Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. If you continue browsing the site, you agree to the use of cookies on this website. See our User Agreement and Privacy Policy.

Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. If you continue browsing the site, you agree to the use of cookies on this website. See our Privacy Policy and User Agreement for details.

Like this presentation? Why not share!

- Chemical Equilibrium by David Richardson 8783 views
- Chem 1 unit 11 presentation by bobcatchemistry 1709 views
- Ap chem unit 14 presentation part 1 by bobcatchemistry 5630 views
- Ap chem unit 15 presentation by bobcatchemistry 5410 views
- Chem unit 12 presentation by bobcatchemistry 1165 views
- Equilibrium student 2014 2 by Abraham Ramirez 4168 views

6,570 views

Published on

AP Chem Unit 13: Chemical Equilibrium

No Downloads

Total views

6,570

On SlideShare

0

From Embeds

0

Number of Embeds

685

Shares

0

Downloads

81

Comments

0

Likes

2

No embeds

No notes for slide

- 1. ChemicalEquilibriumAP Chem Unit 13
- 2. Chemical Equilibrium The Equilibrium Condition The Equilibrium Constant Equilibrium Expressions Involving Pressures Heterogeneous Equilibria Applications of the Equilibrium Constant Solving Equilibrium Problems Le Chatelier’s Principle
- 3. IntroductionTo this point, we have assumed thatreactions proceed to completion, that is,until one of the reactants runs out. Most reactions stop short of completion. In fact, the system reaches chemical equilibrium, the state where the concentrations of all reactants and products remain constant with time.
- 4. IntroductionAny chemical reactions carried out in aclosed vessel will reach equilibrium Some reactions, the equilibrium position favors the products so that the reaction appears to go to completion. The equilibrium position is said to lie “far to the right”. This reflects the direction of the products.
- 5. IntroductionSome reactions only occur to a slightextent. In this case, the equilibrium position is said to lie “far to the left”. This reflects the direction of the reactants.
- 6. The EquilibriumConstant13.1
- 7. The Equilibrium ConditionEquilibrium is not static but a highly dynamicsituation. On a molecular level manymolecules are moving back and forthbetween reactants and products. No net change in concentration of reactants and products.
- 8. The Equilibrium Condition H2O(g) + CO(g) ¬¾ H2(g) + CO2(g) ®
- 9. The Equilibrium Condition H2O(g) + CO(g) ¬¾ H2(g) + CO2(g) ®
- 10. The Equilibrium Condition H2O(g) + CO(g) ¬¾ H2(g) + CO2(g) ®The equilibrium positionlies far to the right. Thisreaction favors theproducts. But thereactants never reach a concentration ofzero.
- 11. The Equilibrium Condition H2O(g) + CO(g)¬¾ H2(g) + CO2(g) ®What would happenif H2O(g) was added tothe system?First, the forward reaction would increase,then the reverse reaction would increase.A new equilibrium would occur.
- 12. Characteristics of ChemicalEquilibriumThe equilibrium position is determined bymany factors: initial concentrations. relative energies of the reactants and products. relative degree of “organization” of the reactants and products.
- 13. The EquilibriumConstant13.2
- 14. Law of Mass ActionThe Law of Mass Action is a general descriptionof the equilibrium condition. jA + kB ¬¾ lC + mD ® l m [C] [D] K= j k [A] [B] The square brackets indicate the concentrations of the the reactants and products at equilibrium. K is the equilibrium constant.
- 15. Practice Problem #1Write the equilibrium expression for thefollowing reaction: 4NH3(g) + 7O2(g) ¬¾ 4NO2(g) + 6H2O(g) ® [NO2 ] [H 2O] 4 6 K= [NH 3 ]4 [O2 ]7
- 16. The Equilibrium ConstantThe value of the equilibrium constant at agiven temperature can be calculated if weknow the equilibrium concentrations of thereaction components. Equilibrium constants are typically given without units.
- 17. Practice Problem #2The following equilibrium concentrationswere observed for the Haber process for thesynthesis of ammonia at 127°C: [NH3] = 3.1 x 10-2 mol/l [N2] = 8.5 x 10-1 mol/l [H2] = 3.1 x 10-3 mol/la) Calculate the value of K at 127°C for thisreaction.3.8 x 104
- 18. Practice Problem #2The following equilibrium concentrationswere observed for the Haber process for thesynthesis of ammonia at 127°C: [NH3] = 3.1 x 10-2 mol/l [N2] = 8.5 x 10-1 mol/l [H2] = 3.1 x 10-3 mol/lb) Calculate the value of the equilibriumconstant at 127°C for the reaction: 2NH3(g) ¬¾ N2(g) + 3H2(g) ®2.6 x 10-5 (the reverse order reaction givesthe reciprocal of K)
- 19. Practice Problem #2The following equilibrium concentrationswere observed for the Haber process for thesynthesis of ammonia at 127°C: [NH3] = 3.1 x 10-2 mol/l [N2] = 8.5 x 10-1 mol/l [H2] = 3.1 x 10-3 mol/lc) Calculate the value of the equilibriumconstant at 127°C for the reaction: 1 2 N2(g) + 2 H 2(g) « NH3(g) 31.9 x 102 (When the coefficients are ½ ofthe balanced equation, new K = K1/2)
- 20. Equilibrium ExpressionSummary The equilibrium expression for a reaction is the reciprocal of that for the reaction written in reverse. When the balanced equation for a reaction is multiplied by a factor n, the equilibrium expression for the new reaction is the original expression raised to the nth power. Knew =Kon K values are customarily written without units. Law of mass action can describe reactions in the solution and gas phase.
- 21. Equilibrium ExpressionSummary Theequilibrium expression and constant for a reaction is the same at a given temperature, regardless of the initial amounts of the reaction components. equilibrium concentrations will not always be the same. See Table 13.1 p600
- 22. Equilibrium ExpressionA set of equilibrium concentrations is calledan equilibrium position. There is only one equilibrium constant for a particular system at a given temperature, but there is an infinite number of equilibrium positions.
- 23. Practice Problem #3These results were Initial Equilibriumcollected for two [SO2] = 2.00 M 1.50 Mexperiments involving [O2] = 1.50 M 1.25 Mthe reaction at 600°C [SO3] = 3.00 M 3.50 Mbetween gaseoussulfur dioxide andoxygen to form Initial Equilibriumgaseous sulfur [SO2] = 0.500 M 0.590 Mtrioxide: [O2] = 0 M 0.0450 MShow that the [SO3] = 0.350 M 0.260 Mequilibrium constantis the same in both 4.36 and 4.32, within experimental error.experiments.
- 24. EquilibriumExpressions InvolvingPressures 13.3
- 25. Pressure EquilibriaSo far we have described equilibria involvinggases in terms of concentrations. Equilibriainvolving gases also can be described withpressure. æ nö PV = nRT, P = ç ÷ RT, P = CRT èVøC represents the molar concentration of the gas. jA + kB ¬¾ lC + mD ® l m l m CC PP Kc = C D j k Kp = C D j k CC A B PP A B
- 26. Practice Problem #4The reaction for the formation of nitrosyl chloride: 2NO(g) + Cl2(g) ¬¾ 2NOCl(g) ®was studied at 25°C. The presurres at equilibriumwere found to be: NOCl =1.2 atm, NO = 5.0 x 10-2atm, Cl2 = 3.0 x 10-1 atm. Calculate the value ofKp for this reaction at 25°C.1.9 x 103
- 27. Kc vs. KpKp = Kc(RT)Δn Δn is the sum of the coefficients of the gaseous products minus the sum of the coefficients of the gaseous reactants jA + kB ¬¾ lC + mD ® Δn=(l + m) – (j + k) more moles of gas = more pressure
- 28. Practice Problem #5Using the value of Kp obtained in Problem#4, calculate the value of K at 25°C for thereaction: 2NO(g) + Cl2(g) ¬¾ 2NOCl(g) ® Kp =1.9 x 103K = 4.6 x 104
- 29. HeterogeneousEquilibria13.4
- 30. Homogenous vs. Heterogeneous Homogenous equlibria is where all the reactants are in the same phase. Typically gases Heterogeneous equilibria involve more than one phase. The position of a heterogeneous equilibrium does not depend on the amounts of pure solids or liquids present. Concentrations of pure solids and liquids cannot change. Concentrations of pure solids and liquids are not included in the equilibrium expression for the reaction
- 31. Example CaCO3(s) ¬¾ CaO(s) + CO2(g) ® [CO2 ]CCaO K= CCaCO3 K = [CO2 ] This simplification only occurs with pure solids or liquids and not solutions or gases.
- 32. Practice Problem #6Write the expressions for K and Kp for thefollowing processes:a) Solid phosphorus pentachloride decomposesto liquid phosphorus trichloride and chloride gas.b) Deep blue solid copper (II) sulfatepentahydrate is heated to drive off water vaporto form white solid copper (II) sulfate.
- 33. Practice Problem #6a) Solid phosphorus pentachloride decomposesto liquid phosphorus trichloride and chloride gas.K = [Cl2] and Kp=PCl2
- 34. Practice Problem #6b) Deep blue solid copper (II) sulfatepentahydrate is heated to drive off water vaporto form white solid copper (II) sulfate.K = [H2O]5 Kp= (PH2O)5
- 35. Applications ofthe EquilibriumConstant13.5
- 36. Applications of the EquilibriumConstantKnowing the equilibrium constant for areaction allows us to predict severalimportant features of the reaction The tendency for the reaction to occur (but not the speed). Whether or not a given set of concentrations represents an equilibrium condition.
- 37. Applications of the EquilibriumConstantIf the reaction is not at equilibrium, we candetermine which way the reaction is moving bytaking the current law of mass action ratio andcomparing it to the equilibrium constant. The ratio of non-equilibrium concentrations gives us the reaction quotient, Q. l m [C] [D] Q= j k [A] [B]
- 38. Applications of the EquilibriumConstantTo determine which direction a system will shift toreach equilibrium, we compare the values of Q andK. Q=K. The system is at equlibrium; no shift will occur. Q>K. The initial concentrations of product to initial reactants is too large. To reach equilibrium, the system must shift left, consuming products and forming reactants. Q<K. The ratio of initial concentrations of products to initial concentrations of reactants is too small. The system must shift to the right to form more products.
- 39. Applications of the EquilibriumConstant
- 40. Practice Problem #7For the synthesis of ammonia at 500°C, theequilibrium constant is 6.0 x 10-2. Predict thedirection in which the system will shift to reachequilibrium in each of the following cases:a) [NH3]=1.0x10-3M, [N2]=1.0x10-5M, [H2]=2.0x10-3Mb) [NH3]=2.00x10-4M, [N2]=1.50x10-5M, [H2]=3.54x10-1Mc) [NH3]=1.0x10-4M, [N2]=5.0M, [H2]=1.0x10-2Ma) Q>K, shift left. b)Q=K, no shift. c)Q<K, shift right.
- 41. Practice Problem #8Dinitrogen tetroxide in its liquid state was used asone of the fuels on the lunar lander for the NASAApollo missions. In the gas phase it decomposes togaseous nitrogen dioxide. N2O4(g) ¬¾ 2NO2(g) ®Consider an experiment in which gaseous N2O4was placed in a flask and allowed to reachequilibrium at a temperature where Kp=0.133. Atequilibrium, the pressure of N2O4 was found to be2.71atm. Calculate the equilibrium pressure ofNO2(g)..600 atm
- 42. The ICE TableWhen initial concentrations and equilibriumconstants are known, but none of the equilibriumpositions are known it is helpful to write an ICEtable. I:The Initial concentrations of products and reactants C: The Change in concentrations needed to reach equilibrium is summarized in terms of variables. E: The Equilibrium values are summarized as a combination of initial and change needed.
- 43. ICE Table Example:Consider the reaction: N2(g) + 3H2(g) ¬¾ 2NH3(g) ® K = 6.0 x 10-2 at 500°C.The initial concentration of N2 is 3.0M and H2 is2.0M. What are the equilibrium positions of thisreaction? Initial (M) Change Equilibrium (M)N2 3.0 -x 3.0 - xH2 2.0 -3x 2.0 – 3xNH3 0.0 +2x 2x
- 44. ICE Table Example: Initial (M) Change Equilibrium (M) ¬¾ ®N2 3.0 -x 3.0 - xH2 2.0 -3x 2.0 – 3xNH3 0.0 +2x 2x N2(g) + 3H2(g) ¬¾ 2NH3(g), K = 6.0 x 10-2 ® [NH 3 ]2 [2x]2 K = .060 = = [N 2 ][H 2 ] [3.0 - x][2.0 - 3x]3 3
- 45. Practice Problem #9At a certain temperature a 1.00 L flask initiallycontained 0.298 mol PCl3(g) and 8.70x10-3 mol ofPCl5(g). After the system had reached equilibrium,2.00 x 10-3 mol Cl2(g) was found in the flask.Gaseous PCl5 decomposes according to thereaction: PCl5(g) ¬¾ PCl3(g) + Cl2(g). Calculate ®the equilibrium concentrations of all species andthe value of K. >
- 46. Practice Problem #9Initial: 0.298 mol PCl3(g). 8.70x10-3 mol of PCl5(g)in 1.00Lequil: 2.00 x 10-3 mol Cl2(g) PCl5(g) ¬¾ PCl3(g) + Cl2(g) ®Equilibrium expression: K = [Cl2 ][PCl3 ]ICE Table: [PCl5 ] Initial (M) Change Equilibrium (M) PCl5(g) 0.298 -2.00 x 10-3 PCl3(g) 8.70x10-3 +2.00 x 10-3 Cl2(g) 0.0 +2.00 x 10-3 2.00 x 10-3k=8.96 x 10-2
- 47. Practice Problem #10Carbon monoxide reacts with steam toproduce carbon dioxide and hydrogen. At700 K the equilibrium constant is 5.10.Calculate the equilibrium concentrations ofall species if 1.00 mol of each component ismixed in a a 1.00L flask. >
- 48. Practice Problem #10Reaction: CO(g) + H2O(g)¬¾ CO2(g) + H2(g), K= 5.10 ®Which way does the equilibrium need to go? Q= [CO2 ][H 2 ] Q=1.00 Q<K, shift right [CO][H 2O]ICE Table: Initial (M) Change Equilibrium (M) CO 1.0 -x 1.0 - x H2O 1.0 -x 1.0 - x CO2 1.0 +x 1.0 + x H2 1.0 +x 1.0 + x
- 49. Practice Problem #10 Initial (M) Change Equilibrium (M) CO 1.0 -x 1.0 - x H2O 1.0 -x 1.0 - x CO2 1.0 +x 1.0 + x H2 1.0 +x 1.0 + x (1.0 - x)(1.0 - x) (1.0 - x) 2 K = 5.10 = = (1.0 + x)(1.0 + x) (1.0 + x)2x = 0.387 mol/L[CO] & [H2O] = .613M, [CO2] & [H2] = 1.387MDouble check K with expression.
- 50. Practice Problem #11Assume that the reaction for the formation ofgaseous hydrogen fluoride from hydrogen andfluorine has an equilibrium constant of 1.15 x 102 ata certain temperature. In a particular experiment,3.000 mol of each component was added to a1.500 L flask. Calculate the equilibriumconcentration of all species. >
- 51. Practice Problem #11Reaction: H2(g) + F2(g) ¬¾ 2HF(g), K=1.15 x 102 ®Which way does the equilibrium need to go? Initial concentrations: 3.000mol/1.500L = 2.00 M Q= [HF]2 Q=1.00 Q<K, shift right [H 2 ][F2 ]ICE Table: Initial (M) Change Equilibrium (M) H2 2.0 -x 2.0 - x F2 2.0 -x 2.0 - x HF 2.0 +2x 2.0 + 2x
- 52. Practice Problem #11 Initial (M) Change Equilibrium (M) H2 2.0 -x 2.0 - x F2 2.0 -x 2.0 - x HF 2.0 +2x 2.0 + 2x (2.000 + 2x)2 K = 1.15x10 2 = (2.000 - x)2x=1.528[H2] & [F2]= 0.472 M[HF] = 5.056 MCheck K with equilibrium values
- 53. SolvingEquilibriumProblems13.6
- 54. Solving Equilibrium ProblemsStrategy:1. Write the balanced equation for the reaction.2. Write the equilibrium expression using law of mass action.3. List the initial concentrations.4. Calculate Q, and determine the direction of the shift needed for equilibrium.
- 55. Solving Equilibrium ProblemsStrategy continued:5. Define the change needed to reach equilibrium, and define the equilibrium concentrations by applying the change to the initial concentrations.6. Substitute the equilibrium concentrations into the equilibrium expression, and solve for the unknown.7. Check your calculated equilibrium concentrations by making sure the give the correct value of K.
- 56. Solving Equilibrium ProblemsTypical systems do not produce anexpression that can be solved by taking thesquare root of both sides. To solve some expressions, we will use the quadratic equation. -b ± b 2 - 4ac x= 2a
- 57. Equilibrium ExampleSuppose for a synthesis of hydrogen fluoride fromhydrogen and fluorine, 3.000 mol H2 and 6.000 molF2 are mixed in a 3.000 L flask. Assume theequilibrium constant for the synthesis reaction atthis temperature is 1.15x102. What are theequilibrium concentrations of each component. >
- 58. Equilibrium Example1. Write the balanced equation for the reaction. H2(g) + F2(g) ¬¾ 2HF(g) ®2. What is the equilibrium expression? [HF]2 K = 1.15x10 2 = [H 2 ][F2 ]3. What are the initial concentrations? [H2] = 3.00mol/3.00 L = 1.000M [F2] = 6.00mol/3.00 L = 2.000M [HF]= 0
- 59. Equilibrium Example4. What is Q? Q does not need to be calculated in this example. Since HF is not present initially, we can assume that the reaction will shift to the right to reach equilibrium.5. What change is required to reach equilibrium? Initial (M) Change Equilibrium (M) H2 1.0 -x 1.0 - x F2 2.0 -x 2.0 - x HF 0.0 +2x 2x
- 60. Equilibrium Example6. What is the value of K? (Use ICE in expression) [HF]2 (2x)2 K= = [H 2 ][F2 ] (1.000 - x)(2.000 - x) collect terms and set = 0 (1.000 - x)(2.000 - x)(1.15x102 ) = 2x 2 ax2 + bx + c = 0 (1.11x102 )x 2 - (3.445x102 )x + (2.30x102 ) = 0 a=1.11x102, b=-3.45x102, c=2.30x102
- 61. Equilibrium Example6. What is the value of K? a=1.11x102, b=-3.45x102, c=2.30x102 -b ± b 2 - 4ac x= 2a Substituting these values give two answers for x: x=2.14 mol/L and 0.968 mol/L Both of these results are not valid; the changes in concentration must be checked for validity
- 62. Equilibrium Example6. What is the value of K? x=2.14 mol/L or 0.968 mol/L [H2] = 1.000 – x, [F2] = 2.000 – x, [HF] = 2x [H2] = 3.2x10-2M, [F2] = 1.032M, [HF] = 1.936M7. Check concentrations by substituting them into the equilibrium expression.
- 63. Practice Problem #12Assume that gaseous hydrogen iodide issynthesized from hydrogen gas and iodine vaporat a temperature where the equilibriumconstant is 1.00x102. Suppose HI at 5.000x10-1atm, H2 at 1.000x10-2 atm, and I2 at 5.000x10-3atm are mixed in a 5.000L flask. Calculate theequilibrium pressures of all species.
- 64. Practice Problem #121. Write the balanced equation2. What is the equilibrium expression?3. What are the initial pressures?4. What is the value of Q?
- 65. Practice Problem #125. What is the change required?6. What is the value of Kp (& equilibrium pressures)?7. Expression check.PHI=4.29x10-1atm, PH2=4.55x10-2atm, PI2=4.05x10-2atm
- 66. Small Equilibrium ConstantsSometimes there are simplifications that can bemade to the math of some equilibriumproblems. When reactions lie far to the left, the equilibrium constants can be very small. Changes in initial concentrations can be negligible and partially disregarded.
- 67. Small K ExampleGaseous NOCl decomposes to form the gasesNO and Cl2. At 35°C the equilibrium constant is1.6x10-5. In an experiment in which 1.0 mol NOClis placed in a 2.0 L flask, what are theequilibrium concentrations?1. What is the equation? 2NOCL(aq) ¬¾ 2NO(g) + Cl2(g) ®2. What is the expression? [NO]2 [Cl2 ] K= 2 = 1.6x10 -5 [NOCl]
- 68. Small K Example3. What are the initial concentrations? [NOCl]=0.50M, [NO]=0, [Cl2] = 04. What is Q? Direction must lie to the right for equilibrium.5. What is the change required? Initial (M) Change Equilibrium (M) NOCl 0.50 -2x 0.50 - 2x NO 0.0 +2x 2x Cl2 0.0 +x x
- 69. Small K Example 6. What is the value of K (& concentrations) Initial Change Equilibrium (M) (M) (2x)2 (x)NOCl 0.50 -2x 0.50 - 2x K = 1.6x10 -5 = (0.50 - 2x)2NO 0.0 +2x 2xCl2 0.0 +x x x must represent a relatively small number. In order to simplify this expression, we can assume that: 0.50 – 2x ≅ .50 Therefore we can simplify the expression: (2x)2 (x) 4x 3 1.6x10 -5 = 2 = (0.50) (.50)2 x=1.0x10-2 x 3 = 1.0x10 -6
- 70. Small K Example[NOCl] = .50 – 2x ≈ 0.50 M[NO] = 2.0 x 10-2 M[Cl2] = 1.0 x 10-2 M7. Check the K expression.
- 71. Le Chatelier’sPrinciple13.7
- 72. Le Chatelier’s PrincipleSeveral factors can control the position of achemical equilibrium. Changes in Concentration Temperature (removal or addition of energy) Pressure
- 73. Le Chatelier’s PrincipleLe Chatelier’s principle states that if a change isimposed on a system at equilibrium, the positionof the equilibrium will shift in a direction thattends to reduce that change. It is important to realize that although changes to the reaction may alter the equilibrium positions, they do not alter the equilibrium constant.
- 74. Le Chatelier’s PrincipleChange in Concentration: If a component (reactant or product) is added to a reaction system at equilibrium (at constant T and P or constant T and V), the equilibrium position will shift in the direction that lowers the concentration of that component. If a component is removed, the opposite effect occurs.
- 75. Practice Problem #13Arsenic can be extracted from its ores by firstreacting the ore with oxygen (called roasting) toform solid As4O6, which is then reduced usingcarbon: As4O6(s) + 6C(s) As4(g) + 6CO(g)Predict the direction ¬¾ shift of the equilibrium of ® theposition in response to each of the followingchanges in concentration.a) Addition of carbon monoxide.b) Addition or removal of carbon or tetrarsenic hexoxide.c) Removal of gaseous arsenic.left shift, no effect, right shift
- 76. Le Chatelier’s PrincipleChange in Pressure: There are three ways to change the pressure of a reaction system involving gaseous components: 1. Add or remove a gaseous reactant or product. 2. Add an inert gas (one not involved in the reaction). 3. Change the volume of the container
- 77. Le Chatelier’s PrincipleChange in Pressure: When an inert gas is added, there is no effect on the equilibrium position. The addition of an inert gas increases the total pressure but has no effect on the concentrations of the reactants or products. The system will remains at the original equilibrium position
- 78. Le Chatelier’s PrincipleChange in Volume: When the volume of the container holding a gaseous system is reduced, the system responds by reducing its own volume. This is done by decreasing the total number of gaseous molecules in the system. A reaction will shift in order to reduce the number of gas molecules.
- 79. Practice Problem #14Predict the shift in equilibrium position that willoccur for each of the following processes whenthe volume is reduced.a. The preparation of liquid phosphorus trichloride by the reaction. P4(s) + 6Cl2(g) ¬¾ 4PCl3(l) ®right shift
- 80. Practice Problem #14Predict the shift in equilibrium position that willoccur for each of the following processes whenthe volume is reduced.b. The preparation of gaseous phosphorus pentachloride according to the equation: PCl3(g) + Cl2(g) ¬¾ PCl5(g) ®right shift
- 81. Practice Problem #14Predict the shift in equilibrium position that willoccur for each of the following processes whenthe volume is reduced.c. The reaction of phosphorus trichloride with ammonia: PCl3(g) + 3NH3(g) ¬¾ P(NH2)3(g) + 3HCl(g) ®no effect
- 82. Le Chatelier’s PrincipleChange in Temperature: If energy is added or removed from a system in equilibrium, the system will shift according to the heat of the reaction. Heat is a product in an exothermic reaction. Heat is a reactant in an endothermic reaction. The effect of temperature on equilibrium changes the value of K because K changes with temperature.
- 83. Practice Problem #15For each of the following reactions, predict howthe value of K changes as the temperature isincreased.a. N2(g) + O2(g) ¾ 2NO(g) ¬® ΔH = 181 kJb. 2SO2(g) + O2(g)¬¾ 2SO3(g) ΔH = -198 kJ ®a) shift right b) shift left
- 84. THE END3 more units to go!!!

No public clipboards found for this slide

×
### Save the most important slides with Clipping

Clipping is a handy way to collect and organize the most important slides from a presentation. You can keep your great finds in clipboards organized around topics.

Be the first to comment