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Chemical Equilibrium Chapter 14 14.1-14.5

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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 and reverse reactions are equal and 2.) the concentrations of the reactants and products remain constant

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Equilibrium There are two types of equilibrium: Physical and Chemical. –Physical Equilibrium H 2 0 (l) ↔ H 2 0 (g) –Chemical Equilibrium N 2 O 4 (g) ↔ 2NO 2

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Physical Equilibrium

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Chemical Equilibrium

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N 2 O 4 (g) ↔ 2NO 2 (g)

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Law of Mass Action Law of Mass Action- For a reversible reaction at equilibrium and constant temperature, a certain ratio of reactant and product concentrations has a constant value (K). The Equilibrium Constant (K)- A number equal to the ratio of the equilibrium concentrations of products to the equilibrium concentrations of reactants each raised to the power of its stoichiometric coefficient.

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Law of Mass Action For the general reaction: K = [C] c [D] d [A] a [B] b aA (g) + bB (g) cC (g) + dD (g)

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Equilibrium Constant N 2 O 4 (g) ↔ 2NO 2 (g)

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Chemical Equilibrium Chemical equilibrium is defined by K. The magnitude of K will tell us if the equilibrium reaction favors the reactants or the products. If K » 1……..favors products If K « 1……..favors reactants

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Equilibrium Constant Expressions Equilibrium constants can be expressed using K c or K p. K c uses the concentration of reactants and products to calculate the eq. constant. K p uses the pressure of the gaseous reactants and products to calculate the eq. constant.

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Equilibrium Constant Expressions Equilibrium Constant Equations K c = [NO 2 ] 2 [N 2 O 4 ] K p = NO 2 P2P2 N2O4N2O4 P aA (g) + bB (g) cC (g) + dD (g)

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Homogeneous Equilibrium Homogeneous Equilibrium- applies to reactions in which all reacting species are in the same phase. N 2 O 4 (g) ↔ 2NO 2 (g) K p = NO 2 P2P2 N2O4N2O4 P In most cases K c K p K c = [NO 2 ] 2 [N 2 O 4 ]

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Equilibrium Constant Expressions Relationship between Kc and Kp K p = K c (RT) n n = moles of gaseous products – moles of gaseous reactants = (c + d) – (a + b)

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Equilibrium Constant Calculations The equilibrium concentrations for the reaction between carbon monoxide and molecular chlorine to form COCl 2 (g) at 740C are [CO] = 0.012 M, [Cl2] = 0.054 M, and [COCl 2 ] = 0.14 M. Calculate the equilibrium constants K c and K p. CO (g) + Cl 2 (g) COCl 2 (g) Kc =Kc = [COCl 2 ] [CO][Cl 2 ] = 0.14 0.012 x 0.054 = 220 K p = K c (RT) n n = 1 – 2 = -1 R = 0.0821T = 273 + 74 = 347 K K p = 220 x (0.0821 x 347) -1 = 7.7

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Equilibrium Constant Calculations The equilibrium constant K p for the reaction is 158 at 1000K. What is the equilibrium pressure of O 2 if the P NO = 0.400 atm and P NO = 0.270 atm? K p = 2 P NO P O 2 P NO 2 2 POPO 2 = K p P NO 2 2 2 POPO 2 = 158 x (0.400) 2 /(0.270) 2 = 347 atm

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Heterogeneous Equilibrium Heterogeneous Equilibrium- results from a reversible reaction involving reactants and products that are in different phases. Can include liquids, gases and solids as either reactants or products. Equilibrium expression is the same as that for a homogeneous equilibrium. Omit pure liquids and solids from the equilibrium constant expressions.

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Heterogeneous Equilibrium Constant CaCO 3 (s) CaO (s) + CO 2 (g) [CaCO 3 ] = constant [CaO] = constant K p = P CO 2 The concentration of solids and pure liquids are not included in the expression for the equilibrium constant.

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Heterogeneous Equilibrium Constant

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Equilibrium Constant Calculations Consider the following equilibrium at 295 K: The partial pressure of each gas is 0.265 atm. Calculate K p and K c for the reaction. NH 4 HS (s) NH 3 (g) + H 2 S (g) K p = P NH 3 H2SH2S P= 0.265 x 0.265 = 0.0702 K p = K c (RT) n K c = K p (RT) - n n = 2 – 0 = 2 T = 295 K K c = 0.0702 x (0.0821 x 295) -2 = 1.20 x 10 -4

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Multiple Equilibria Multiple Equilibria- Product molecules of one equilibrium constant are involved in a second equilibrium process. A + B C + D C + D E + F A + B E + F KcKc ‘ KcKc ‘‘ KcKc K c =KcKc ‘‘ KcKc ‘ x ‘ [C][D] [A][B] K c = ‘ ‘ [E][F] [C][D] [E][F] [A][B] K c =

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Writing Equilibrium Constant Expressions The concentrations of the reacting species in the condensed phase are expressed in M. In the gaseous phase, the concentrations can be expressed in M or in atm. The concentrations of pure solids, pure liquids and solvents do not appear in the equilibrium constant expressions. The equilibrium constant is a dimensionless quantity. In quoting a value for the equilibrium constant, you must specify the balanced equation and the temperature. If a reaction can be expressed as a sum of two or more reactions, the equilibrium constant for the overall reaction is given by the product of the equilibrium constants of the individual reactions. 14.2

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What does the Equilibrium Constant tell us? We can: –Predict the direction in which a reaction mixture will proceed to reach equilibrium –Calculate the concentration of reactants and products once equilibrium has been reached

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Predicting the Direction of a Reaction The K c for hydrogen iodide in the following equation is 53.4 at 430ºC. Suppose we add 0.243 mol H 2, 0.146 mol I 2 and 1.98 mol HI to a 1.00L container at 430ºC. Will there be a net reaction to form more H 2 and I 2 or HI? H 2 (g) + I 2 (g) → 2HI (g) [HI] 0 2 [H 2 ] 0 [I 2 ] 0 K c = [1.98] 2 [0.243] [0.146] K c = K c = 111

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Reaction Quotient The reaction quotient (Q c ) is calculated by substituting the initial concentrations of the reactants and products into the equilibrium constant (K c ) expression. IF Q c > K c system proceeds from right to left to reach equilibrium Q c = K c the system is at equilibrium Q c < K c system proceeds from left to right to reach equilibrium

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Reaction Quotient

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Calculating Equilibrium Concentrations If we know the equilibrium constant for a reaction and the initial concentrations, we can calculate the reactant concentrations at equilibrium. ICE method ReactantsProducts Initial (M): Change (M): Equilibrium (M):

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Calculating Equilibrium Concentrations At 1280ºC the equilibrium constant (K c ) for the reaction is 1.1 x 10 -3. If the initial concentrations are [Br 2 ] = 0.063 M and [Br] = 0.012 M, calculate the concentrations of these species at equilibrium. Br 2 (g) 2Br (g) Let x be the change in concentration of Br 2 Br 2 (g) 2Br (g) Initial (M) Change (M) Equilibrium (M) 0.063 -x-x 0.063 - x 0.012 +2x 0.012 + 2x

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Calculating Equilibrium Concentrations [Br] 2 [Br 2 ] K c = (0.012 + 2x) 2 0.063 - x = 1.1 x 10 -3 4x 2 + 0.048x + 0.000144 = 0.0000693 – 0.0011x 4x 2 + 0.0491x + 0.0000747 = 0 ax 2 + bx + c = 0 -b ± b 2 – 4ac 2a2a x = x = -0.00178 x = -0.0105

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Calculating Equilibrium Concentrations Br 2 (g) 2Br (g) Initial (M) Change (M) Equilibrium (M) 0.0630.012 -x-x+2x 0.063 - x0.012 + 2x At equilibrium, [Br] = 0.012 + 2x = -0.009 M At equilibrium, [Br 2 ] = 0.063 – x = 0.0648 M or 0.00844 M

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Calculating Equilibrium Concentrations Express the equilibrium concentrations of all species in terms of the initial concentrations and a single unknown x, which represents the change in concentration. Write the equilibrium constant expression in terms of the equilibrium concentrations. Knowing the value of the equilibrium constant, solve for x. Having solved for x, calculate the equilibrium concentrations of all species.

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Factors that Affect Chemical Equilibrium Chemical Equilibrium represents a balance between forward and reverse reactions. Changes in the following will alter the direction of a reaction: –Concentration –Pressure –Volume –Temperature

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Le Châtlier’s Principle Le Châtlier’s Principle- if an external stress is applied to a system at equilibrium, the system adjusts in such a way that the stress is partially offset as the system reaches a new equilibrium position. Stress???

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Changes in Concentration Increase in concentration of reactants causes the equilibrium to shift to the ________. Increase in concentration of products causes the equilibrium to shift to the ________.

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Changes in Concentration ChangeShift in Equilibrium Increase in [Products]left Decrease in [Products]right Increase in [Reactants]right Decrease in [Reactants]left

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Changes in Concentration FeSCN 2+ (aq) ↔ Fe 3+ (aq) + SCN - (aq) a.) Solution at equilibrium b.) Increase in SCN - (aq) c.) Increase in Fe 3+ (aq) d.) Increase in FeSCN 2+ (aq)

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Changes in Volume and Pressure Changes in pressure primarily only concern gases. Concentration of gases are greatly affected by pressure changes and volume changes according to the ideal gas law. PV = nRT P = (n/V)RT

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Changes in Pressure and Volume ChangeShift in Equilibrium Increase in PressureSide with fewest moles Decrease in PressureSide with most moles Increase in VolumeSide with most moles Decrease in VolumeSide with fewest moles

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Changes in Pressure and Volume

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Changes in Temperature Equilibrium position vs. Equilibrium constant A temperature increase favors an endothermic reaction and a temperature decrease favors and exothermic reaction. ChangeEndo. RxExo. Rx Increase TK decreasesK increases Decrease TK increasesK decreases

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Changes in Temperature Consider: N 2 O 4 (g) ↔ 2NO 2 (g) The forward reaction absorbs heat; endothermic heat + N 2 O 4 (g) ↔ 2NO 2 (g) So the reverse reaction releases heat; exothermic 2NO 2 (g) ↔ N 2 O 4 (g) + heat Changes in temperature??

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Effect of a catalyst How would the presence of a catalyst affect the equilibrium position of a reaction?

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Chemical Equilibrium Chapter 14

Chemical Equilibrium Chapter 14

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