Reaction Rates & Equilibrium

Unit 12 - Chapter 18

Reaction Rate

• Reaction rate – how fast reactants disappear and how fast product appears

A B

13.1

rate = -[A]

t

rate = [B]

t

time

Reaction Rate• Reaction Rate = ∆ [A]

∆ t• Example:

CO(g) + NO2(g) CO2(g) + NO(g)

- at t = 4.0 min, [CO2] = .12 mol/L

- at t = 8.0 min, [CO2] = .24 mol/L- reaction rate = .24 mol/L - .12 mol/L

8.0 min – 4.0 min = 0.030 mol/L . min

• Unit for reaction rate = conc. with some time unit• Products have a (+) rate• Reactants have a (-) rate

Collision Theory of Kinetics• Kinetics is the study of the factors that affect the speed

of a reaction and the mechanism by which a reaction proceeds.

• In order for a reaction to take place, the reacting molecules must collide into each other.

• Once molecules collide they may react together or they may not, depending on two factors -Whether the collision has enough energy to "break the bonds

holding reactant molecules together";Whether the reacting molecules collide in the proper

orientation for new bonds to form.

Effective Collisions• Collisions in which these two conditions are met

(and therefore the reaction occurs) are called effective collisions.

• The higher the frequency of effective collisions the faster the reaction rate.

• When two molecules have an effective collision, a temporary, high energy (unstable) chemical species is formed - called an activated complex It is a transition state between reactant and product It has a very short lifetime (10-13 s)Has to form for product to be formed

Activated Complex• The difference in potential energy between the reactant

molecules and the activated complex is called the activation energy, Ea

• This is the minimum amount of energy that particles must have in order to react.

• The larger the activation energy, the slower the reaction• The energy to overcome the activation energy comes

from the kinetic energy of the collision being converted into potential energy, or from energy available in the environment, i.e. heat.

• Different reactions have different activated complexes and therefore different activation energies

Energy Diagram

• Energy of products is lower than energy of reactantsenergy lost, exothermic, -∆H

• Energy of products is higher than energy of reactantsenergy gained, endothermic, +∆H

What is this called?

Factors Affecting Reaction Rate1. Nature of the Reactants

Cl2(g) + CH4(g) CH3Cl(g) + HCl(g)

• Cl2 Cl + Cl (fast)• Cl + CH4 CH3Cl + H (slow)• H + Cl HCl (very fast)

individual steps = elementary steps all steps together = reaction mechanism the slowest step determines the rate of the reaction

• called the rate determining step• Intermediates – product in one step, reactant in another

Factors Affecting Reaction Rate2. Concentration• The larger the concentration of reactant molecules,

the faster the reaction will go. Increases the frequency of reactant molecule

collisions

3. Particle Size (Surface Area) more particles on the surface = more particles available for

collisions more collisions = more act. complex = more product smaller particles give you more surface area

Factors Affecting Reaction Rate4. Agitation

this puts more liquid/gas particles in contact with the solid = ↑ collisions = ↑ act. complex = ↑ product

5. Pressure↑ pressure by ↓ volume – puts particles closer together

= ↑ collisions = ↑ act. complex = ↑ product

• All of these factors are similar, in terms of explanation, to concentration!!!

Factors Affecting Reaction Rate

6. Temperaturemost effective at speeding up a reaction↑ temp. = ↑ KE (particles moving faster)particles move faster leading to more collisionsthe collisions are also harderthese harder collisions contain the needed

energy to overcome the Ea

therefore the reaction rate will increase

Factors Affecting Reaction Rate

7. Catalystsubstance that speeds up a reaction, but isn’t

used up in the reactionprovides a “different pathway” that requires

lower Ea

lower Ea = more collisions having the proper amount of energy = ↑ act. complex = ↑ product

Reaction Dynamics

• If the products of a reaction are removed from the system as they are made, then a chemical reaction will proceed until the limiting reactants are used up.

• However, if the products are allowed to accumulate; they will start reacting together to form the original reactants - called the reverse reaction.

• We show this reverse reaction by using a double arrow (H2(g) + I2(g) 2HI(g))

Reaction Dynamics

• The forward reaction slows down as the amounts of reactants decreases because the reactant concentrations are decreasing

• At the same time the reverse reaction speeds up as the concentration of the products increases.

• Eventually the forward reaction is using reactants and making products as fast as the reverse reaction is using products and making reactants. This is called chemical equilibrium. rateforward = ratereverse Note: This equilibrium is dynamic

Chemical Equilibrium• Dynamic Equilibrium can only be reached in a

closed system!!• When a system reaches equilibrium, the amounts

of reactants and products in the system stays constant the forward and reverse reactions still continue, but

because they go at the same rate the amounts of materials don't change.

• There is a mathematical relationship between the amounts of reactants and products at equilibrium

Equilibrium Expression

• Capital letters (A,B,C,D) – reactants or products• Lowercase letter (a,b,c,d) – coefficients from the equation• NOTE – products on top, reactants on bottom• In this expression, Keq is a number called the equilibrium

constant. ratio of product concentration to reactant concentration at

equilibrium• Do not include solids or liquids, only solutions and gases• The value of Keq depends on temp. of the reaction – if

temp. changes then the value of Keq changes.

aA + bB cC + dD = [C]c[D]d

[A]a[B]bKeq

Example – Determine the value of the Equilibrium Constant for the Reaction 2 SO2(g) + O2(g) 2 SO3(g)

Determine the Equilibrium Expression

Plug the equilibrium concentrations into to Equilibrium Expression

Solve the Equation3.503.00SO3

1.251.50O2

1.502.00SO2

[Equilibrium][Initial]Chemical

4.36 1.25)((1.50)

(3.50)

]O[][SO

][SO K

2

2

22

2

23

So what does this Keq value tell us???

Position of Equilibrium

• The size of the equilibrium constant shows whether products or reactants are favored at equilibrium.

• Keq > 1, products are favored at equilibrium

• Keq < 1, reactants are favored at equilibrium

Determine the Equilibrium Expression

Plug the equilibrium concentrations and Equilibrium Constant into the Equilibrium Expression

Solve the Equation?3.00SO3

1.251.50O2

1.502.00SO2

[Equilibrium][Initial]Chemical

3.51 12.3 ][SO

12.3 (1.25)x (1.50)x 4.36 ][SO

4.36 1.25)((1.50)

][SO

]O[][SO

][SO K

3

223

2

23

22

2

23

Example – If the value of the Equilibrium Constant for the Reaction 2 SO2 + O2 2 SO3 is 4.36,

Determine the Equilibrium Concentration of SO3

More Equilibrium Practice1. Write the equilibrium constant expression for the

following reaction.

3H2(g) + N2(g) ↔ 2NH3(g)

2. An analysis of an equilibrium mixture for this reaction in a 1.0 L flask at 300oC gave the following results: 0.15 mol H2, 0.25 mol N2 and 0.10 mol NH3. Calculate the Keq for this reaction.

3. 2BrCl(g) ↔ Cl2(g) + Br2(g)

The equilibrium constant for this reaction is 11.1. The equilibrium mixture contains 4.00 mol Cl2 and 4.00 moles of Br2. How many moles of BrCl are present?

Le Châtelier’s Principle

• Le Châtelier's Principle guides us in predicting the effect various changes have on the position of equilibrium

• it says that if stress is applied to a system in dynamic equilibrium, the system will change to relieve the stress.The position of equilibrium moves to counteract the

change.• Three common stressors:

ConcentrationTemperaturePressure

Concentration Changes and Le Châtelier’s Principle

• A + B ↔ C + DAdding a reactant – equilibrium shifts rightRemoving a reactant – equilibrium shifts leftAdding a product – equilibrium shifts leftRemoving a product – equilibrium shifts right

• Only affects a reaction involving gases with an unequal number of mole of reactants & products.

• Increasing the pressure on the system causes the position of equilibrium to shift toward the side of the reaction with the fewer gas molecules

• Decreasing pressure causes a shift toward the side with more gas molecules

• Example3H2(g) + N2(g) 2NH3(g) + 92kJ

↑ Pressure – shifts to the right↓ Pressure – shift to the left

Changing Pressure and Le Châtelier’s Principle

• Increasing the temperature causes the reaction to shift away from the heat.

• For exothermic reactions -Think of heat as a product of the reactionTherefore shift the position of equilibrium toward the

reactant side • For endothermic reactions -

Think of heat as a reactantThe position of equilibrium will shift toward the products

• Cooling an exothermic or endothermic reaction will have the opposite effects.

Changing Temperature and Le Châtelier’s Principle

Examples – Le Chatelier’s PrincipleWhat effect do the following changes have on the equilibrium position for the following reaction?1. PCl5(g) + heat ↔ PCl3(g) + Cl2(g)

a. addition of Cl2

b. increase in pressure

c. removal of heat

d. removal of PCl3 as formed

2. C(s) + H2O(g) + heat ↔ CO(g) + H2(g)

a. Lowering the temperature

b. Increasing the pressure

c. Removal of H2 as formed

Examples – Le Chatelier’s Principle

3. At 425 K –

Fe3O4(s) + 4H2(g) ↔ 3Fe(s) + 4H2O(g)

How would the equilibrium concentration of H2O be affected by the following:

a. Adding more H2

b. Adding more Fe(s)

c. Decreasing the pressure

d. Adding a catalyst