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EQUILIBRIUM
means - balance EQUILIBRIUM a state in which opposing forces or GENERAL CONCEPTS influences are balanced
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CHEMICAL REACTTION REVERSIBLE CHEMICAL REACTIONS
A substance changes into another can go in both directions substance the reactants can change into the products and the products can change back into the reactants
reactants products
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CHEMICAL EQUILIBRIUM CHEMICAL EQUILIBRIUM
the forward and reverse reactions are proceeding at the same rate and reaction never cease the concentration of reactants and products are rarely equal the concentrations of all species remain constant over time
N2O4 N2O4 NO2 NO2 http://schoolbag.info/chemistry/central/136.html http://www.kchemistry.com/the_keq_family.htm 5 6 5 6
CONCENTRATION CHANGES MASS ACTION EXPRESION
Relationship between the concentrations of the reactants and products for any chemical system at equilibrium is known as: REACTION QUOTIENT, Q. IT IS RELATIONSHP BETWEEN THE CONCENTRATION OF THE REACTANTS AND The decomposition of N O (g) into NO (g). 2 4 2 PRODUCTS FOR ANY CHEMICAL system AT Any
The concentrations of N2O4 and NO2 change relatively quickly at time first, but eventually stop changing with time
when equilibrium is reached. 7 8 7 8
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THE NUMERICAL VALUE Equilibrium among OF THE MASS ACTION EXPRESSION H2, I2, HI
→ is equal Q at equilibrium H2 (g) + I2 (g) 2HI(g) [F] f [G]g → Different amounts of the dD + eE fF + gG d e = Kc [D] [E] reactants and products are placed in a 10.0 L reaction The form is always “products over vessel at 440oC where the reactants” raised to the appropriate gases establish equilibrium.
powers When equilibrium is The exponents in the mass action reached, different amounts of reactants and products expression are the same as remain. the stoichiometric coefficients.
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EQUILIBRIUM AMONG H2, I2, HI EQUILIBRIUM AMONG H2, I2, HI
➢ the mass action expression = reaction quotient, Q at equilibrium The reaction quotient, Q, can be evaluated [HI]2 at any concentrations. 2 Expt. [H2 ] [I2 ] [HI] [HI] [H2 ][I2 ] Q = [H ][I ] I 0.0222 0.0222 0.156 49.4 2 2 II 0.0350 0.0450 0.280 49.8 At equilibrium (and 440oC) for the this III 0.0150 0.0135 0.100 49.4 reaction the reaction quotient has the value 49.5 (an unit less number). IV 0.0442 0.0442 0.311 49.5
Average = 49.5 11 12 11 12
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EQUILIBRIUM AMONG H2, I2, HI QUIZ
The value 49.5 is called the equilibrium For the following reaction find the formula of
constant, Kc, and characterizes the system KC: 2 [HI] o 3H2 + N2 ↔ 2NH3 Kc = = 49.5 (at 440 C) [H2 ][I2 ] a) b) This relationship is called the equilibrium law for the system. For chemical equilibrium to exist, the reaction c) quotient Q must be equal to the equilibrium
constant Kc. 13 14 13 14
QUIZ SOLUTION
Given the equilibrium concentrations for the reaction below, what is the value of the equilibrium constant in terms of concentrations?
KC = 6 2A(aq) + 3 B(aq) ↔ C(aq) + 2 D(aq)
At equlibrium: [A]= 1 M; [C]= 3M
[B]= 2 M; [D] = 4 M 15 16 15 16
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QUIZ OPERATIONS PERFORMED ON EQUILIBRIUM EXPRESSIONS
For the reaction below, the equilibrium Changing the direction of equilibrium constant is equal to one. Which of the these Multiplying coefficients sets of equilibrium concentrations is possible? Adding chemical equilibria 2A(aq) + B(aq) ↔ 4C(aq) A) [A]= 2 M, [B]= 1 M, [C]= 1 M B) [A]= 2 M, [B]= 2 M, [C]= 2 M C) [A]= 1 M, [B]= 1 M, [C]= 2 M D) [A]= 1 M, [B]= 1 M, [C]= 1 M
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CHANGING THE DIRECTION OF Multiplying the coefficients by a factor EQUILIBRIUM
➢ When the coefficients in an equation are multiplied by a factor, the equilibrium constant ➢ When the direction of an equilibrium is reversed, is raised to a power equal to that factor; the new equilibrium constant is the reciprocal of the original: → [PCl5 ] → [PCl ] PCl + Cl PCl K = PCl + Cl PCl K = 5 3 2 5 c [PCl ][Cl ] 3 2 5 c [PCl ][Cl ] 3 2 3 2 2 2 → " [PCl3 ] [Cl2 ] 2 → ' [PCl3 ][Cl2 ] 1 2PCl 2PCl + 2Cl K = = K PCl PCl + Cl K = = 5 3 2 c 2 c 5 3 2 c [PCl5 ] [PCl5 ] Kc
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Adding chemical equilibriums EXAMPLE
➢When chemical equilibria are added, their equilibrium constants are multiplied Using the equilibrium constants from Table below calculate the value of K for the reaction 2 [N O] CaCO (s) + H+(aq) → Ca2+(aq) + HCO –(aq) 2N + O → 2N O K = 2 3 3 2 2 2 c1 2 [N2 ] [O2 ] 4 → [NO2 ] 2+ 2– –7 2N O + 3O 4NO K = CaCO3(s) → Ca (aq) + CO3 (aq) K1 = 5.01 x 10 2 2 2 c2 [N O]2[O ]3 2 2 – + 2– –11 HCO3 (aq) → H (aq) + CO3 (aq) K2 = 5.01 x 10 Adding gives : 4 → [NO2 ] 2N2 + 4O2 4NO2 Kc3 = 2 4 = Kc1 Kc2 [N2 ] [O2 ]
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SOLUTION: The magnitude of K and the position of equilibrium. The net reaction is the sum of reaction 1 and the reverse of reaction 2:
K = 5.01 x 10–7 2+ 2– 1 CaCO3(s) → Ca (aq) + CO3 (aq)
+ 2– – H (aq) + CO3 (aq) → HCO3 (aq)
+ 2+ – A large amount of product and very little reactant at CaCO3(s) + H (aq) → Ca (aq) + HCO3 (aq) equilibrium gives K>>1 (large K). When K 1 , approximately equal amounts of reactant and product are present at equilibrium. When K<<1, mostly reactant and very little product are present at equilibrium. 23 24 23 24
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LARGE KC VALUES SMALL KC
Large KC value means that reaction proceeds Small KC value means that reaction proceeds almost completely to the rights (products are almost completely to the lefts (reactants are favoured, products predominate over reactants) favoured, reactants predominate over products) e.g. e.g.
2H2 + O2↔ 2H2O N2 + O2 ↔2NO
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INTERMEDIATE KC QUIZ Look at the following equilibrium constants. Do -2 2 When KC has a value between 10 and 10 it these equilibrium reactions lie to the left, to the tells that at equilibrium all components of right, or in the middle? reaction are present in significant amounts, A) H CO +H O↔HCO - + H O+ and that the equilibrium reaction is fairly evenly 2 3 2 3 3 K = 4.2 . 10-7 balanced (neither lying to the left nor to the C . 12 right). B) 2NO + O2 ↔ 2NO2 Kc = 2.25 10 o C) CH + H O↔ CO + 3H K = 9.4 .10-1 e.g. H2 + I2 ↔ 2HI at 400 C KC = 50.5 4 2 2 c - + (reactants and products are both present in D) CH3COOH + H2O↔ CH3COO + H3O . -5 approximately equal amounts). Kc = 1.8 10
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PREDICTING THE DIRECTION OF A REACTION 1. Q< KC
Q (reaction quotient)
Kc (equilibrium constant)
aA+bB → cC + dD [C] and [D] increase [A] and [B] (must be 2 values: Kc (equilibrium constant) decrease formed) Q (reaction quotient) (must be Greater than at consumed) equilibrium
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2. Q> KC 3. Q=KC
Q (reaction quotient) The reaction quotient is the same as the K (equilibrium constant) c equilibrium constant. ➢ [C] and [D] The reaction is already at equilibrium, decrease; no changes in the balance between the forward must be [A] and [B] consumed and back reactions. increase must be Less than at formed equilibrium
Reaction proceeds from the right to the left 31 32 31 32
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PROBLEM TO SOLVE SOLVING The concentration of the three substances in the following reaction:
H2 + I2 ↔ 2HI -1 are as follows: [H2] = 1.5 mol L ’ [I ] = 0.5 mol L-1 2 Q = 1.3 [HI] = 1 mol L-1
KC = 50.5. 1.3 < 50.5 In which direction would the reaction need to proceed for equilibrium to be reached? The reaction must proceed from the left to the right (forward reaction) for equilibrium to be reached 33 34 33 34
EQUILIBRIUM CONSTANT IN TERMS TO SELF-CHECK OF PARTIAL PRESSURES The concentration of the three substances in the following reactions: The gas law can be used to write the equilibrium constant in terms of partial H2 + I2 ↔ 2HI -1 pressures are as follows: [H2] = 0.05 mol L ’ PV = nRT -1 [I2] = 1.05 mol L n [HI] = 1.9 mol L-1 P = RT = MRT molarity V KC = 50.5. In which direction would the reaction need to Equilibrium constants written in terms of proceed for equilibrium to be reached? partial pressures are given the symbol Kp
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EQUILIBRIUM CONSTANT KC AND KP EQUILIBRIUM CONSTANT KC AND KP
The two different forms of the equilibrium → N2 (g) + 3H 2 (g) 2NH 3 (g) constants can be related [NH ]2 P2 PV = nRT K = 3 and K = NH3 c [N ][H ]3 P P P3 n 2 2 N2 H 2 P = RT = MRT so that V The size of the equilibrium constant gives ng a measure of how the reaction proceeds. KP = Kc (RT ) where
General statements can be made about ng = (moles of gaseous products) the equilibrium constant (either K or K ). c P -(moles of gaseous reactants)
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QUIZ QUIZ CONTINUATION
Nitrogen gas and hydrogen gas combine to For the same process at 25oC, K = 3.5 x108. form ammonia in the Haber Bosch process. C What is K ? What equation can be used to convert K to K p C p -2 for the Haber Bosch process Kp= KC (RT) . 8 . . -2 . 5 Kp =(3.5 10 ) (0.082 298) = 5.8 10 N2(g) + 3H2(g) ↔ 2NH3(g) Δn Kp = Kc (RT)
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HOMOGENEOUS AND HETEROGENEOUS REACTIONS In a homogeneous reactions, all the reactants and products are in the same phase. Heterogeneous reactions involve more than one phase e.g. Thermal decomposition of sodium bicarbonate (baking soda)
heat 2NaHCO3 (s) ⎯ ⎯⎯ → Na2CO3 (s) + H 2O(g) + CO2 (g)
Heterogeneous reactions can come to
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HETEROGENOUS EQUILIBRIUM HETEROGENOUS EQUILIBRIUM
In general, equilibrium constant expressions are presented in terms of molar concentration If NaHCO3 is placed in a sealed or gas phase pressures. container, heterogeneous equilibrium is For pure liquids or solids, experiments have established → shown that: 2NaHCO3 (s) Na2CO3 (s) + CO2 (g) + H 2O(g) The position of the equilibrium state of a The equilibrium constant is: system does not depend on the amount of [Na CO ][CO ][H O] liquid or solid in the reaction, provided that 2 3 2 2 = K [NaHCO ]2 some exists. 3 The equilibrium law involving pure Pure liquids and solids do not appear in the liquids and pure solids can be simplified equilibrium expression. 43 44 43 44
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HETEROGENOUS EQUILIBRIUM HETEROGENOUS EQUILIBRIUM
The equilibrium law for a heterogeneous reaction is written without concentration terms for pure solids or pure liquids. For a pure liquid or solid, the ratio of amount of substance to volume of substance is constant Kc = [CO2 ][H2O] The concentration of a substance in a solid is constant. Doubling the number of moles doubles the volume, but the ratio of moles to volume remains the The equilibrium constants found in tables same. represent all the constants combined
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QUIZ Solution
Write the equilibrium constant expression What is the equilibrium constant expression Kp the normal way for gases, in terms of the partial for the following reaction? pressures of each gas. CO2(g) + H2(g) < - > CO(g) + H2O(l) The liquid water on the right hand side of the equation will not have a term in the expression,
so the answer is:
K = PCO/PCO2PH2
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FREE GIBBS ENERGY AND EQUILIBRIUM APPLICATION OF VAN’T HOFF ISOTERM 1.
9 o ΔG is related to equilibrium constant Kc through CO + Cl2 ↔COCl2 K = 4.57 x10 at 100 C van’t Hoff isotherm ΔG = -RT ln K ΔG = - (8.31 J K -1mol-1 ) ((273.15 + 100)K) (ln 4.57 x109) = - 69.0 kJ mol-1 ΔG = - RT ln Kc ΔG = 0 at equilibrium because both reactant and product concentrations remain constant. ΔG is negative, so reaction occurs spontaneously, reaction lies to the right (the products predominate over the reactants) and it is energetically favourable
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LE CHÂTELIER’S PRINCIPLE APPLICATION OF VAN’T HOFF ISOTERM 2.
-31 o N2 + O2 ↔2NO K = 4.7 x10 at 25 C If an outside influence upsets an equilibrium, the ΔG = -RT ln K system undergoes a change in the direction that counteracts the disturbing influence and, if possible, ΔG = - (8.31 J K -1mol-1 ) ((273.15 + 25)K) (ln 4.7 returns the system to equilibrium -31 -1 x10 ) = +173.0 kJ mol common “stresses” ΔG is positive, so reaction does not occur ➢ Adding or removing a product or reactant spontaneously, reaction lies to the left (very few ➢ Temperature products form) and it is energetically ➢ Pressure unfavourable ➢ Catalyst ➢ Inert gas
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Adding or removing a product or reactant REACTANTS OR PRODUCTS ADD/ REMOVE A + 2B ↔ C + D The equilibrium shifts to remove reactants increase the concentration of A. or products that have been added. It means that the position of equilibrium will The equilibrium shifts to replace reactants move so that or products that have been removed. the concentration of A decreases again - by reacting it with B and turning it into C and D.
The position of equilibrium moves to the right.
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REACTANTS OR PRODUCTS ADD/ REMOVE TEMPERATURE
Decreasing A A + 2B ↔ C + D Increasing the temperature shifts a reaction in a direction that produces an endothermic (heat-absorbing) change. The position of equilibrium will move so that the concentration of A increases again. It means that more C and D will react to replace Decreasing the temperature shifts a the A that has been removed. reaction in a direction that produces an exothermic (heat-releasing) change.
The position of equilibrium moves to the left. https://www.youtube.com/watch?v=RjFW3smI1fY https://www.youtube.com/watch?v=R0z8Ya-_kI4 55 56 55 56
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TEMPERATURE TEMPERATURE
A + 2B ↔ C + D ΔH= -250 KJ MOL-1 This shows that 250 kJ is evolved (hence the negative sign) when 1 mole of A reacts completely with 2 moles of B. For reversible reactions, the value is always given as if the reaction was one-way in the forward direction.
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TEMPERATURE PRESSURE/ VOLUME
This only applies to reactions involving gases:
A(g) + 2B(g) ↔ C(g) + D(g) Increasing pressure The more molecules are present in the container, the Suppose the system is at equilibrium at 300°C, higher the pressure will be. and the temperature is increasing to 500°C. The system can reduce the pressure by reacting in such a way as to produce fewer molecules. How can the reaction counteracts the change it has been made? How can it cool itself down again?
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PRESSURE ADDING CATALYST OR INERT GAS A(G) + 2B(G) ↔ C(G) + D(G) Decreasing pressure ➢ Catalysts have no effect on the position of equilibrium Producing more molecules. In this case, the Catalysts change how fast a system achieves position of equilibrium will move towards the equilibrium, not the relative distribution of reactants left-hand side of the reaction. and products https://www.youtube.com/watch?v=YMqyG9Q ➢ Adding an inert gas at constant volume G6oc If the added gas cannot react with any reactants or products it is inert towards the substances in the https://www.youtube.com/watch?v=L6GfhqoCz equilibrium 8Y No concentration changes occur, so Q still equals K and no shift in equilibrium occurs
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TO SUM UP A(G) + 2B(G) ↔ C(G) + D(G) + H QUIZ Change Effect Increasing (reactants)favours the forward reaction Decreasing (reactants)favours the back reaction What properties can be changes to shift the Concentration Increasing (products)favours the back reaction reaction equilibrium? Decreasing (products) favours the forward reaction A) volume B) temperature Increasing the pressure favours the reactions which C) changing the reactant or product Pressure yields the smaller number of molecules Decreasing the pressure favours the reactions concentration which yields the largest number of molecules D) catalyst Increasing the temperature favours the Temperature endothermic reaction E) inert gas Decreasing the temperature favours the exothermic the true answers are……. reactions 63 64 63 64
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QUIZ QUIZ
Le Chatelier's principle states that Changes in pressure will only affect substances a) if a chemical system at equilibrium is stressed, that are in the ______state. the system will adjust to increase the stress a. gaseous b) if a chemical system at equilibrium is stressed, b. liquid the system will adjust to reduce the stress c. solid c) if a chemical system at equilibrium is stressed, the system will not adjust
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EXAMPLE EXAMPLE
Predicting the Effect of Changing Volume on Gas- NH + 2O ↔ HNO + H O Phase Reactions: 3(G) 2(G) 3(L) 2 (L) Predict whether a decrease in the volume of the container will drive an equilibrium system for each Decreased volume shifts the system to the side reaction toward more products, toward more reactants, or neither. of the reaction that has fewer moles of gas.
a. NH3(g) + 2O2(g) ↔ HNO3(l) + H2O(l)
b. CO2(g) + CF4(g) ↔ 2COF2(g)
c. C(s) + H2O(g) ↔ CO(g) + H2(g)
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EXAMPLE EXAMPLE
CO2(G) + CF4(G) ↔ 2COF2(G)
C(S) + H2O(G) ↔ CO(G) + H2(G) This reaction has the same number of moles of gaseous reactants and products, Be careful with this one. Although there are the same number of moles of reactants and products, one of the reactants is a solid.
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EQULIBRIUM CALCULATIONS EQULIBRIUM CALCULATIONS
N2O4(g) ↔ 2NO2(g) Equilibrium calculations can be divided into If 0.0350 mol N2O4(g) is placed in a 1 L flask at two main categories: 25oC, at equilibrium the following concentration 1) Calculating equilibrium constants from known are present : equilibrium concentrations or partial pressures. [N2O4 ] = 0.0292 mol/L 2) Calculating one or more equilibrium [NO2 ] = 0.0116 mol/L, and concentrations or partial pressures using the 2 2 [NO2 ] (0.0116) −3 known value of Kc or KP Kc = = = 4.6110 [N2O4 ] (0.0292)
Calculating the equilibrium constant this way is easy. 71 72 71 72
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EQULIBRIUM CALCULATIONS EQULIBRIUM CALCULATIONS
➢ Example: Ethyl acetate, CH3CO2C2H5, is produced from acetic acid and ethanol by the More commonly, a set of initial conditions reaction and an equilibrium constant are used for → calculation of reagents and products CH 3CO2 H(l) + C2 H5OH (l) CH 3CO2C2 H5 (l) + H 2O(l) concentration at equilibrium. o At 25 C, Kc= 4.10 for this reaction. Suppose 0.100 mol of ethyl acetate and 0.150 mol of water are The Initial, Change, Equilibrium or “ICE” placed in a 1.00 L reaction vessel. table is a useful way to summarize the What are the concentrations of all species at problem. equilibrium? ANALYSIS: Use an ICE table and the equilibrium constant to find the concentrations. 73 74 73 74
EQULIBRIUM CALCULATIONS EQULIBRIUM CALCULATIONS
All species are in the liquid phase. This can be solved by putting it in Let x = concentration that reacts, then quadratic form: → For an equation in the form CH 2CO2 H + C2 H5OH CH 3CO2C2 H5 + H 2O − b b2 − 4ac I (M ) 0 0 0.100 0.150 ax2 + bx + c = 0, x = 2a C (M ) + x + x - x - x For this system : E (M ) x x 0.100- x 0.150- x 0.0150-0.250x + x2 K = = 4.10 or (0.100- x)(0.150- x) c x2 Kc = 4.10 = 2 x 3.10 x2 + 0.250 x - 0.0150 = 0 2 0.0150- 0.250x + x and x- = −0.121 and x+ = 0.0401 = 2 x 75 76 75 76
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EQULIBRIUM CALCULATIONS EQULIBRIUM CALCULATIONS - SIMPLIFICATION
Negative concentrations are not Sometime simplifications can be made allowed, so Example: Nitrogen and oxygen react to form nitrogen monoxide
x = 0.0401 and at equilibrium → N2 (g ) + O2 (g ) 2NO(g ) [CH 3CO2 H ] = x = 0.0401 M with K = 4.8x10-31. [C2 H5OH ] = x = 0.0401 M c o In air at 25 C and 1 atm, the N2 concentrations [CH 3CO2C2 H5 ] = 0.100 − x = 0.060 M and O2 are initially 0.033 M and 0.00810 M. [H 2O] = 0.150 − x = 0.011 M What are the equilibrium concentrations? ANALYSIS: The equilibrium constant is very small, A similar procedure can be used to very little of the reactants will be converted into products calculate partial pressures using KP 77 78 77 78
EQULIBRIUM CALCULATIONS - SIMPLIFICATION EQULIBRIUM CALCULATIONS - SIMPLIFICATION
N + O → 2NO 2 2 Substituting:
I (M ) 0.033 0.00810 0 [N2]=0.033-x=0.033 M C(M ) − x − x + x [O2]=0.00810-x=0.00810 M [NO]=2x=1.60x10-17 M E(M ) 0.033- x 0.00810- x 2x −31 Kc = 4.810 (2x)2 = , (0.033- x)(0.00810- x) 4x2 or x = 8.0110-18 (0.033)(0.00810) 79 80 79 80
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EQUILIBRIA IN SOLUTIONS OF WEAK ACIDS THE SAME CALCULATION AND BASES
Assuming that the change of the reactant All weak acids behave the same way in concentrations is insignificant: aqueous solution: they partially ionize. 2 −31 2푥 4.8 ∙ 10 = In terms of the “general” weak acid HA, 0.033 ∙ 0.0081 this process can be written as: −18 푥 = 5.66 ∙ 10 → + − HA + H 2O H3O + A [NO]=2x=1.132x10-17 M Following the procedures [H O+ ][A− ] [H + ][A− ] K = 3 a [HA] [HA]
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WEAK ACIDS QUIZ
Ka is called the acid ionization constant. + What is the H3O concentration and pH of a This is often reported as the pKa 0.10 M solution of hypochlorous acid (HOCl)? Hint: K = 3.5 x 10-8 pKa = −logKa a + - HOCl + H2O = H3O + OCl list the Ka and pKa for a number of acids
A “large” pKa, means a “small” value of Ka and only a “small” fraction of the acid molecules ionizes
A “small” pKa ,means a “large” value of Ka and a “large” fraction of the acid molecules ionizes
83 http://www.nauticus.org/chemistry/chemacidbase.html 84 83 84
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SOLUTION ANSWER
+ -5 - x = [H3O ] = 5.9 x 10 = [OCl ] (also) + -5 pH = -log[H3O ] = -log(5.9 x 10 ) = 4.23 [HOCl] = 0.10 - 5.9 x 10-5 = 0.10 M
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WEAK BASES QUIZ Weak bases behave in a similar manner like acids in water What is the OH- concentration and pH of a For the “general” base B: 0.05 M solution of NH ? B + H O→HB + + OH − 3 2 -5 Hint: Kb = 1.8 x 10 the base ionization constant is [HB + ][OH − ] K = b [B]
Values of Kb and pKb for a number of weak bases are listed in special Tables. Where, like for acids:
pKb = −logKb 87 88 87 88
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CONJUGATE ACID-BASE PAIR CONJUGATE ACID-BASE PAIR
For the weak acid: [H O+ ][A− ] There is an interesting relationship → + − 3 HA + H 2O H3O + A Ka = between the acid and base ionization [HA] constants for a conjugate acid-base pair for the conjugate base: [HA][OH − ] A− + H O→ HA + OH − K = 2 b [A− ] the product is [H O+ ][A− ] [HA][OH − ] K K = 3 a b [HA] [A− ] + − 89 90 = [H3O ][OH ] = Kw 89 90
CONJUGATE ACID-BASE PAIR RELATIVE STRENGTHS OF CONJUGATE ACID-BASE PAIRS
Thus, for any conjugate acid-base pair: The stronger the acid is, the Ka Kb = Kw and weaker the o conjugate base. pKa + pKb = pKw =14.00 (at 25 C) The weaker the acid, the stronger the conjugate base. Most tables of ionization constants only give values for the molecular member of Very strong acids the conjugate acid-base pair. ionize 100% and their conjugate The ionization constant of the ion bases do not react member of the conjugate acid-base pair to any measurable extent. is then calculated as needed. 91 92 91 92
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COMMON CALCULATION COMMON CALCULATION Example: Morphine is very effective at relieving intense pain and it The primary goal is usually to determine the is a weak base. What is the Kb, pKb, and percentage equilibrium concentration for all species in ionization of morphine if a 0.010 M solution has a pH of the mass action expression 10.10? ANALYSIS: The reaction can be represented as: The percentage ionization of the acid or base [BH + ][OH − ] is defined as: B(aq) + H O→ BH + (aq) + OH − K = 2 b [B] moles ionized per liter x 2 percentage ionization = 100% I 0.010 0 0 = moles available per liter 0.010- x This, and the pH, are often used or C - x + x + x requested in equilibrium calculations E 0.010- x x x 93 At equilibrium, [OH-] = x = 10-pOH 94 93 94
SOLUTIONS
Use pOH = 14.00 – pH, substituting: [OH − ] =10−(14.00−10.10) =1.310−4 M , then x2 (1.310−4 )2 K = = b 0.010 − x (0.010 −1.310−4 ) −6 =1.610 so pKb = 5.80, and x % ionization = 100% 0.010 =1.3%
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SIMPLIFYING ASSUMPTIONS SIMPLIFYING ASSUMPTIONS?
Example: Calculate the pH of a 0.0010 M solution -4 of dimethylamine for which Kb=9.6x10 .
ANALYSIS: 400* Kb>0.0010 M, so use of the The quadratic equation is usually used to quadratic equation is indicated solve equilibrium problems – time consuming SOLUTION: Set the problem up
Simplifying assumptions : → + − −4 B(aq) + H 2O BH (aq) + OH (aq) Kb = 9.610 the initial concentration of a weak acid or base [BH + ][OH - ] I 0.0010 0 0 = in pure water is more that 400 times the [B] ionization constant x2 C − x + x + x = CO >400* K (0.0010 − x) E 0.0010− x x x
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Put in standard form x2 K = b (0.0010 − x) BUFFERS 2 (0.0010 − x)Kb = x Minimizes pH changes when a small 2 0 = x + Kb x − 0.0010Kb amount of strong acid or base is added to = x2 + 9.610−4 x − 9.610−7 certain solutions Solve for x and the equilibrium concentrations Buffer solution usually contains two solutes, Only positive solutions are allowed, so one providing a weak acid and the other a − 9.610−4 + (9.610−4 )2 − 4(1)(−9.610−7 ) weak base x = 2(1) If the weak acid is molecular, then the = 6.110−4 M and conjugate base can be supplied as a soluble salt of the acid [BH + ] = [OH − ] = x = 6.110−4 M [B] = (0.0010 − x)M = 3.910−4 M 99 100 99 100
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HOW IT'S WORKING? HOW IT'S WORKING?
Consider the general buffer made so that Consider the general buffer made so that both acid HA and salt A- are present in both acid HA and salt A- are present in solution solution 1. When base (OH-) is added, 2. When acid (H+) is added, the weak acid react with added base: the salt A- react with added acid
− − − + HA(aq) +OH (aq) → A (aq) + H2O A (aq) + H (aq) → HA(aq)
Net result: small changes in pH Net result: small changes in pH
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CALCULATIONS INVOLVING BUFFER SOLUTIONS Example: What is the pH of a buffer made by
It is worth remembering that for buffers: adding 0.10 mol NH3 and 0.11 mol NH4Cl to the initial concentration of both the weak 2.0 L of solution? The Kb for ammonia is acid and its conjugate base can be used 1.8x10-5 as though they were equilibrium values ANALYSIS: This is a buffer, initial concentrations can be used as equilibrium values: + − → + − [NH 4 ][OH ] molar concentrations or moles can be NH 3 (aq) + H 2O NH 4 (aq) + OH (aq) Kb = [NH 3 ] used in the Ka (or Kb) (the same units must be 0.10 mol used for both members of the pair) [NH ] = = 0.050M 3 2.0 L 0.11 mol [NH + ] = = 0.055M (from NH Cl) 4 2.0 L 4 103 104 103 104
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WHAT DETERMINES THE PH OF THE BUFFER?
SOLUTION: Solve for [OH-] and use this to For the general weak acid HA: + − calculate the pH → + − [H ][A ] HA(aq) H (aq) + A (aq) K = a [HA]
+ − − rearranging gives [NH 4 ][OH ] −5 (0.055)[OH ] Kb = =1.810 = + [HA] + mol HA [NH ] (0.050) [H ] = Ka or [H ] = Ka 3 [A− ] mol A− − −5 (0.050) −5 [OH ] =1.810 =1.6410 Thus both : 0.055 pOH = 4.79 and pH = 9.21 the value of Ka the ratio of the molarities (or the ratio of moles) affect the pH.
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WHAT DETERMINES THE PH OF THE BUFFER? − [A ]initial Example: A solution of 0.20 M in acetic acid (HAc) and 0.10 pH = pKa + log or [HA]initial M in sodium acetate (NaAc) were mixed. What is the pH of new solution? [salt] the Henderson-Hasselbalch equation pH = pK + log a [acid] it's worth remembering !!! the Henderson-Hasselbalch equation it's worth remembering !!!
the pH is mostly determined by the pKa of the acid;(In buffer, the concentration ratio is usually near 1) it's worth remembering !!!
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ICE TABLE
Compare x to the Ka value. In our sample problem, the concentrations of the weak acid and its conjugate base -2 -5 are 10 and the Ka is 10 . The difference is greater than 100 so both of the x quantities may be ignored.
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BUFFER CAPACITY
The goal of a buffer is to keep the pH of a solution within a narrow range Generally, the pH change in an experiment must be limited to about 0.1 pH unit Buffer capacity, ß, =the effectiveness of a buffer + -5 pH = -log[H3O ] = -log[3.6 x 10 ] = 4.44 A buffer’s capacity is determined by the magnitudes of the molarities of its components.
111 For a useful buffer : pH = pKa 1 112 111 112
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BUFFER CAPACITY Example: 0.02 mol of HCl was added to a buffer Δ푛 훽 = made from 0.10 mol HA (pKa=7.20) and 0.15 Δ 푝퐻 mol NaA in 2.0 L with no volume change. What is the pH change? ANALYSIS: This buffer problem is “best” solved in terms of moles. HCl is a strong acid. The H+ it contributes to solution increases the amount of HA present at the expense of A-. SOLUTION: The pH before addition of HCl was:
[A− ] 0.15 mol pH = pK + log = 7.20 + log = 7.38 a [HA] 0.10 mol http://adsorption.org/awm/utils/buf/BuCap.htm 113 114 113 114
AND IF THERE WAS NO BUFFER After the HCl ionizes and reacts: − [A ] final = (0.15− 0.02) mol = 0.13 mol
[HA] final = (0.10 + 0.02) mol = 0.12 mol If the HCl had been added to pure water, the pH change the pH of the new solution is: would have been much larger: 0.13 mol pH = 7.20 + log = 7.23, and the pH change is 0.12 mol 0.02 pH = − log − 7.00 = −5.00 pH = pH − pH = 7.23− 7.38 = −0.15 finial initial 2.0 Δ푛 0.02 훽 = = = 0.13 Δ 푝퐻 0.15 The pH change was greater than –0.1 unit. The buffer effectively resisted the pH change, however.
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HOW TO CALCULATE BUFFER CAPACITY
Using the following data Thus, To the first beaker 0.125 milimoles of HCl was introduced (n = M x V) 푛 퐻퐶푙 푖푛푡푟표푑푢푐푒푑 0.125 So the 훽 = = = 0.081 ∆ 푝퐻 4.64 −3.1 To two beakers containing the same volume of e.g. In similar way it is possible to calculate buffer buffer solution of pH 4.4 the following solutions were added: capacity in beaker 2 or when different buffer a) to the first 0.25 mL of 0.5 M HCl pH is used b) to the second 0.25 mL of 0.5 M NaOH
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WHEN BUFFER HAS THE HIGHEST ß?
Buffer Solution pH Δ n solution Before After Δ pH of base ß No adding adding introduce value NaOH NaOH d 1 3.3 4.04 0.74 0.025 0.034 2 3.96 4.3 0.34 0.025 0.074 3 4.54 4.85 0.31 0.025 0.081 4 5.03 5.38 0.35 0.025 0.071 5 6.03 6.66 0.63 0.025 0.040 The highest ß value is observed for buffer solution 3, of pH 4.54
In experiment acetic buffer was used. The pKA of acetic acid is 4.76. Thus, the statement that ß is the highest when buffer
pH = pKA±1.0 is the true.
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THANK YOU FOR YOUR KIND ATTENTION!!!
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