Copyright © by Holt, Rinehart and Winston. All rights reserved. Ch 17 and 18 reaction kinetics and equilibrium
Copyright © by Holt, Rinehart and Winston. All rights reserved. Warm up 1. What is Enthalpy? 2. What is reaction equilibrium? Turn in warm ups
Copyright © by Holt, Rinehart and Winston. All rights reserved. Reaction Kinetics: Ch. 17 Understanding chemical reactions that occur at different rates
Copyright © by Holt, Rinehart and Winston. All rights reserved. reaction kinetics: area of study concerned with reaction rates and mechanics reaction rate: change in concentration of reactants per unit time during a reaction. units: M/s Rates are measured in a unit of something per time interval (Molarity/seconds) Br 2 (aq) + HCOOH (aq) 2Br - (aq) + 2H + (aq) + CO 2 (g) time
Copyright © by Holt, Rinehart and Winston. All rights reserved. Properties used to measure reaction rates Rates are measured by the rate of formation of a product or disappearance of a reactant How concentration of reactants or products changes over time. Ex: observe change in color. Measure pressure change use gas laws to calculate the concentrations. Measure temperature change
Copyright © by Holt, Rinehart and Winston. All rights reserved. As the concentration of the reactants decreases over time, the product concentration increases General Equation for the Rate of Reaction 2Br − (aq) + H 2 O 2 (aq) + 2H 3 O + (aq) → Br 2 (aq) + 4H 2 O(l)
Copyright © by Holt, Rinehart and Winston. All rights reserved. A B rate = - [A] tt rate = [B] tt time Chapter 17Which molecules are the reactant and which are the products?
Copyright © by Holt, Rinehart and Winston. All rights reserved. Collision Theory: In order for reactions to occur between substances, their particles must collide with enough energy and in the correct orientation. Reactant molecules must collide with a favorable orientation and with enough energy to merge the valence electrons and disrupt the bonds of the molecules to form to the products. Number of collisions per unit time determines how fast a reaction can take place.
Copyright © by Holt, Rinehart and Winston. All rights reserved. Particle Collisions
Copyright © by Holt, Rinehart and Winston. All rights reserved. 1. Surface area 2. Temperature 3. Concentration 4. Presence of a catalyst 4 Major Rate-Influencing Factors Chapter 17
Copyright © by Holt, Rinehart and Winston. All rights reserved. An increase in surface area increases the rate of reactions. Because the reaction rate depends on the area of contact of the reaction substances. 1. Surface Area: An increase in temperature increases the reaction rate since the average kinetic energy of the particles increases; greater number of effective collisions. 2. Temperature
Copyright © by Holt, Rinehart and Winston. All rights reserved. Generally, an increase in the concentration of one or more of the reactants will increase the reaction rate The reaction rate depend on concentrations of the reacting species. example: A substance that oxidizes in air (18% O 2 ) oxidizes more vigorously in pure oxygen. 3. Concentration
Copyright © by Holt, Rinehart and Winston. All rights reserved. A catalyst will lower amount of energy required for a reaction to take place (activation energy) Ex: enzymes, shaking, another chemical, etc 4. Presence of a Catalyst E a is the initial input of energy needed to overcome the repulsion forces between molecules as come close together
Copyright © by Holt, Rinehart and Winston. All rights reserved. A catalyst lowers the energy barrier and the reaction proceeds at a fast rate
Copyright © by Holt, Rinehart and Winston. All rights reserved. [Br 2 ] Absorption Endothermic rxn: overall absorption of energy: ΔH of reaction is +, ΔH of environment is - (feels cold) Exothermic rxn: Overall release of energy: ΔH is positive (feels hot)
Copyright © by Holt, Rinehart and Winston. All rights reserved. Using a calorimeter to calculate energy change
Copyright © by Holt, Rinehart and Winston. All rights reserved. ΔH soln or heat of solution: enthalpy change associated with the process of a solute dissolving in a solvent ΔH soln = J/g (Joules/gram solute amount of heat change (q), can be calculated using a calorimeter and the equation: q= m × s × ΔT Joules= unit of energy s=4.18J/g °C m = total mass of the solution (solute plus solvent), S = specific heat constant of the solution ΔT = observed temperature change ex: 5.0g solute dissolved in 10g water with a temp change of 12.0 °C. What is the heat of solution?
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end chapter 17
Copyright © by Holt, Rinehart and Winston. All rights reserved. Warm ups #5 (May-June) 1. What units are used to represent energy? 2. If 10grams of solute is dissolved in 20ml of water and the change is temperature is -15 C, what is the heat of the solution? q=m solution x s x ΔT (s=4.18J/gC) 3. What is the enthalpy of the solution? ΔH solu = J/g solute 4. Is this an exothermic or endothermic rxn? New warm ups
Copyright © by Holt, Rinehart and Winston. All rights reserved. Chemical Equilibrium: Ch 18
Copyright © by Holt, Rinehart and Winston. All rights reserved. reversible reaction: the products can react to re-form the reactants written with a double arrow chemical equilibrium: when the rate of the forward reaction equals the rate of the reverse reaction and the concentrations of its products and reactants remain unchanged
Copyright © by Holt, Rinehart and Winston. All rights reserved. Equilibrium, a Dynamic State, continued Forward reaction: products of the forward reaction favored, lies to the right Reverse reaction: products of the reverse reaction favored, lies to the left Neither reaction is favored
Copyright © by Holt, Rinehart and Winston. All rights reserved. The Equilibrium Expression Initially, the concentrations of C and D are zero and those of A and B are maximum. Over time the rate of the forward reaction decreases as A and B are used up. The rate of the reverse reaction increases as C and D are formed. When these two reaction rates become equal, equilibrium is established. Chapter 18
Copyright © by Holt, Rinehart and Winston. All rights reserved. Reaction Rate Over Time for an Equilibrium System Chapter 18
Copyright © by Holt, Rinehart and Winston. All rights reserved. Le Châtelier’s principle: if a system at equilibrium is subjected to a stress, the equilibrium is shifted in the direction that tends to relieve the stress. Changes in pressure, concentration, or temperature can alter the equilibrium position and thereby change the relative amounts of reactants and products. N 2 (g) + 3H 2 (g) 2NH 3 (g)
Copyright © by Holt, Rinehart and Winston. All rights reserved. Predicting direction of equilibrium shift: There must first be a difference in number of molecules on each side of equation for there to be an affect with pressure. Pressure: pressure change only affects gases in equilibrium Increased pressure causes the system to reduce the total pressure by reducing the number of molecules by forming bonds. N 2 (g) + 3H 2 (g) 2NH 3 (g)
Copyright © by Holt, Rinehart and Winston. All rights reserved. N 2 (g) + 3H 2 (g) 2NH 3 (g) Increasing pressure tents to form bonds, decrease number of molecules Increasing heat tends to break bond, increase number of molecules
Copyright © by Holt, Rinehart and Winston. All rights reserved. High pressure favors the reverse reaction. Why? Low pressure favors the formation of CO 2. Example:
Copyright © by Holt, Rinehart and Winston. All rights reserved. Predicting direction of equilibrium shift: Temperature: Heat speeds up a reaction, to a point Adding heat favors the endothermic reaction, (forms bonds), removing heat favors the exothermic reaction (breaks bonds). N 2 (g) + 3H 2 (g) 2NH 3 (g)
Copyright © by Holt, Rinehart and Winston. All rights reserved. Example of temperature change: The synthesis of ammonia by the Haber process is exothermic. A high temperature favors the decomposition of ammonia, the endothermic reaction. At low temperatures, the forward reaction is too slow to be commercially useful. The temperature used represents a compromise between kinetic and equilibrium requirements.
Copyright © by Holt, Rinehart and Winston. All rights reserved. Is the forward reaction exothermic or endothermic? How does adding heat shift equilibrium? 1.2C(s)+O 2 (g) ⇌ 2CO(g)+heat 2.heat+6CO 2 (g)+6H 2 O(l) ⇌ C 6 H 12 O 6 (aq)+6O 2 (g ) 3.H 2 O(l) ⇌ H 2 O(g) Practice 1:
Copyright © by Holt, Rinehart and Winston. All rights reserved. Changes Affect an Equilibrium System can often be tracked by changing color, or formation/ disappearance of a precipitate Heating adds energy allowing bonds to form!
Copyright © by Holt, Rinehart and Winston. All rights reserved. Predicting direction of equilibrium shift: Concentration: Adding more reactant will force the production of more product to reach equilibrium, and vice versa (adding product forces more reactant) N 2 (g) + 3H 2 (g) 2NH 3 (g) An increase in the concentration of A creates a stress. To relieve the stress, some of the added A reacts with B to form products C and D. The equilibrium is reestablished with a higher concentration of A than before the addition and a lower concentration of B.
Copyright © by Holt, Rinehart and Winston. All rights reserved. Practice 2 1.What will happen to the equilibrium when more SO 2 (g) is added to the following system? 2SO 2 (g)+O 2 (g) ⇌ 2SO 3 (g) 2.What will happen to the equilibrium of the above reaction when the volume of the system is decreased? 3.What will happen to the equilibrium when the temperature of the system is decreased? N 2 (g)+O 2 (g) ⇌ 2NO(g) ΔH solution =-180.5kJ
Copyright © by Holt, Rinehart and Winston. All rights reserved. Practice 2 1.What will happen to the equilibrium when more SO 2 (g) is added to the following system? 2SO 2 (g)+O 2 (g) ⇌ 2SO 3 (g) 2.What will happen to the equilibrium of the above reaction when the volume of the system is decreased? 3.What will happen to the equilibrium when the temperature of the system is decreased? N 2 (g)+O 2 (g) ⇌ 2NO(g) ΔH solution =-180.5kJ Shift to right Increase pressure, shift to right (less molecules) Endothermic, remove heat forces shift to exothermic = left
Copyright © by Holt, Rinehart and Winston. All rights reserved. Catalysts have no effect on relative equilibrium amounts, they only affect the rate at which equilibrium is reached.
Copyright © by Holt, Rinehart and Winston. All rights reserved. Common-Ion Effect common-ion effect: phenomenon in which the addition of an ion common to two solutes brings about precipitation example: sodium sulfate is added to saturated barium sulfate solution. Na 2 SO 4 ⇌ 2Na + (aq) + SO 4 2- (aq) BaSO 4 (s) ⇌ Ba 2+ (aq) + SO 4 2- (aq) Equilibrium shifts left, forming precipitate, reducing ionization
Copyright © by Holt, Rinehart and Winston. All rights reserved. In general, the addition of a salt with an ion common to the solution of a weak electrolyte reduces the ionization of the electrolyte.
Copyright © by Holt, Rinehart and Winston. All rights reserved. Practice 3: X(g) +2Y(g) ⇌ Z(g) + energy as heat What will each of the following do to the equilibrium of the reaction 1.Remove Z by continuous condensation 2.Decrease the volume of the container 3.Add energy as heat 4.Add extra Y 5.Add a catalyst 6.Introduce an inert gas 1.Shift right 2.Shift right 3.Shift left 4.Shift right 5. no shift 6.no shift
Copyright © by Holt, Rinehart and Winston. All rights reserved. Warm up Predict the shift in equilibrium of the following: H 2(g) + I 2(g) ⇌ 2HI (g) ΔH solution =- 11J/mol 1. When pressure is decreased? 2. When I 2 is added. 3. When H 2 is removed. 4. When temperature is increased. 5. BaSO 4 (s) ⇌ Ba 2+ (aq) + SO 4 2- (aq) when pressure is increased.
Copyright © by Holt, Rinehart and Winston. All rights reserved. 1.NaCl (s) ⇌ Na + (aq) + Cl - (aq) 1.HIn (aq) H + (aq) + In - (aq) Yellow Blue 3.Co(H 2 O) 6 +2 (aq) + 4Cl - (aq) CoCl 4 -2 (aq) + 6H 2 O (l) ∆H = -50kJ/mol Pink Purple / Blue Adding silver nitrate causes silver to combine with chlorine, removing chlorine from solution. This causes equilibrium to shift _________ creating more ____________ La Châtelier’s Equilibrium Lab
Copyright © by Holt, Rinehart and Winston. All rights reserved. Equilibrium Constant: K, ration of product to reactant concentration (aka: equilibrium expression) After equilibrium is reached, the individual concentrations of reactants and products undergo no further change if conditions remain the same. A ratio of their concentrations should also remain constant. Chapter 18
Copyright © by Holt, Rinehart and Winston. All rights reserved. Rules of Equilibrium Constant K If K is small, the reactants are favored. If K is large the products are favored. Pure solids and liquids are omitted because their concentrations cannot change. K is changed by temperature but not concentrations Only the concentrations of substances that can actually change are included in K. Example 1 N 2 (g) + 3H 2 (g) ⇌ 2NH 3 (g) Example 2 NH 3 (aq) + H 2 O (l) ⇌ NH 4 + (aq) +OH - (aq)
Copyright © by Holt, Rinehart and Winston. All rights reserved. Equilibrium Constants
Copyright © by Holt, Rinehart and Winston. All rights reserved. Determining K eq for Reaction at Chemical Equilibrium
Copyright © by Holt, Rinehart and Winston. All rights reserved. Sample A: pg 595 An equilibrium mixture of N 2, O 2, producing NO gases at 1500 K is determined to consist of 6.4 10 –3 mol/L of N 2, 1.7 10 –3 mol/L of O 2, and 1.1 10 –5 mol/L of NO. What is the equilibrium constant for the system at this temperature? Given: [N 2 ] = 6.4 10 –3 mol/L [O 2 ] = 1.7 10 –3 mol/L [NO] = 1.1 10 –5 mol/L 2. White chemical equilibrium expression Solution: 1. Balance the chemical equation:
Copyright © by Holt, Rinehart and Winston. All rights reserved. Sample Problem A Solution, continued 2. White chemical equilibrium expression Solution: 1. Balance the chemical equation:
Copyright © by Holt, Rinehart and Winston. All rights reserved. Rules of Equilibrium Constant K If K is small, the reactants are favored. If K is large the products are favored. Pure solids and liquids are omitted because their concentrations cannot change. K is changed by temperature but not concentrations Only the concentrations of substances that can actually change are included in K.
Copyright © by Holt, Rinehart and Winston. All rights reserved. Equilibrium problems: put in classwork section of notebook Pg 595 Practice A #1 & 3 Answer on page 922
Multiple Choice 1. A chemical reaction is in equilibrium when A. forward and reverse reactions have ceased. B. the equilibrium constant equals 1. C. forward and reverse reaction rates are equal. D. No reactants remain. Test Preparation Chapter 18
2. Which change can cause the value of the equilibrium constant to change? A. temperature B. concentration of a reactant C. concentration of a product D. None of the above Test Preparation Multiple Choice Chapter 18
3.Consider the following reaction: The equilibrium constant expression for this reaction is A. C. B. D. Test Preparation Multiple Choice Chapter 18
4. Consider the following equation for an equilibrium system: Which concentration(s) would be included in the denominator of the equilibrium constant expression? A. Pb(s), CO 2 (g), and SO 2 (g) B. PbS(s), O 2 (g), and C(s) C. O 2 (g), Pb(s), CO 2 (g), and SO 2 (g) D. O 2 (g) Test Preparation Multiple Choice Chapter 18
5.If an exothermic reaction has reached equilibrium, then increasing the temperature will A. favor the forward reaction. B. favor the reverse reaction. C. favor both the forward and reverse reactions. D. have no effect on the equilibrium. Test Preparation Multiple Choice Chapter 18
6.Le Châtelier’s principle states that A.at equilibrium, the forward and reverse reaction rates are equal. B.stresses include changes in concentrations, pressure, and temperature. C.to relieve stress, solids and solvents are omitted from equilibrium constant expressions. D.chemical equilibria respond to reduce applied stress. Test Preparation Multiple Choice Chapter 18
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