO3(g) increases the

number of collisions among product molecules.• This example as well as many others can be

analyzed more simply by Le Chatelier’s Principle.

• Imagine you squeeze a balloon on one side.• Air moves toward the other side of the balloon

causing the it to bulge.

• As a result, the pressure is reduced on the side where you are squeezing.

• The air moves in a way that relieves the stress caused by increased pressure.

Stress Relief

Change in Concentration• Shift due to increase in concentration of a reactant.

o If the concentration of a reactant is increased, the reaction will shift in a way that reduces it’s concentration.

• Shift due to decrease in concentration of a product.o If the concentration of a product is decreased, the reaction will

shift in a way that increases it’s concentration.

• Increase conc. 4NH3 (g) + 5O2(g) ↔ 4NO(g) + 6H2O(g) + heat

↑NH3

(what happens to everything else?)

• Decrease conc.4NH3 (g) + 5O2(g) ↔ 4NO(g) + 6H2O(g) + heat

↓O2

( what happens to everything else?)

Example:

Change in Temperature

• Shift due to an increase in temperature.o If the temperature increases, the reaction will shift

in a way that uses heat.

• Shift due to a decrease in temperature.o If the temperature decreases, the reaction will

shift in a way that releases heat.

Example:• Increase temp.4NH3 (g) + 5O2(g) ↔ 4NO(g) + 6H2O(g) + heat

↑heat

• Decrease temp.4NH3 (g) + 5O2(g) ↔ 4NO(g) + 6H2O(g) + heat

↓heat

Change due to Pressure (gases only!!)• CO2(g) ↔ CO2(aq) – If ↑P shifts right (more dissolved)– If ↓P shifts left (more gas) *when soda bottle

opens• If both sides are gases: N2(g) + 3H2(g) ↔ 2NH3(g)

- ↑P favors side with less molecules ()- ↓P favors side with more molecules ()

• If both sides have the same # of molecules:H2(g) + Cl2(g) ↔ 2HCl (g)

- Pressure has no effect

Change due to Catalyst

• Affects forward and reverse reaction– Equilibrium is reached more quickly

Equilibrium Analysis

• The reaction N2(g) + 3H2(g) = 2NH3(g) is at equilibrium.• Nitrogen and hydrogen are added to the reaction

container:o What happens to the concentration of ammonia?

It increases.o How do the concentrations of reactants and products at

the new equilibrium compare to those at the original equilibrium?

o How does the ratio of the concentrations of the reactants to products at the new equilibrium compare to those at the original equilibrium?

They’re both higher.

It is the same. (This is predicted by Le Chatelier’s principle .)

Equilibrium Expression

• Fraction of concentrations of reactants and products– Expressed as moles/L– Equals Keq (equilibrium constant)

Steps to write equilibrium expression• 1- write balance equation N2(g) + 3H2(g) ↔ 2NH3(g) + heat• 2- products as numerators; reactants as denominators• 3- put brackets around each formula, brackets represent

molar concentration• 4- write coefficients of each substance as powers of the

concentration • 5- set it equal to Keq

• **only ΔT will affect Keq

• When Keq is large- products are favored

• When Keq is small- reactants are favored

[NH3]2 [N2][H2]3 Keq =

• Step 1: Write a balanced equation.N2(g) + 3H2(g) = 2NH3(g)

• Step 2: Write the equilibrium expression.

• Step 3: Substitute values into the equilibrium expression and solve.

Solving Equilibrium ExpressionsA reaction vessel contains 0.10 M nitrogen gas and 0.10 M hydrogen gas in equilibrium with ammonia? What is the concentration of the ammonia?

[NH3]2 [N2][H2]3 6.7 × 105 =

[NH3]2 (0.10M)(0.10M)3 6.7 × 105 =

67= [NH3]2

[NH3] = 8.2M