Chapter 18

Chapter 18Rates of Reaction

Collision Theory

The speed of a chemical reactions can vary from instantaneous (strike a match) to extremely slow (coal)

Speed is measured as a change in distance in a given interval of time. Rate = distance/time

Rate is a measure of the speed of any change that occurs within an interval of time.

In chemistry, the rate of chemical change (the reaction rate) is usually expressed as the amount of reactant changing per unit time.

Collision TheoryAccording to the collision theory, atoms, ions, and

molecules can react to form products when they collide with one another, provided that the colliding particles have enough kinetic energy.

Particles lacking the necessary kinetic energy to react, bounce apart unchanged when they collide.

To illustrate the collision theory, If soft balls of clay are thrown together with great force, they will stick tightly together. (analogous to colliding particles of high energy that react)

Balls of clay thrown together gently, don’t stick to one another. (analogous to colliding particles of low energy that fail to react)

Collision Theory

If you roll clay into a rope and begin to shake one end more and more vigorously, eventually it will break.

If enough energy is applied to a molecule, the bonds holding it together can break.

The minimum energy that colliding particles must have in order to react is called the activation energy.

When two reactant particles with the necessary activation energy collide, a new entity called the activated complex may form.

An activated complex is an unstable arrangement of atoms that forms momentarily at the peak of the activation energy barrier.

Activated Complex

Collision TheoryThe lifetime of an activated complex is typically about

10-13 s. The reactants either re-form or the products form.

Both cases are equally likely, thus the activated complex is sometimes called the transition state.

High activation energies explain the slow reaction of some natural substances at room temperature.

The collisions are not great enough to break the bonds, thus the reaction rate is essentially zero or very slow.

Factors Affecting Reaction RatesEvery chemical reaction proceeds at its own rate. Some

fast, some slow under the same conditions.

By varying the conditions of a reaction, you can modify the rate of almost any reaction.

The rate of a chemical reaction depends upon:

• temperature

• concentration

• particle size

• the use of a catalyst.

TemperatureUsually, raising the temperature speeds up reactions,

while lowering the temperature slows down reactions.

At higher temperatures, the motions of the reactant particles are faster and more chaotic than they are at lower temperatures.

Increasing the temperature increases both the frequency of collisions and the number of particles that have enough KE to slip over the activation energy barrier to become products.

An increase in temperature causes products to form faster.

ConcentrationThe number of particles in a given volume affects the rate

at which reactions occur.

Cramming more particles into a fixed volume increases the frequency of collisions.

Increased collision frequency leads to a higher reaction rate.

Particle SizeSurface area plays an important role in determining the

rate of reaction.

The smaller the particle size, the larger the surface area for a given mass of particles.

An increase in surface area increases the amount of the reactant exposed for reaction, which increases the collision frequency and the reaction rate.

One way to increase the surface area of solid reactants is to dissolve them. In solution, particles are separated and more accessible to other reactants.

You can also increase the surface area by grinding it into a fine powder.

CatalystsIncreasing the temperature is not always the best way to

speed up a reaction. A catalyst is often better.

A catalyst is a substance that increases the rate of a reaction without being used up during the reaction.

Catalysts permit reactions to proceed along a lower energy path.

The activation energy barrier for a catalyzed reaction is lower than that of a uncatalyzed reaction.

With a lower activation energy barrier, more reactants have the energy to form products within a given time.

Because a catalyst is not consumed during a reaction, it does not appear as a reactant or product in the chemical equation.

CatalystsEnzymes are biological catalysts that increase the rates

of biological reactions.

For example, without catalysts, digesting protein would take years.

An inhibitor is a substance that interferes with the action of a catalyst.

The inhibitor reduces the amount of functional catalyst available.

Reactions slow or even stop when a catalyst is poisoned by an inhibitor.

End of Section 18.1

Reversible ReactionsA reversible reaction is one in which the conversion of

reactants to products and the conversion of products to reactants occur simultaneously.

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

The double arrow tells you that this reaction is reversible.

When the rates of the forward and reverse reactions are equal, the reaction has reached a state of balance called chemical equilibrium.

At chemical equilibrium, no net change occurs in the actual amounts of the components of the system.

The amount of SO3 in the equilibrium mixture is the maximum amount that can be produced by this reaction under the conditions of the reaction.

Reversible ReactionsChemical equilibrium is a dynamic state.

Both the forward and reverse reactions continue, but because their rates are equal, no net change occurs in their concentrations.

Even though the rates are equal at equilibrium, the concentrations of the components on both side of the equation are not necessarily the same.

The relative concentrations of the reactants and products at equilibrium constitute the equilibrium position of a reaction.

The equilibrium position indicates whether the reactants or products are favored.

Reversible ReactionsA B Product is favored. Equilibrium mixture

contains more product than reactant.

A B Reactant is favored. Equilibrium mixture contains more reactant than product.

In principle, almost all reactions are reversible to some extent under the right conditions.

In practice, one set of components is often so favored that the other set cannot be detected.

If one set of components (reactants) is completely converted to new substance (products), you can say that the reaction has gone to completion, or is irreversible.

Reversible ReactionsYou can mix chemicals expecting to get a reaction but no

products can be detected, you can say that there is no reaction.

Reversible reactions occupy a middle ground between the theoretical extremes of irreversibility and no reaction.

A catalyst speeds up both the forward and the reverse reactions equally.

The catalyst lowers the activation energy of the reaction by the same amount in both the forward and reverse directions.

Catalysts do not affect the amounts of reactants and products present, just the time it takes to get to equilibrium.

Factors Affecting Equilibrium

Changes of almost any kind can disrupt the balance of equilibrium

When the equilibrium of a system is disturbed, the system makes adjustment to restore equilibrium.

The equilibrium position of the restored equilibrium is different from the original equilibrium position.

The amount of products and reactants may have increased or decreased. This is called a shift in the equilibrium system.

Factors Affecting Equilibrium

LeChatelier’s principle states that if a stress is applied to a system in equilibrium, the system changes in a way that reflects the stress.

Stresses that upset the equilibrium include:• Changes to the concentration of reactants or products• Changes to temperature• Changes in pressure

Change in Concentration


Adding a product to a reaction at equilibrium pushes a reversible reaction in the direction of reactants.

Removing a product always pushes a reversible reaction in the direction of products.

Farmers use this to increase yield of eggs

If product are continually removed, the reaction will shift equilibrium to produce more product until the reactants are all used up. (will never reach equilibrium)

Changes in TemperatureIncreasing the temperature causes the equilibrium

position of a reaction to shift in the direction that absorbs heat.

The heat absorption reduces the applied temperature stress.

add heat direction of shift

2SO2 (g) + O2 (g) 2SO3 (g) + heat remove heat direction of shift

Heat can be considered a product, just like SO3.

Cooling, pulls equilibrium to right, and product yield increases. Heating pushed equilibrium to left and product yield decreases.

Changes in PressureA change in pressure affects only gaseous equilibria that

have an unequal number of moles of reactants and products.

add Pressure

direction of shift

N2 (g) + 3H2 (g) 2NH3 (g) reduce pressure direction of shift

When pressure is increased for gases at equilibrium, the pressure momentarily increases because the same number of molecules is contained in a smaller volume.

System immediately relieves some of the pressure by reducing the number of gas molecules.


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

Removing a product always pushes a reversible reaction in the direction of products.

Farmers use this to increase yield of eggs

If product are continually removed, the reaction will shift equilibrium to produce more product until the reactants are all used up. (will never reach equilibrium)

If an external stress is applied to a system at equilibrium, the system adjusts in such a way that the stress is partially offset as the system reaches a new equilibrium position.

Le Châtelier’s Principle

• Changes in Concentration

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

AddNH3

Equilibrium shifts left to offset stress

14.5


• Changes in Concentration continued

Change Shifts the Equilibrium

Increase concentration of product(s) leftDecrease concentration of product(s) right

Decrease concentration of reactant(s)Increase concentration of reactant(s) right

left14.5

aA + bB cC + dD

AddAddRemove Remove


• Changes in Volume and Pressure

A (g) + B (g) C (g)

Change Shifts the Equilibrium

Increase pressure Side with fewest moles of gasDecrease pressureSide with most moles of gas

Decrease volumeIncrease volume Side with most moles of gas

Side with fewest moles of gas

14.5

End of Section 18.2

Equilibrium is a state in which there are no observable changes as time goes by.

Chemical equilibrium is achieved when:

• the rates of the forward and reverse reactions are equal

• the concentrations of the reactants and products remain constant

The Concept of Equilibrium

The Concept of Equilibrium

As the reaction progresses• [A] decreases to a constant,

• [B] increases from zero to a constant.

• When [A] and [B] are constant, equilibrium is achieved.

A B

The Equilibrium Constant• No matter the starting composition of reactants and

products, the same ratio of concentrations is achieved at equilibrium.

• For a general reaction

the equilibrium constant expression is

where Keq is the equilibrium constant. The square brackets indicate the concentrations of the species.

aA + bB(g) pP + qQ

ba

qp

eqKBA

QP

The Equilibrium Constant Expression

For the general reaction:

aA + bB gG + hH

The equilibrium expression is:Each concentration is simply raised to

the power of its coefficient

Products in numerator.

Reactants in denominator.

[G]g[H]h

Kc = [A]a[B]b

N2O4 (g) 2NO2 (g)

= 4.63 x 10-3K = [NO2]2

[N2O4]

aA + bB cC + dD

K = [C]c[D]d

[A]a[B]bLaw of Mass Action

K >> 1

K << 1

Lie to the right Favor products

Lie to the left Favor reactants

Equilibrium Will

14.1

uncatalyzed catalyzed

Catalyst lowers Ea for both forward and reverse reactions.

Catalyst does not change equilibrium constant or shift equilibrium.

• Adding a Catalyst• does not change K• does not shift the position of an equilibrium system• system will reach equilibrium sooner



Change Shift Equilibrium

Change Equilibrium Constant

Concentration yes no

Pressure yes no

Volume yes no

Temperature yes yes

Catalyst no no

14.5

Write the equilibrium expression for Keq for the following reactions:

Write the equilibrium-constant expression, Kc for

Calculation of the Equilibrium Constant

At 454 K, the following reaction takes place:

3 Al2Cl6(g) = 2 Al3Cl9(g)

At this temperature, the equilibrium concentration of Al2Cl6(g) is 1.00 M and the equilibrium concentration of Al3Cl9(g) is 1.02 x 10-2 M. Compute the equilibrium constant at 454 K.

Ionic Compounds Solubility Rules

• Ksp is the equilibrium constant between an ionic solute and its ions in a saturated solution.

• A very small Ksp indicates that only a small amount of solid will dissolve in water.

• Ksp is equal to the product of the concentration of the ions in the equilibrium, each raised to the power of its coefficient in the equation.

• The smaller Ksp the lower the solubility of the compound.

Ksp Solubility Product Constant

Solubility Products

• Ksp is not the same as solubility.

• Solubility is generally expressed as the mass of solute dissolved in 1 L (g/L) or 100 mL (g/mL) of solution, or in mol/L (M).

The Solubility Product ConstantKsp

Example:

AgBr(s) Ag+(aq) + Br-

(aq)Ksp = [Ag+][Br-]

Example: Ca(OH)2(s) Ca2+(aq) + 2OH-(aq)

Ksp = [Ca2+][OH-]2

Example: Ag2CrO4(s) 2Ag+(aq) + CrO4

2-(aq)

Ksp = [Ag+]2[CrO42 -]

What is the concentration of lead ions and chromate ions in a saturated lead chromate solution at 25ºC (Ksp = 1.8 x 10-14)

PbCrO4 (s) Pb2+ (aq) + CrO42- (aq)

Ksp = [Pb2+] x [CrO42-] = 1.8 x 10-14

At equilibrium [Pb2+] = [CrO42-]

Ksp = [Pb2+] x [Pb2+] = 1.8 x 10-14

[Pb2+]2 = 1.8 x 10-14

[Pb2+] = [CrO42-] = 1.8 x 10-14

[Pb2+] = [CrO42-] = 1.3 x 10-7 M

1. NaCl has a solubility of 35.7 g/100 mL. What is the molar solubility and Ksp? (Ans: 6.10 M, 37.2)

2. CaCl2 has a solubility of 74.5 g/100 mL. What is the molar solubility and Ksp? (Ans: 6.71 M, 1.21 X 103)

Ksp and Solubility

Ksp and Solubility

1. A saturated soln of AgCl is found to have an eq. concentration of Ag+ 1.35 X 10-5 M. Calculate Ksp. (Ans: 1.82 X 10-10.)

2. A saturated soln of MgF2 is prepared. At eq. the concentration of Mg2+ is measured to be 0.0012 M. Calculate Ksp. (Ans: 7.0 X 10-9)

Explaining the Common Ion Effect

The presence of a common ion in a solution will lower the solubility of a salt.

• LeChatelier’s Principle:

The addition of the common ion will shift the solubility equilibrium backwards. This means that there is more solid salt in the solution and therefore the solubility is lower

CaF2(s) Ca2+(aq) + 2F-(aq)

a) Add Ca2+ (shifts to reactants)

b) Add F- (shifts to reactants)

The Ksp of silver bromide is 5.0 x 10-13. What is the bromide ion concentration of a 1.00L saturated solution of AgBr to which 0.020 mol of AgNO2 is added?

AgBr (s) Ag+ (aq) + Br- (aq)

Ksp = [Ag+] x [Br-] = 5.0 x 10-13

[Ag+] x [X] = 5.0 x 10-13

X = 5.0 x 10-13 / [Ag+]

X = 5.0 x 10-13 / 0.020

X = 2.5 x 10-11

What is the concentration of sulfide ion in a 1.0 L solution of iron (II) sulfide to which 0.04 mol of iron (II) nitrate is added. The Ksp of FeS is 8 x 10-19

FeS (s) Fe2+ (aq) + S2- (aq)

Ksp = [Fe2+] x [Sr2-] = 8 x 10-19

[Fe2+] x [X] = 8 x 10-19

X = 8 x 10-19 / [Fe2+]

X = 8 x 10-19 / 0.04

X = 2.0 x 10-17

49

Will a ppt form?

Q = Reaction Quotient = product of a concentration of the ions

Q < Ksp Shifts to prod (no ppt)

Q = Ksp Eq. (ppt)

Q > Ksp Shifts to reac.(ppt)

50

Will a ppt form if a solution is made from 0.50L of 0.002M Ba(NO3)2 and mix it with 0.50L of 0.008M Na2SO4? Ksp of BaSO4 is 1.1 x 10-10

Q = [Ba(NO3)2] x [Na2SO4] = (0.002/2)(0.008/2)

Q = 4 x 10-6

A precipitate will form because 4 x 10-6 is larger than 1.1 x 10-10

Will a ppt form?

End of Section 18.3

End of Chapter 17

Education

Chapter 18