I never let my schooling get in the way of my education.
|
|
- Bathsheba Primrose Hubbard
- 6 years ago
- Views:
Transcription
1 Chemistry NT I never let my schooling get in the way of my education. Mark Twain Chem NT Chemical Equilibrium Module Describing Chemical Equilibrium The Equilibrium Constant Equilibrium Constant for Sums of Reactions Heterogeneous Equilibria Oscillating patterns formed by a reaction far from equilibrium
2 Chemical Equilibrium When compounds react, they eventually form a mixture of products and unreacted reactants, in a dynamic equilibrium. A dynamic equilibrium consists of a forward reaction, in which substances react to give products, and a reverse reaction, in which products react to give the original reactants. Chemical equilibrium is the state reached by a reaction mixture when the rates of the forward and reverse reactions have become equal. 4 Chemical Equilibrium Much like water in a U-shape tube, there is constant mixing back and forth through the lower portion of the tube. reactants products It s as if the forward and reverse reactions were occurring at the same rate. The system appears to static (stationary) when, in reality, it is dynamic(in constant motion). 5 Chemical Equilibrium Consider the gaseous reaction of carbon monoxide and hydrogen to produce methane and steam. CO (g) + H (g) CH 4 (g) + HO(g) Suppose you put.000 mol CO and.000 mol H into a 0.00-L container at 00. The rate of the forward reaction, which depends on reactant concentrations, is large at first but steadily decreases. 6
3 Chemical Equilibrium Consider the gaseous reaction of carbon monoxide and hydrogen to produce methane and steam. CO (g) + H (g) CH 4 (g) + HO(g) Suppose you put.000 mol CO and.000 mol H into a 0.00-L container at 00. The rate of the reverse reaction starts at zero but steadily increases. 7 Chemical Equilibrium Consider the gaseous reaction of carbon monoxide and hydrogen to produce methane and steam. CO (g) + H (g) CH 4 (g) + HO(g) Suppose you put.000 mol CO and.000 mol H into a 0.00-L container at 00. The forward rate decreases and the reverse rate increases until they eventually become equal and the reaction mixture has reached equilibrium. 8 Chemical Equilibrium Consider the gaseous reaction of carbon monoxide and hydrogen to produce methane and steam. CO (g) + H (g) CH 4 (g) + HO(g) Suppose you put.000 mol CO and.000 mol H into a 0.00-L container at 00. The next figure illustrates this approach to equilibrium. 9
4 The experiment begins with.00 mol CO and.000 mol H in a 0.00-L vessel. Note that the amounts of substances become constant at equilibrium. 0 The forward rate is large at first but steadily decreases, whereas the reverse rate starts at zero and steadily increases. Eventually, both rates become equal (at equilibrium). Chemical Equilibrium Another example is the Haber process for producing ammonia from N and H which does not go to completion. N (g) + H(g) NH (g) It establishes an equilibrium state where all three species are present. 4
5 Chemical Equilibrium Another example is the Haber process for producing ammonia from N and H which does not go to completion. N + H NH A Problem To Consider Applying Stoichiometry to an Equilibrium Mixture. Suppose we place.000 mol N and.000 mol H in a reaction vessel at 450 o C and 0.0 atmospheres of pressure. The reaction is: N (g) + H(g) NH (g) What is the composition of the equilibrium mixture if it contains mol NH? 4 A Problem To Consider Using the information given, set up the following table. Starting.000 Change -x Equilibrium.000-x N (g) + H(g) NH (g).000 -x.000-x 0 +x x = mol The equilibrium amount of NH was given as mol. Therefore, x = mol NH (x = mol). 5 5
6 A Problem to Consider Using the information given, set up the following table. Starting.000 Change -x Equilibrium.000-x N (g) + H(g) NH (g).000 -x.000-x 0 +x x = mol Equilibrium amount of N = = mol N 6 A Problem to Consider Using the information given, set up the following table. Starting.000 Change -x Equilibrium.000-x N (g) + H(g) NH (g).000 -x.000-x 0 +x x = mol Equilibrium amount of H =.000-( x 0.040) =.880 mol H 7 A Problem to Consider Using the information given, set up the following table. Starting.000 Change -x Equilibrium.000-x N (g) + H(g) NH (g).000 -x.000-x 0 +x x = mol Equilibrium amount of NH = x = mol NH 8 6
7 The Equilibrium Constant Every reversible system has its own position of equilibrium under any given set of conditions. The ratio of products produced to unreacted reactants for any given reversible reaction remains constant under constant conditions of pressure and temperature. The numerical value of this ratio is called the equilibrium constant for the given reaction. 9 The Equilibrium Constant The equilibrium-constant expression for a reaction is obtained by multiplying the concentrations of products, dividing by the concentrations of reactants, and raising each concentration to a power equal to its coefficient in the balanced chemical equation. aa + bb cc + dd For the general equation above, the equilibriumconstant expression would be: c d [C] [D] c = a [A] [B] b 0 The Equilibrium Constant The equilibrium-constant expression for a reaction is obtained by multiplying the concentrations of products, dividing by the concentrations of reactants, and raising each concentration to a power equal to its coefficient in the balanced chemical equation. aa + bb cc + dd The molar concentration of a c d [C] [D] substance is denoted by writing c = a b its formula in square brackets. [A] [B] 7
8 The Equilibrium Constant The equilibrium constant c, is the value obtained for the equilibrium-constant expression when equilibrium concentrations are substituted. Since c products reactan ts A large c indicates high concentrations of product at equilibrium. The Equilibrium Constant The equilibrium constant c, is the value obtained for the equilibrium-constant expression when equilibrium concentrations are substituted. Since c products reactan ts A small c indicates a large concentration of unreacted reactants at equilibrium. The Equilibrium Constant The law of mass action states that the value of the equilibrium constant expression c is constant for a particular reaction at a given temperature, whatever equilibrium concentrations are substituted. Consider the equilibrium established in the Haber process. N (g) + H(g) NH (g) 4 8
9 The Equilibrium Constant The equilibrium-constant expression would be: [NH c = [N ][H ] ] Note that the stoichiometric coefficients in the balanced equation have become the powers to which each concentration is raised. N (g) + H(g) NH (g) 5 A Problem To Consider Write the equilibrium-constant expression c for catalytic methanation. CO(g) + H (g) CH4(g) + HO(g) Write the concentrations of products in the top (numerator) of the equilibrium constant expression, and write the concentrations of reactants in the bottom (denominator). 6 A Problem To Consider Write the equilibrium-constant expression c for catalytic methanation. CO(g) + H (g) CH4(g) + HO(g) Raise each concentration term to the power equal to the coefficient of the substance in the chemical equation. 7 9
10 A Problem To Consider Write the equilibrium-constant expression c for catalytic methanation. CO(g) + H (g) CH4(g) + HO(g) [CH4][H c = [CO][H O] ] 8 A Problem To Consider Write the equilibrium-constant expression c for the reverse of the previous reaction. CH4 (g) + HO(g) CO(g) + H(g) c = [CO][H] [CH ][H O] 4 9 A Problem To Consider Our previous look at the Haber process led us to the equilibrium-constant expression below. N (g) + H(g) NH (g) [NH c = [N ][H ] ] 0 0
11 A Problem To Consider Write the equilibrium-constant expression c when the equation for the previous reaction is written. N (g) + H(g) NH (g) Whether coefficients are fractional or not, they still become exponents in the equilibrium-constant expression. c = [NH] [N ] [H ] Equilibrium: A inetics Argument If the forward and reverse reaction rates in a system at equilibrium are equal, then it follows that their rate laws would be equal. Consider the decomposition of N O 4, dinitrogen tetroxide. NO4(g) NO (g) If we start with some dinitrogen tetroxide and heat it, it begins to decompose to produce NO. Equilibrium: A inetics Argument If the forward and reverse reaction rates in a system at equilibrium are equal, then it follows that their rate laws would be equal. Consider the decomposition of N O 4, dinitrogen tetroxide. NO4(g) NO (g) However, once some NO is produced it can recombine to form N O 4.
12 Equilibrium: A inetics Argument NO4(g) NO (g) Call the decomposition of N O 4 the forward reaction and the formation of N O 4 the reverse reaction. These are elementary reactions, and you can immediately write the rate law for each. k f k r Rate(forward) = kf [NO4] Rate = k [NO (reverse) r ] Here k f and k r represent the forward and reverse rate constants. 4 Equilibrium: A inetics Argument k f NO4(g) NO (g) k r Ultimately, this reaction reaches an equilibrium state where the rate of the forward and reverse reactions are equal, therefore: k = f [NO 4] kr[no ] rate forward = rate reverse 5 Equilibrium: A inetics Argument NO4(g) NO (g) Combining the constants you can identify the equilibrium constant c, as the ratio of the forward and reverse rate constants. k f k r k k c = f = r [NO ] [N O ] 4 6
13 Obtaining Equilibrium Constants for Reactions Equilibrium concentrations for a reaction must be obtained experimentally and then substituted into the equilibrium-constant expression in order to calculate c. This c can then be used to define any equilibrium composition for that particular reaction. 7 Obtaining Equilibrium Constants for Reactions Consider the reaction below. CO (g) + H (g) CH 4(g) + HO(g) Suppose we started with initial concentrations of CO and H of 0.00 M and 0.00 M, respectively. 8 Obtaining Equilibrium Constants for Reactions Consider the reaction below. CO (g) + H (g) CH 4(g) + HO(g) When the system finally settled into equilibrium we determined the equilibrium concentrations to be as follows. Reactants [CO] = 0.06 M [H ] = 0.89 M Products [CH 4 ] = M [H O] = M 9
14 Obtaining Equilibrium Constants for Reactions Consider the reaction below. CO (g) + H (g) CH 4(g) + HO(g) The equilibrium-constant expression for this reaction is: [CH4][H O] c = [CO][H ] 40 Obtaining Equilibrium Constants for Reactions Consider the reaction below. CO (g) + H (g) CH 4(g) + HO(g) If we substitute the equilibrium concentrations, we obtain: (0.087M)(0.087M) = (0.06M)(0.89M) c =.9 4 Obtaining Equilibrium Constants for Reactions Consider the reaction below. CO (g) + H (g) CH 4(g) + HO(g) Note that regardless of initial concentrations (whether they be reactants or products) the law of mass action dictates that the reaction will always settle into an equilibrium where the equilibrium-constant expression equals c. 4 4
15 Obtaining Equilibrium Constants for Reactions Consider the reaction below. CO (g) + H (g) CH 4(g) + HO(g) For example, if we repeat the previous experiment, only this time, starting with initial concentrations of products: [CH 4 ] initial = M and [H O] initial = M 4 Obtaining Equilibrium Constants for Reactions Consider the reaction below. CO (g) + H (g) CH 4(g) + HO(g) We find that these initial concentrations result in the following equilibrium concentrations. Reactants [CO] = 0.06 M [H ] = 0.89 M Products [CH 4 ] = M [H O] = M 44 Obtaining Equilibrium Constants for Reactions Consider the reaction below. CO (g) + H (g) CH 4(g) + HO(g) Substituting these values into the equilibriumconstant expression, we obtain the same result. (0.087M)(0.087M) = (0.06M)(0.89M) c =.9 Whether we start with reactants initially or products, the system establishes the same ratio. 45 5
16 The Equilibrium Constant p In discussing gas -phase equilibria, it is often more convenient to express concentrations in terms of partial pressures rather than molarities. It can be seen from the ideal gas equation that partial pressure of a gas is proportional to its molarity. n P = ( ) RT = MRT V 46 The Equilibrium Constant p If we express a gas -phase equilibria in terms of partial pressures, we obtain p. Consider the reaction below. CO (g) + H (g) CH 4(g) + HO(g) The equilibrium-constant expression in terms of partial pressures becomes: p = P P CH 4 CO P P HO H 47 The Equilibrium Constant p In general, the numerical value of p is different than that of c. From the relationship n/v=p/rt, one can show Dn p = c(rt) where Dn is the sum of the moles of gaseous products in a reaction minus the sum of the moles of gaseous reactants. 48 6
17 A Problem To Consider Consider the reaction SO (g) + O (g) SO (g) The c for the reaction is.8 x 0 at 000. Calculate the p for the reaction at this temperature. Therefore, Dn = - 49 A Problem To Consider Consider the reaction SO (g) + O (g) SO (g) Since Dn p = c(rt) and from the equation we see that n = -, we can simply substitute the given reaction temperature and the value of R ( L. atm/mol. ) to obtain p. 50 A Problem To Consider Consider the reaction SO (g) + O (g) SO (g) Since Dn p = c(rt).8 0 ( Latm p = 000 ) p =.4 mol - 5 7
18 Equilibrium Constant for the Sum of Reactions Similar to the method of combining reactions we saw using Hess law in chapter 6, we can combine equilibrium reactions whose c s are known to obtain the c for the overall reaction. Just as with Hess law, when we reversed reactions or took multiples of them prior to adding them together, we had to manipulate the H s to reflect what we had done. The rules are a bit different for manipulating c s. 5 Equilibrium Constant for the Sum of Reactions. If you reverse a reaction, invert the value of c. For example, aa + bb cc + dd c [C] [D] forward = a [A] [B] d b If we reverse : cc + dd aa + bb a [A] [B] reverse = = c [C] [D] forward b d 5 Equilibrium Constant for the Sum of Reactions. If you multiply each of the coefficients in an equation by the same factor (,, ), raise the c to the same power(,, ). For example, A + B C + D [C][D] initial = [A][B] 54 8
19 Equilibrium Constant for the Sum of Reactions. If you multiply each of the coefficients in an equation by the same factor (,, ), raise the c to the same power(,, ). If we triple the reaction: A + B C + D [C] [D] tripled = ( initial ) = [A] [B] 55 Equilibrium Constant for the Sum of Reactions. If you divide each of the coefficients in an equation by the same factor (,, ) take the corresponding root of the c (i.e., square root, cube root, ). For example, A + B C + D [C][D] initial = [A][B] 56 Equilibrium Constant for the Sum of Reactions. If you divide each of the coefficients in an equation by the same factor (,, ) take the corresponding root of the c (i.e., square root, cube root, ). If we halve the reaction: A + B C+ [C] [D] halved = = (initial) = [A] [B] D initial 57 9
20 Equilibrium Constant for the Sum of Reactions 4. When you finally combine (that is, add) the individual equations together, take the product of the equilibrium constants to obtain the overall c. A + B C + D For example, C + D E + F A + B E + F overall 58 Equilibrium Constant for the Sum of Reactions 4. When you finally combine (that is, add) the individual equations together, take the product of the equilibrium constants to obtain the overall c. Simplifying: [C][D][E][F] overall = [A][B][C][D] [E][F] overall = = [A][B] 59 Equilibrium Constant for the Sum of Reactions For example, nitrogen and oxygen can combine to form either NO(g) or N O (g) according to the following equilibria. () N (g) + O (g) NO(g) c = 4. x 0 - () N (g) + O (g) N O(g) c =.4 x 0-8 Using these two equations, we can obtain the c for the formation of NO(g) from N O(g) () N O(g) + O (g) NO(g) c =? 60 0
21 Equilibrium Constant for the Sum of Reactions To combine equations () and () to obtain equation (), we must first reverse equation (). When we do we must also take the reciprocal of its c value. () (g) O (g) NO(g) c = 4. x 0 - () () N + N O(g) N(g) + O (g) c = N O(g) + O (g) NO(g) c(overall) = (4. 0 ) ( ) = Heterogeneous Equilibria A heterogeneous equilibrium is an equilibrium that involves reactants and products in more than one phase. The equilibrium of a heterogeneous system is unaffected by the amounts of pure solids or liquids present, as long as some of each is present. The concentrations of pure solids and liquids is considered to be and therefore, do not appear in the equilibrium expression. 6 Heterogeneous Equilibria Consider the reaction below. C(s) + H O(g) CO(g) + H (g) The equilibrium-constant expression contains terms for only those species in the homogeneous gas phase H O, CO, and H. c = [CO][H] [H O] 6
22 Operational Skills Applying stoichiometry to an equilibrium mixture Writing equilibrium-constant expressions Obtaining the equilibrium constant from reaction composition Determining equilibrium constants for sums of reactions Time for a few review questions
Chemical Equilibrium
Chemical Equilibrium Chemical Equilibrium When compounds react, they eventually form a mixture of products and unreacted reactants, in a dynamic equilibrium. A dynamic equilibrium consists of a forward
More informationChemical Equilibrium-A Dynamic Equilibrium
CHAPTER 14 Page 1 Chemical Equilibrium-A Dynamic Equilibrium When compounds react, they eventually form a mixture of products and (unreacted) reactants, in a dynamic equilibrium Much like water in a U-shape
More information2.0 Equilibrium Constant
2.0 Equilibrium Constant When reactions are reversible and chemical equilibrium is reached, it is important to recognize that not all of the reactants will be converted into products. There is a mathematical
More informationCHEMICAL EQUILIBRIA: GENERAL CONCEPTS
CHEMICAL EQUILIBRIA: GENERAL CONCEPTS THE NATURE OF THE EQUILIBRIUM STATE: Equilibrium is the state where the concentrations of all reactants and products remain constant with time. (in stoichiometry,
More informationCHEMICAL EQUILIBRIUM. Chapter 15
Chapter 15 P a g e 1 CHEMICAL EQUILIBRIUM Examples of Dynamic Equilibrium Vapor above a liquid is in equilibrium with the liquid phase. rate of evaporation = rate of condensation Saturated solutions rate
More informationThe. Equilibrium. Constant. Chapter 15 Chemical Equilibrium. The Concept of Equilibrium. The Concept of Equilibrium. A System at Equilibrium
The Concept of Chapter 15 Chemical AP Chemistry 12 North Nova Education Centre 2017 Chemical equilibrium occurs when a reaction and its reverse reaction proceed at the same rate. The Concept of As a system
More informationDr. Valverde s AP Chemistry Class
AP* Chemistry Dr. Valverde s AP Chemistry Class Chapter CHEMICAL 13 Review: EQUILIBRIA: Chemical Equilibrium GENERAL CONCEPTS THE NATURE OF THE EQUILIBRIUM STATE: Equilibrium is the state where the rate
More informationChapter 15 Chemical Equilibrium. Equilibrium
Chapter 15 Chemical The Concept of Chemical equilibrium occurs when a reaction and its reverse reaction proceed at the same rate. The Concept of As a system approaches equilibrium, both the forward and
More informationChem 1B Dr. White 1 Chapter 13: Chemical Equilibrium Outline Chemical Equilibrium. A. Definition:
Chem 1B Dr. White 1 Chapter 13: Chemical Equilibrium Outline 13.1. Chemical Equilibrium A. Definition: B. Consider: N 2 O 4 (g, colorless) 2NO 2 (g, brown) C. 3 Main Characteristics of Equilibrium 13.2-13.4.
More informationChemical Equilibrium
Chemical Equilibrium Concept of Equilibrium Equilibrium Constant Equilibrium expressions Applications of equilibrium constants Le Chatelier s Principle The Concept of Equilibrium The decomposition of N
More informationCHEMISTRY. Chapter 15 Chemical Equilibrium
CHEMISTRY The Central Science 8 th Edition Chapter 15 Chemical Kozet YAPSAKLI The Concept of Chemical equilibrium is the point at which the concentrations of all species are constant. Chemical equilibrium
More informationChapter 15. Chemical Equilibrium
Chapter 15. Chemical Equilibrium 15.1 The Concept of Equilibrium Consider colorless frozen N 2 O 4. At room temperature, it decomposes to brown NO 2. N 2 O 4 (g) 2NO 2 (g) At some time, the color stops
More informationCh#13 Outlined Notes Chemical Equilibrium
Ch#13 Outlined Notes Chemical Equilibrium Introduction A. Chemical Equilibrium 1. The state where the concentrations of all reactants and products remain constant with time 2. All reactions carried out
More information1.6 Chemical equilibria and Le Chatelier s principle
1.6 Chemical equilibria and Le Chatelier s principle Reversible reactions: Consider the reaction: Mg(s) + H2SO4(aq) MgSO4(aq) + H2(g) The reaction stops when all of the limiting reagent has been used up.
More informationChapter 15: Chemical Equilibrium: How Much Product Does a Reaction Really Make?
Chapter 15: Chemical Equilibrium: How Much Product Does a Reaction Really Make? End-of-Chapter Problems: 15.1-15.10, 15.13-15.14, 15.17-15.91, 15.94-99, 15.10-15.103 Example: Ice melting is a dynamic process:
More informationChapter 6: Chemical Equilibrium
Chapter 6: Chemical Equilibrium 6.1 The Equilibrium Condition 6. The Equilibrium Constant 6.3 Equilibrium Expressions Involving Pressures 6.4 The Concept of Activity 6.5 Heterogeneous Equilibria 6.6 Applications
More informationChapter Fifteen. Chemical Equilibrium
Chapter Fifteen Chemical Equilibrium 1 The Concept of Equilibrium Dynamic Equilibrium Opposing processes occur at equal rates Forward and reverses reaction proceed at equal rates No outward change is observed
More informationChapter 15 Equilibrium
Chapter 15. Chemical Equilibrium Common Student Misconceptions Many students need to see how the numerical problems in this chapter are solved. Students confuse the arrows used for resonance ( )and equilibrium
More informationSection 10. Rates of Reactions Goal: Learn how temperature, concentration, and catalysts affect the rate of reaction. Summary
Chapter 10 Reaction Rates and Chemical Equilibrium Section 10. Rates of Reactions Goal: Learn how temperature, concentration, and catalysts affect the rate of reaction. Summary The rate of a reaction is
More informationChemical Equilibrium. Professor Bice Martincigh. Equilibrium
Chemical Equilibrium by Professor Bice Martincigh Equilibrium involves reversible reactions Some reactions appear to go only in one direction are said to go to completion. indicated by All reactions are
More informationChapter 9. Chemical Equilibrium
Chapter 9. Chemical Equilibrium 9.1 The Nature of Chemical Equilibrium -Approach to Equilibrium [Co(H 2 O) 6 ] 2+ + 4 Cl- [CoCl 4 ] 2- + 6 H 2 O Characteristics of the Equilibrium State example) H 2 O(l)
More informationEquilibrium. Forward and Backward Reactions. Hydrogen reacts with iodine to make hydrogen iodide: H 2 (g) + I 2 (g) 2HI(g)
Equilibrium Forward and Backward Reactions Hydrogen reacts with iodine to make hydrogen iodide: H 2 (g) + I 2 (g) 2HI(g) forward rate = k f [H 2 ][I 2 ] 2HI(g) H 2 (g) + I 2 (g) backward rate = k b [HI]
More informationChapter 15. Chemical Equilibrium
Chapter 15. Chemical Equilibrium 15.1 The Concept of Equilibrium Consider colorless frozen N 2 O 4. At room temperature, it decomposes to brown NO 2. N 2 O 4 (g) 2NO 2 (g) At some time, the color stops
More information1.0 L container NO 2 = 0.12 mole. time
CHEM 1105 GAS EQUILIBRIA 1. Equilibrium Reactions - a Dynamic Equilibrium Initial amounts: = mole = 0 mole 1.0 L container = 0.12 mole moles = 0.04 mole 0 time (a) 2 In a 1.0 L container was placed 4.00
More informationChapter 13. The Concept of Equilibrium. A System at Equilibrium. The Concept of Equilibrium. Chemical Equilibrium. N 2 O 4 (g) 2 NO 2 (g)
PowerPoint to accompany The Concept of Equilibrium Chapter 13 Chemical Equilibrium Figure 13.1 Chemical equilibrium occurs when a reaction and its reverse reaction proceed at the same rate. The Concept
More informationThe N 2 O 4 -NO 2 Equilibrium
Chemical Equilibria William L Masterton Cecile N. Hurley Edward J. Neth cengage.com/chemistry/masterton Chapter 1 Gaseous Chemical Equilibrium For a gaseous chemical equilibrium, more than one gas is present:
More information6. Which expression correctly describes the equilibrium constant for the following reaction? 4NH 3 (g) + 5O 2 (g) 4NO(g) + 6H 2 O(g)
1. Which of the following can we predict from an equilibrium constant for a reaction? 1. The extent of a reaction 2. Whether the reaction is fast or slow 3. Whether a reaction is exothermic or endothermic
More informationChapter 15 Chemical Equilibrium
Equilibrium To be in equilibrium is to be in a state of balance: Chapter 15 Chemical Equilibrium - Static Equilibrium (nothing happens; e.g. a tug of war). - Dynamic Equilibrium (lots of things happen,
More informationJanuary 03, Ch 13 SB equilibrium.notebook
Ch 13: Chemical Equilibrium exists when 2 opposing reactions occur simultaneously at the same rate (dynamic rather than static) Forward rate = reverse rate https://www.youtube.com/watch?v=wld_imyqagq The
More informationFor the reaction: A B R f = R r. Chemical Equilibrium Chapter The Concept of Equilibrium. The Concept of Equilibrium
Chemical Equilibrium Chapter 15.1-4 This is the last unit of the year, and it contains quite a lot of material. Do not wait until the end of the unit to begin studying. Use what you have learned about
More informationReaction Rate. Products form rapidly. Products form over a long period of time. Precipitation reaction or explosion
Reaction Rate Products form rapidly Precipitation reaction or explosion Products form over a long period of time Corrosion or decay of organic material Chemical Kinetics Study of the rate at which a reaction
More informationChapter 15 Chemical Equilibrium
Chapter 15 Chemical Chemical 15.1 The Concept of 15.2 The Constant (K) 15.3 Understanding and Working with Constants 15.4 Heterogeneous Equilibria 15.5 Calculating Constants 15.6 Applications of Constants
More informationAP* Chapter 13. Chemical Equilibrium
AP* Chapter 13 Chemical Equilibrium Section 13.1 The Equilibrium Condition Chemical Equilibrium The state where the concentrations of all reactants and products remain constant with time. On the molecular
More informationQuantity Relationships in Chemical Reactions
Chapter 10 Relationships in Chemical Reactions Section 10.1 Conversion Factors from a Chemical Equation Goal 1 The coefficients in a chemical equation give us the conversion factors to get from the number
More informationChapter 13. Chemical Equilibrium
Chapter 13 Chemical Equilibrium Section 13.1 The Equilibrium Condition Chemical Equilibrium The state where the concentrations of all reactants and products remain constant with time. On the molecular
More informationC h a p t e r 13. Chemical Equilibrium
C h a p t e r 13 Chemical Equilibrium Chemical equilibrium is achieved when: the rates of the forward and reverse reactions are equal and the concentrations of the reactants and products remain constant
More informationCH. 12 STOICHIOMETRY
CH. 12 STOICHIOMETRY Balanced Chemical Equations Used to calculate: How much of each reactant is needed How much product will form If you know one quantity you can calculate the rest. Quantity may be in
More informationChapter 15 Equilibrium
Chapter 15. Chemical Equilibrium Common Student Misconceptions Many students need to see how the numerical problems in this chapter are solved. Students confuse the arrows used for resonance ( )and equilibrium
More informationLe Châtelier s Principle. 19 Copyright Pearson Education, Inc., or its affiliates. All Rights Reserved. Equilibrium: Le Châtelier s Principle
Factors Affecting : Le Châtelier s Principle Pressure Factors Affecting : Le Châtelier s Principle Pressure When volume decreases, the pressure increases. systems in which some reactants and products are
More informationChapter 15 Equilibrium
Chapter 15. Chemical Equilibrium 15.1 The Concept of Equilibrium Chemical equilibrium is the point at which the concentrations of all species are constant. A dynamic equilibrium exists when the rates of
More informationChapter 14 Chemical Equilibrium
Chapter 14 Chemical Equilibrium Fu-Yin Hsu Chemical reaction The speed of a chemical reaction is determined by kinetics. The extent of a chemical reaction is determined by thermodynamics. 14.1 Fetal Hemoglobin
More informationChapter 6: Chemical Equilibrium
Chapter 6: Chemical Equilibrium 6.1 The Equilibrium Condition 6.2 The Equilibrium Constant 6.3 Equilibrium Expressions Involving Pressures 6.4 The Concept of Activity 6.5 Heterogeneous Equilibria 6.6 Applications
More informationEquilibrium. Introduction
Equilibrium Introduction From kinetics, we know that reactants sometimes collide to give products. But why can t products collide to go back to reactants? Theoretically, all chemical reactions are reversible.
More informationEQUILIBRIUM CONSTANT, K eq or K. The Law of Chemical Equilibrium: (Guldberg & Waage, 1864)
1 EQUILIBRIUM CONSTANT, K eq or K The Law of Chemical Equilibrium: (Guldberg & Waage, 1864) States that: At equilibrium, there is a constant ratio between the concentration of the products and the concentration
More informationUNIT 11 Practice Test Page 1 of 13 Equilibrium
UNIT 11 Practice Test Page 1 of 13 Do NOT write on this test. $0.10/page lost or damaged fee. 1. In which of the following does the reaction go farthest to completion? A. K = 10 5 B. K = 10 5 C. K = 1000
More informationChem 116 POGIL Worksheet - Week 7 Kinetics to Equilibrium
Chem 116 POGIL Worksheet - Week 7 Kinetics to Equilibrium Why? Most chemical reactions are reversible. This means that once products are formed, they can react to reform the reactants. If we allow a reaction
More information8. The table below describes two different reactions in which Reaction 1 is faster. What accounts for this observation? Reaction 1 Reaction 2.
Public Review - Rates and Equilibrium June 2005 1. What does X represent in the diagram below? (A) activation energy for the forward reaction (B) activation energy for the reverse reaction (C) heat of
More information91166 Demonstrate understanding of chemical reactivity Collated questions on equilibria
(2017:2) 91166 Demonstrate understanding of chemical reactivity Collated questions on equilibria The addition of a small amount of iron to a mixture of nitrogen and hydrogen gases helps to speed up the
More informationChemical Kinetics and
Chemical Kinetics and Equilibrium Part 2: Chemical Equilibrium David A. Katz Department of Chemistry Pima Community College Tucson, AZ USA The Concept of Equilibrium Kinetics applies to the speed of a
More informationChapter 13: Chemical Equilibrium
Chapter 13: Chemical Equilibrium May 5 2:04 PM 13.1 The Equilibrium Condition When you finish this section you will be able to list some characteristics of reactions at equilibrium. Chemical equilibrium
More informationA reversible reaction is a chemical reaction where products can react to form the reactants and vice versa.
Chemistry 12 Unit II Dynamic Equilibrium Notes II.1 The Concept of Dynamic Equilibrium A reversible reaction is a chemical reaction where products can react to form the reactants and vice versa. A reversible
More informationThis is Chemical Equilibrium, chapter 15 from the book Principles of General Chemistry (index.html) (v. 1.0).
This is Chemical Equilibrium, chapter 15 from the book Principles of General Chemistry (index.html) (v. 1.0). This book is licensed under a Creative Commons by-nc-sa 3.0 (http://creativecommons.org/licenses/by-nc-sa/
More informationChapter 13: Chemical Equilibrium
Chapter 13: Chemical Equilibrium 13.1 The Equilibrium Condition Equilibrium: a state in which no observable changes occur H 2 O (l) H 2 O (g) Physical equilibrium: no chemical change. N 2(g) + 3H 2(g)
More informationChemistry 132 NT. Reaction Rates. Chem 132 NT. We are what we repeatedly do. Excellence is, then, not an act, but a habit.
Chemistry 132 NT We are what we repeatedly do. Excellence is, then, not an act, but a habit Aristotle 1 Reaction Rates Module 1 Reaction Rates Definition of Reaction Rate Experimental Determination of
More informationCollision Theory. Unit 12: Chapter 18. Reaction Rates. Activation Energy. Reversible Reactions. Reversible Reactions. Reaction Rates and Equilibrium
Collision Theory For reactions to occur collisions between particles must have Unit 12: Chapter 18 Reaction Rates and Equilibrium the proper orientation enough kinetic energy See Both In Action 1 2 Activation
More informationStoichiometry. Please take out your notebooks
Stoichiometry Please take out your notebooks Stoichiometry stochio = Greek for element metry = measurement Stoichiometry is about measuring the amounts of elements and compounds involved in a reaction.
More informationPractice Test F.1 (pg 1 of 7) Unit F - General Equilibrium Kp and Kc Name Per
Practice Test F. (pg of 7) Unit F - General Equilibrium Kp and Kc Name Per This is practice - Do NOT cheat yourself of finding out what you are capable of doing. Be sure you follow the testing conditions
More informationUnit 8: Equilibrium Unit Review
1. Predict the effect of increasing pressure on the position of equilibrium in the following systems: a. CH 4 (g) + 2H 2 O(g) CO 2 (g) + 4H 2 (g) b. N 2 O 5 (g) + NO(g) 3NO 2 (g) c. NO(g) + NO 2 (g) N
More informationChemical Equilibrium. Chapter
Chemical Equilibrium Chapter 14 14.1-14.5 Equilibrium Equilibrium is a state in which there are no observable changes as time goes by. Chemical equilibrium is achieved when: 1.) the rates of the forward
More informationCHEMICAL EQUILIBRIA. Dynamic Equilibrium Equilibrium involves reversible reactions which do not go to completion.
CHEMICAL EQUILIBRIA Dynamic Equilibrium Equilibrium involves reversible reactions which do not go to completion. If we consider a reaction between A and B to form C and D which is reversible. When A and
More informationDynamic Equilibrium Illustrated
שו וי מ שק ל Equilibrium Reactants Products In an equilibrium, the forward and reverse processes continue to occur but at equal rates! The reactant and product concentrations remain constant We are usually
More information15/04/2018 EQUILIBRIUM- GENERAL CONCEPTS
15/04/018 EQUILIBRIUM- GENERAL CONCEPTS When a system is at equilibrium, the forward and reverse reactions are proceeding at the same rate. The concentrations of all species remain constant over time,
More informationEQUILIBRIUM GENERAL CONCEPTS
017-11-09 WHEN THE REACTION IS IN EQUILIBRIUM EQUILIBRIUM GENERAL CONCEPTS The concentrations of all species remain constant over time, but both the forward and reverse reaction never cease When a system
More information15.1 The Concept of Equilibrium
Lecture Presentation Chapter 15 Chemical Yonsei University 15.1 The Concept of N 2 O 4 (g) 2NO 2 (g) 2 Chemical equilibrium occurs when a reaction and its reverse reaction proceed at the same rate. The
More informationAP Chapter 13: Kinetics Name
AP Chapter 13: Kinetics Name Warm-Ups (Show your work for credit) Date 1. Date 2. Date 3. Date 4. Date 5. Date 6. Date 7. Date 8. AP Chapter 13: Kinetics 2 Warm-Ups (Show your work for credit) Date 1.
More informationQuestions 1-3 relate to the following reaction: 1. The rate law for decomposition of N2O5(g) in the reaction above. B. is rate = k[n2o5] 2
Questions 1-3 relate to the following reaction: 2N2O5(g) 4NO2(g) + O2(g) 1. The rate law for decomposition of N2O5(g) in the reaction above A. is rate = k[n2o5] B. is rate = k[n2o5] 2 C. is rate = [NO2]
More informationThe Equilibrium State. Chapter 13 - Chemical Equilibrium. The Equilibrium State. Equilibrium is Dynamic! 5/29/2012
Chapter 13 - Chemical Equilibrium The Equilibrium State Not all chemical reactions go to completion; instead they attain a state of equilibrium. When you hear equilibrium, what do you think of? Example:
More informationThe Concept of Equilibrium
Chemical Equilibrium The Concept of Equilibrium Sometimes you can visually observe a certain chemical reaction. A reaction may produce a gas or a color change and you can follow the progress of the reaction
More informationStudy Guide for Module 13 An Introduction to Equilibrium
Chemistry 1020, Module 13 Name Study Guide for Module 13 An Introduction to Equilibrium Reading Assignment: Section 12.1 and Chapter 13 of Chemistry, 6th Edition by Zumdahl. Guide for Your Lecturer: 1.
More informationChapter 9. Table of Contents. Stoichiometry. Section 1 Introduction to Stoichiometry. Section 2 Ideal Stoichiometric Calculations
Stoichiometry Table of Contents Section 1 Introduction to Stoichiometry Section 2 Ideal Stoichiometric Calculations Section 3 Limiting Reactants and Percentage Yield Section 1 Introduction to Stoichiometry
More informationCHEMICAL EQUILIBRIUM Chapter 13
1 CHEMICAL EQUILIBRIUM Chapter 13 Pb 2+ (aq) + 2 Cl (aq) PbCl 2 (s) 1 Objectives Briefly review what we know of equilibrium Define the Equilibrium Constant (K eq ) and Reaction Quotient (Q) Determining
More informationChapter 15. Chemical Equilibrium
1 Chapter 15. 15.1 The Concept of Equilibrium 1,2,3 Consider colorless frozen N 2 O 4. At room temperature, it decomposes to brown NO 2. N 2 O 4 (g) 2NO 2 (g) At some time, the color stops changing and
More informationChemistry 142 (Practice) MIDTERM EXAM II November. Fill in your name, section, and student number on Side 1 of the Answer Sheet.
Chemistry 4 (Practice) MIDTERM EXAM II 009 November (a) Before starting, please check to see that your exam has 5 pages, which includes the periodic table. (b) (c) Fill in your name, section, and student
More informationChemical Equilibria. OCR Chemistry A H432
Chemical Equilibria Chemical equilibrium is a dynamic equilibrium. Features of a dynamic equilibrium, which can only be established in a closed system (nothing added or removed): - rates of forward and
More informationChapter 12 Stoichiometry
12.2 Chemical Calculations > Chapter 12 Stoichiometry 12.1 The Arithmetic of Equations 12.22 Chemical Calculations 12.3 Limiting Reagent and Percent Yield 1 Copyright Pearson Education, Inc., or its affiliates.
More informationTECHNICAL SCIENCE DAS12703 ROZAINITA BT. ROSLEY PUSAT PENGAJIAN DIPLOMA UNVERSITI TUN HUSSEIN ONN MALAYSIA
TECHNICAL SCIENCE DAS12703 ROZAINITA BT. ROSLEY PUSAT PENGAJIAN DIPLOMA UNVERSITI TUN HUSSEIN ONN MALAYSIA ii TABLE OF CONTENTS TABLE OF CONTENTS... i LIST OF FIGURES... iii Chapter 1... 4 SOLUTIONS...
More informationSection 7.2: Equilibrium Law and the Equilibrium Constant Tutorial 1 Practice, page (a) 2 CO 2 (g) #!!"
Section 7.: Equilibrium Law and the Equilibrium Constant Tutorial Practice, page 4. (a) CO (g) #!!"! CO(g) + O (g) Products: CO(g); O (g) Reactant: CO (g) [CO [O Equilibrium law equation: [CO (b) Cl (g)
More informationGas Phase Equilibrium
Gas Phase Equilibrium Chemical Equilibrium Equilibrium Constant K eq Equilibrium constant expression Relationship between K p and K c Heterogeneous Equilibria Meaning of K eq Calculations of K c Solving
More informationGestão de Sistemas Energéticos 2017/2018
Gestão de Sistemas Energéticos 2017/2018 Exergy Analysis Prof. Tânia Sousa taniasousa@tecnico.ulisboa.pt Conceptualizing Chemical Exergy C a H b O c enters the control volume at T 0, p 0. O 2 and CO 2,
More informationCHAPTER 3: CHEMICAL EQUILIBRIUM
CHAPTER 3: CHEMICAL EQUILIBRIUM 1 LESSON OUTCOME Write & explain the concepts of chemical equilibrium Derive the equilibrium constant Kc or Kp Solving the problem using the ICE table 2 Equilibrium is a
More informationEquilibrium means that the rxn rates are equal. evaporation H20(l) condensation
Reversible reactions Most chemical reactions are reversible they can occur backwards as well as forwards reactants products Consider an open container of water (non-equilibrium) A closed water bottle is
More informationChemical Equilibrium. A state of no net change in reactant & product concentrations. There is a lot of activity at the molecular level.
Chemical Equilibrium A state of no net change in reactant & product concentrations. BUT There is a lot of activity at the molecular level. 1 Kinetics Equilibrium For an elementary step in the mechanism:
More informationHow would we know if equilibrium has been reached? Or if it will be reach?
1 How would we know if equilibrium has been reached? Or if it will be reach? The reaction quotient, Q, or trial KC, enables us to determine this information. The reaction quotient is determined by using
More informationEQUILIBRIUM. Opposing reactions proceed at equal rates Concs. of reactants & products do not change over time
EQUILIBRIUM Opposing reactions proceed at equal rates Concs. of reactants & products do not change over time Examples: vapor pressure above liquid saturated solution Now: equilibrium of chemical reactions
More information(g) 2NH 3. (g) ΔH = 92 kj mol 1
1 The uses of catalysts have great economic and environmental importance For example, catalysts are used in ammonia production and in catalytic converters (a) Nitrogen and hydrogen react together in the
More informationThe Equilibrium State
15.1 The Equilibrium State All reactions are reversible and under suitable conditions will reach a state of equilibrium. At equilibrium, the concentrations of products and reactants no longer change because
More information14.1 Factors That Affect Reaction Rates
14.1 Factors That Affect Reaction Rates 1) 2) 3) 4) 14.2 Reaction Rates How does increasing the partial pressures of the reactive components of a gaseous mixture affect the rate at which the compounds
More informationChapter 15: Chemical Equilibrium. Chem 102 Dr. Eloranta
Chapter 15: Chemical Equilibrium Chem 102 Dr. Eloranta Equilibrium State in which competing processes are balanced so that no observable change takes place as time passes. Lift Gravity Sometimes called
More informationThermodynamic and Stochiometric Principles in Materials Balance
Thermodynamic and Stochiometric Principles in Materials Balance Typical metallurgical engineering problems based on materials and energy balance NiO is reduced in an open atmosphere furnace by excess carbon
More informationChapter 7: Stoichiometry in Chemical Reactions
Chapter 7: Stoichiometry in Chemical Reactions Mini Investigation: Precipitating Ratios, page 315 A. ZnCl 2 (aq) + Na 2 CO 3 (aq) ZnCO 3 (s) + 2 NaCl(aq) 3 AgNO 3 (aq) + Na 3 PO 4 (aq) Ag 3 PO 4 (s) +
More informationChapter 4. Chemical Quantities and Aqueous Reactions
Chapter 4 Chemical Quantities and Aqueous Reactions Stoichiometry The study of the numerical relationship between chemical quantities in a chemical reaction Making Pizza The number of pizzas you can make
More information(b) Increase in pressure. (1)
1 This question is about the equilibrium reaction between hydrogen and carbon dioxide. H 2 (g) + O 2 (g) H 2 O(g) + O(g) H = +40 kj mol 1 What effect would the following changes have on the rate of reaction
More informationOFB Chapter 7 Chemical Equilibrium
OFB Chapter 7 Chemical Equilibrium 7-1 Chemical Reactions in Equilibrium 7-2 Calculating Equilibrium Constants 7-3 The Reaction Quotient 7-4 Calculation of Gas-Phase Equilibrium 7-5 The effect of External
More informationAS Paper 1 and 2 Kc and Equilibria
AS Paper 1 and 2 Kc and Equilibria Q1.When one mole of ammonia is heated to a given temperature, 50 per cent of the compound dissociates and the following equilibrium is established. NH 3(g) ½ N 2 (g)
More informationAP Chem Chapter 14 Study Questions
Class: Date: AP Chem Chapter 14 Study Questions 1. A burning splint will burn more vigorously in pure oxygen than in air because a. oxygen is a reactant in combustion and concentration of oxygen is higher
More informationAP Chemistry - Notes - Chapter 12 - Kinetics Page 1 of 7 Chapter 12 outline : Chemical kinetics
AP Chemistry - Notes - Chapter 12 - Kinetics Page 1 of 7 Chapter 12 outline : Chemical kinetics A. Chemical Kinetics - chemistry of reaction rates 1. Reaction Rates a. Reaction rate- the change in concentration
More informationName Chem 6 Section #
Equilibrium Constant and its Meaning 1. Write the expressions for K eq for the following reactions. a) CH 4 (g) + 2 H 2 S(g) CS 2 (g) + 4 H 2 (g) b) 2 N 2 O 5 (g) 4 NO 2 (g) + O 2 (g) c) 3 O 2 (g) 2 O
More informationChemical Equilibrium Basics
Chemical Equilibrium Basics Reading: Chapter 16 of Petrucci, Harwood and Herring (8th edition) Problem Set: Chapter 16 questions 25, 27, 31, 33, 35, 43, 71 York University CHEM 1001 3.0 Chemical Equilibrium
More informationChemistry Chapter 16. Reaction Energy
Chemistry Reaction Energy Section 16.1.I Thermochemistry Objectives Define temperature and state the units in which it is measured. Define heat and state its units. Perform specific-heat calculations.
More informationWrite equilibrium law expressions from balanced chemical equations for heterogeneous and homogeneous systems. Include: mass action expression.
Equilibrium 1 UNIT 3: EQUILIBRIUM OUTCOMES All important vocabulary is in Italics and bold. Relate the concept of equilibrium to physical and chemical systems. Include: conditions necessary to achieve
More information