Reaction Rates and Equilibrium M.Elizabeth 2011. Collision Theory Used to Explain Reaction Rates Atoms, ions, and molecules can form a chemical bond when.

Reaction Rates and EquilibriumM.Elizabeth2011

Collision Theory Used to Explain Reaction Rates

•Atoms, ions, and molecules can form a chemical bond when they collide, provided the particles have enough kinetic energy.

•Particles lacking the necessary kinetic energy to react still collide, but simply bounce apart.

•Activation energy - the minimum energy colliding particles must have in order to react.

Chemical Reactions•ordinarily occur as a result of collisions between

reacting particles. Consider the reaction: • CO(g) + NO2(g) ----> CO2(g) + NO(g) rate = k(conc CO)(conc

NO2(g))

▫ doubling the [CO], holding [NO2] constant, the number of collisions in a given time doubles.

▫ doubling the [NO2] , holding CO constant, has the same effect.

• the number of collisions per unit time is directly proportional to the concentration of CO or NO2.

•The fact that the rate is directly proportional to these concentrations indicates that reaction occurs as a direct result of collisions between CO and NO2 molecules.

NOT EVERY COLLISION LEADS TO REACTION!!!!!

• It is possible to calculate the rate at which molecules collide with each other by using the kinetic theory. Consider a mixture of CO and NO2 at 700 K and a concentration of 0.10 mol/L

▫every molecule would collide with about a billion other molecules in one second!

▫if every collision resulted in a reaction, then the whole mixture would be reacted in a fraction of a second.

▫the actual reaction takes about 20 seconds.

Effective Collisions• In order for collisions to be effective, there

must be considerable force in the collisions. The slower moving molecules do not have enough kinetic energy to react when they collide...they bounce off one another and retain their identity.

•Only those molecules moving at high speed have enough energy for collisions to result in a reaction.

•Every reaction requires a certain minimum energy for the reaction to occur--it is called activation energy, Ea, and is expressed in kJ.

Activated Complex-an unstable, high energy species • forward reaction

exothermic (∆H<0), Ea is smaller than Ea

1

• forward reaction endothermic (∆ H>0), Ea is larger than Ea

1

• if ∆ H = +200 kJ, then Ea = Ea

1 + DH = Ea

1 + 200 kJ

Factors that Affect Reaction RatesCollision Theory 1. Temperature 2. Concentration 3. Particle size 4. Catalyst

Factors that Affect Reaction Rates Collision Theory

Temperature •Increasing temperature increases the number of particles that have enough kinetic energy to react when they collide. Concentration changes (amt per vol)•Cramming more particles into a fixed volume increases the collision frequency.

Factors that Affect Reaction Rates Collision TheoryParticle Size •the smaller the particle size, the larger the surface area for a given mass of particles. Decreasing particle size will increase the rate of reaction. Catalyst •A catalyst is a substance that increases the rate of a reaction without being used up itself in the reaction.

Effects of Catalyst on Activation Energy •Enzymes are

biological catalyst usually made of proteins.

•Speed reactions by lowering the activation energy of the reaction.

Chemical Equilibrium

Dynamic (in constant motion) Reversible

•Chemical equilibrium occurs when the forward and reverse reaction are taking place at the same rate. •There is no net change in the actual amounts of the components of the system.

Equilibrium Example

Rate vs EquilibriumAt Equilibrium:

RATES ARE EQUAL• the concentrations

of reactants and products are constant.

[ ]’s = 0•The forward and

reverse reactions continue after equilibrium is attained.

Kinetics Reaction Rate Orders• the order of reaction with respect to a certain

reactant, is defined as the power to which its concentration term in the rate equation is raised.

•For example, 2A + B → C▫r = k[A]2[B]1 the reaction order with respect to A

would be 2 and with respect to B would be 1, the total reaction order would be 2 + 1 = 3.

•Reaction orders can be determined only by experiment. The reaction order is not necessarily related to the stoichiometry of the reaction, unless the reaction is elementary. Complex reactions may or may not have reaction orders equal to their stoichiometric coefficients

The Reaction Quotient, QIn general, all reacting chemical systems are In general, all reacting chemical systems are characterized by their characterized by their REACTION QUOTIENT, REACTION QUOTIENT, QQ..

When the When the system is at system is at equilibriumequilibrium, Q = K, Q = K

Reactions Review•“Systems”: two reactions that differ only in

direction

•Any reversible reaction H2 + I2 ↔ 2HI

noted by the double arrow; ↔Two reactions: only difference is the

DirectionH2 + I2 ↔ 2HI

Reactant products 2HI ↔ H2 + I2

Left Right

Reversible ReactionsH2 + I2 ↔ 2HI

• the products may react back to original reactants.

•“closed system”: ONLY if all reactant are present

• If one piece is completely gone it has ”gone to competition” and no longer reversible

• Examples: Reversible Reactions.• Unopened Soda• Breathing• Rechargeable batteries • Color changing shirt

Blue to pinkCo(H2O)4Cl2 + 2 H2O Co(H2O)6Cl2

The product is hexa-coordinated with water and the reactant is tetra-coordinated with water. Notice that the net charge on the left and right sides of the first equation is zero.

Properties of an Equilibrium

Pink to blueCo(H2O)6Cl2 → Co(H2O)4Cl2 + 2 H2 H22OO

Blue to pinkCo(H2O)4Cl2 + 2 H2O → Co(H2O)6Cl2

Equilibrium systems areDYNAMIC (in constant motion)REVERSIBLE can be approached from either direction

Reaction at Equilibrium

A and B are _________. C and D are _______.

Factors Affecting Equilibrium•Changes in temperature, pressure, and concentration affect equilibrium.

•The outcome is governed by LE CHÂTELIER’S PRINCIPLE“...if a system at equilibrium is disturbed, the system tends to shift its equilibrium position to counter the effect of the disturbance.”

Writing and Manipulating Keq Solids NEVER appear in equilibrium

expressions.

S(s) + O2(g) ---> SO2(g)

Liquids NEVER appear in equilibrium expressions.

NH3(aq) + H2O(liq) ---> NH4+(aq) + OH-(aq)

Henri Le Châtelier

•Henri Le Châtelier•1850-1936•Studied mining

engineering.•Interested in glass

and ceramics.

A + B C + D

Change in Concentration

•What happens when there is an increase in reactant or product? •What happens when there is a decrease in reactant or product?

A + B C + D

Change in ConcentrationStress ShiftIncrease in A or B forwardIncrease in C or D reverseDecrease in A or B reverseDecrease in C or D forward

Changes in Pressure

Only affects equilibrium system with an unequal number of moles of gaseous reactants and products •Decrease in pressure shifts the reaction in the direction that produces the larger number of moles of gas. •An increase in pressure shifts the reaction in the direction that produces the smaller number of moles of gas.

Product or Reactant Favored?

•Keq greater than 1 Products favored•Keq less than 1 Reactants favored

Endothermic vs Exothermic

•Exothermic - Heat is a product. (-ΔH)

•Endothermic - Heat is a reactant. (+ΔH)

Changes in Temperature

•An increase in temperature favors endothermic reactions.

•A decrease in temperature favors exothermic reactions.

Changes in Temperature

•+ΔH = Endothermic↑ temp favors the forward rxn.

•-ΔH = Exothermic↑ temp favors the reverse rxn.

Temperature Effects on Equilibrium

N2O4 (colorless) + heat 2 NO2 (brown)

Equilibrium and Catalysts•Add catalyst = no change equilibrium concentration

•A catalyst only affects the RATE of approach to equilibrium

Catalytic exhaust system.

Applying a Stress to aSystem at Equilibrium

•N2 + 3H2 2NH3 + HeatΔH = -92 kJ/mol rxn

•Increase temperature

•Increase pressure

•Add a catalyst No Change

•Adding H2

•Removing NH3

Work on Equilibrium Practice Problems

The Equilibrium ConstantFor any type of chemical equilibrium of the type: a A + b B ↔ c C + d DKeq is a CONSTANT (at a given Temp)

Equilibrium ConstantN2 (g) + O2 (g) ↔ 2NO (g)

Equilibrium Constant

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

Magnitude of Keq

•Varies only with temperature

•Is constant at a given temperature

•Is independent of the initial concentrations

Magnitude of Keq

A + B ↔ C + D•Keq > 1 mostly products•Keq < 1 mostly reactants•Keq ~ 1 equal amounts of products and reactants

The Meaning of Keq•Keq is used to tell if a reaction will favor

products or reactants.

•For N2(g) + 3 H2(g) 2 NH3(g)

•Concentration of products is much greater than that of reactants at equilibrium.

•The reaction strongly favors products

The Meaning of Keq•AgCl(s) ↔ Ag+(aq) + Cl (aq)•Keq = 1.8 x 10-5

Concentration ofproducts is much lessthan that of reactants atequilibrium.The reaction stronglyfavors reactants

The Meaning of Keq

AgCl(s) ↔ Ag+(aq) + Cl (aq)Keq = 1.8 x 10-5

Reactant Favored

The Reverse reactionAg+(aq) + Cl-(aq) ↔ AgCl(s)Keq = 1.8 x 105

is product-favored.

Calculation of Keq – Learning Check

•For the reaction 2HI(g) ↔ H2(g) +I2(g), at 448•C,

•The equilibrium concentrations are ▫HI =0.0040 M, ▫H2 =0.0075 M, and ▫I2 = 0.000043 M.

•Calculate the equilibrium constant at this temperature.▫Write the equilibrium expression

(Prod/React)▫Fill in the values and solve

Calculation of Keq2HI(g) ↔ H2(g) + I2(g),

•Equil 0.0040 0.0075 0.000043

Calculation of Keq Learning Check

N2 + 3H2 2NH3

•When equilibrium is established in a 5.00 L container the equilibrium concentrations are: N2, 3.01 mol; H2, 2.10 mol; NH3, 0.565 mol.•Calculate Keq

▫Write the equation for Keq▫Calculate molar concentration (M)▫Fill in and solve.

Calculation of Keq

•[N2] = 3.01 mol/ 5.00L•[H2] = 2.10 mol/ 5.00L•[NH3] = .565 mol/ 5.00L

•[N2] = .602 M•[H2] = .420 M•[NH3] = .113 M

N2 + 3H2 2NH3

Calculation of Keq

N2 + 3H2 ↔ 2NH3•Equil 0.602 0.420 0.113

Uses of Keq

We can us the Keq to find equilibrium concentrations at that temperature.

Factors that Affect Equilibria•Once a reaction has reached equilibrium, it remains at equilibrium until it is disturbed by some change in conditions.