# Reaction Rates and Equilibrium

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Reaction Rates and Equilibrium
M.Elizabeth 2011

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 Ea1 forward reaction endothermic (∆ H>0), Ea is larger than Ea1 if ∆ H = +200 kJ, then Ea = Ea1 + DH = Ea kJ

Factors that Affect Reaction Rates Collision 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 Theory
Particle 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 Equilibrium At Equilibrium: RATES ARE EQUAL
the concentrations of reactants and products are constant. D [ ]’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, Q
In general, all reacting chemical systems are characterized by their REACTION QUOTIENT, Q. When the system is at equilibrium, 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 Direction H2 + I2 ↔ 2HI Reactant products 2HI ↔ H2 + I2 Left Right

Reversible Reactions H2 + 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 pink Co(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. Inclusion of 4[H2O] assumes additional information that the reactant is hexa-coordinated with water and the product is tetra-coordinated with water. Notice that the net charge on the left and right sides of the first equation is zero. Blue to pink Co(H2O)4Cl2 + 2 H2O  Co(H2O)6Cl2

Properties of an Equilibrium
Pink to blue Co(H2O)6Cl2 → Co(H2O)4Cl H2O Blue to pink Co(H2O)4Cl H2O → Co(H2O)6Cl2 Equilibrium systems are DYNAMIC (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?

Change in Concentration
A + B  C + D Change in Concentration Stress Shift Increase in A or B forward Increase in C or D reverse Decrease in A or B reverse Decrease 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 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 a System at Equilibrium
N2 + 3H2  2NH3 + Heat ΔH = -92 kJ/mol rxn Increase temperature Increase pressure Add a catalyst No Change Adding H2 Removing NH3

Work on Equilibrium Practice Problems

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

Equilibrium Constant N2 (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 For N2(g) + 3 H2(g)  2 NH3(g)
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 Concentration of products is much less
AgCl(s) ↔ Ag+(aq) + Cl (aq) Keq = 1.8 x 10-5 Concentration of products is much less than that of reactants at equilibrium. The reaction strongly favors reactants

The Meaning of Keq AgCl(s) ↔ Ag+(aq) + Cl (aq) Keq = 1.8 x 10-5 Reactant Favored The Reverse reaction Ag+(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 = M, H2 = M, and I2 = M. Calculate the equilibrium constant at this temperature. Write the equilibrium expression (Prod/React) Fill in the values and solve

Calculation of Keq 2HI(g) ↔ H2(g) + I2(g),
Equil

Calculation of Keq Learning Check
N2 + 3H2  2NH3 When equilibrium is established in a L container the equilibrium concentrations are: N2, 3.01 mol; H2, mol; NH3, 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
N2 + 3H2  2NH3 [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 We must convert concentrations to Molarity = mol/L

Calculation of Keq N H2 ↔ 2NH3 Equil

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.