Presentation on theme: "14 - 1 Chemical Equilibrium Introduction to Chemical Equilibrium Equilibrium Constants and Expressions Calculations Involving Equilibrium Constants Using."— Presentation transcript:
14 - 1 Chemical Equilibrium Introduction to Chemical Equilibrium Equilibrium Constants and Expressions Calculations Involving Equilibrium Constants Using Le Châtelier’s Principle Some Important Equilibria
14 - 2 Chemical equilibrium For the general chemical reaction: aA + bB cC + dD equilibrium If A and B are brought together, a point is reached where there is no further change to the system - equilibrium. A chemical equilibrium is a dynamic equilibrium. Species are constantly being formed but with no net change in concentration
14 - 3 Chemical equilibrium Different types of arrows are used in chemical equations associated with equilibria. Single arrow Assumes that the reaction proceeds to completion as written. Two single-headed arrows Used to indicate a system in equilibrium. Two single-headed arrows of different sizes. May be used to indicate when one side of an equilibrium system is favored.
14 - 4 Chemical equilibrium Homogeneous equilibria Equilibria that involve only a single phase.Examples. All species in the gas phase H 2 (g) + I 2 (g) 2HI (g) All species are in solution. HC 2 H 2 O 2 (aq) H + (aq) + C 2 H 3 O 2 - (aq)
14 - 5 Chemical equilibrium Basic steps. For the general reaction: A + B C We can view the reaction as occurring in three steps. Initial mixing Kinetic region Equilibrium region
14 - 6 Chemical equilibrium Initial mixing. When A and B are first brought together, there is no C present. The reaction proceeds as A + B C This is just at the very start of the reaction. Things change as soon as some C is produced.
14 - 7 Chemical equilibrium Kinetic region. As soon as some C has been produced, the reverse reaction is possible. A + B C Overall, we still see an increase in the net concentration of C. As we approach equilibrium, the rate of the forward reaction becomes slower.
14 - 8 Chemical equilibrium Equilibrium region. A point is finally reached where the forward and reverse reactions occur at the same rate. A + B C There is no net change in the concentration of any of the species.
14 - 9 Chemical equilibrium Concentration Time C B A Equilibrium Region Kinetic Region
14 - 10 Equilibrium constants and expressions For the general chemical reaction: aA + bB eE + fF The equilibrium can be expressed as: K c = [E] e [F] f [A] a [B] b K c Equilibrium constant for a homogeneous equilibrium. [ ] n Molar concentrations raised to the power of their coefficients in the balanced chemical equation.
14 - 11 Equilibrium constants and expressions Hetrogeneous equilibria Equilibria that involve more than one phase. CaCO 3 (s) CaO (s) + CO 2 (g) do not Equilibrium expressions for these types of systems do not include the concentrations of the pure solids (or liquids). K c = [CO 2 ]
14 - 12 Equilibrium constants and expressions Hetrogeneous equilibria We don’t include the pure solids and liquids because their concentrations do not vary. Their values end up being included in the K value. CO 2 CaO & CaCO 3 As long as the temperature is constant and some solid is present, the amount of solid present has no effect on the equilibrium.
14 - 13 Writing an equilibrium expression Write a balanced equation for the equilibrium. Put the products in the numerator and the reactants in the denominator. Omit pure solids and liquids from the expression Omit solvents if your solutes are dilute (<0.1M). The exponent of each concentration should be the same as the coefficient for the species in the equation.
14 - 14 Writing an equilibrium expression Example. What would be the equilibrium expression for the following? (NH 4 ) 2 CO 3 (s) 2 NH 3 (g) + CO 2 (g) + H 2 O (g) K c = [NH 3 ] 2 [CO 2 ] [H 2 O] (NH 4 ) 2 CO 3 is a pure solid so is not included K c. We use [NH 3 ] 2 because the coefficient for NH 3 (g) in the equation is 2.
14 - 15 Equilibrium and rate of reaction Chemical reactions tend to go to equilibrium providing that reaction takes place at a significant rate. There is no relationship between the magnitude of the equilibrium constant and the rate of a reaction. Example. Example. 2H 2 (g) + O 2 (g) 2H 2 O (g) K c = 2.9 x 10 31 = However, the reaction will take years to reach equilibrium at room temperature. [H 2 O] [H 2 ] 2 [O 2 ]
14 - 16 Determining equilibrium constants Equilibrium constants can be found by experiment. If you know the initial concentrations of all of the reactants, you only need to measure the concentration of a single species at equilibrium to determine the K c value. Lets consider the following equilibrium: H 2 (g) + I 2 (g) 2HI (g)
14 - 17 Determining equilibrium constants H 2 (g) + I 2 (g) 2HI (g) Assume that we started with the following initial concentrations at 425.4 o C. H 2 (g) 0.005 00 M I 2 (g) 0.012 50 M HI (g) 0.000 00 M At equilibrium, we determine that the concentration of iodine is 0.007 72 M
14 - 18 Determining equilibrium constants The equilibrium expression for our system is: K c = Based on the chemical equation, we know the equilibrium concentrations of each species. I 2 (g) = the measured amount = 0.007 72 M That means that 0.004 78 mol I 2 reacts to produce HI in 1.00 L of solution. [HI] 2 [H 2 ] [I 2 ]
14 - 19 Determining equilibrium constants I 2 (g) =0.007 72 M HI (g) =0.004 78 M = 0.009 56 M H 2 (g) =0.005 00 M - 0.004 78 M = 0.000 22 M K c == = 54 2 mol HI 1 mol I 2 [HI] 2 [H 2 ] [I 2 ] (0.009 56) 2 (0.000 22)(0.007 72)
14 - 20 Partial pressure equilibrium constants At constant temperature, the pressure of a gas is proportional to its molarity. PV = nRT Remember, for an ideal gas: PV = nRT and molarity is: M = mole / liter or n/V so: P = R T M whereR is the gas law constant T is the temperature, K.
14 - 21 Partial pressure equilibrium constants For equilibria that involves gases, partial pressures can be used instead of concentrations. aA (g) + bB (g) eE (g) + fF (g) K p = K p is used when the partial pressures are expressed in units of atmospheres. p E e p F f p A a p B b
14 - 22 Partial pressure equilibrium constants In general, K p = K c, instead K p = K c (RT) n g n g is the number of moles of gaseous products minus the number of moles of gaseous reactants. n g = (e + f) - (a + b)
14 - 23 Partial pressure equilibrium constants For the following equilibrium, K c = 1.10 x 10 7 at 700. o C. What is the K p ? 2H 2 (g) + S 2 (g) 2H 2 S (g) K p = K c (RT) n g T = 700 + 273 = 973 K R= 0.08206 n g = ( 2 ) - ( 2 + 1) = -1 atm L mol K
14 - 24 Partial pressure equilibrium constants K p = K c (RT) n g = 1.10 x 10 7 (0.08206 ) (973 K) = 1.378 x10 5 atm L mol K [ ]
14 - 25 Equilibrium calculations We can predict the direction of a reaction by calculating the reaction quotient. Reaction quotient, Q For the reaction: aA + bB eE + fF Q has the same form as K c with one important difference. Q can be for any set of concentrations, not just at equilibrium. Q = [E] e [F] f [A] a [B] b
14 - 26 Reaction quotient Any set of concentrations can be given and a Q calculated. By comparing Q to the K c value, we can predict the direction for the reaction. Q < K c Q < K c Net forward reaction will occur. Q = K c Q = K c No change, at equilibrium. Q > K c Q > K c Net reverse reaction will occur.
14 - 27 Reaction quotient example For an earlier example H 2 (g) + I 2 (g) 2HI (g) we determined the K c to be 54 at 425.4 o C. If we had a mixture that contained the following, predict the direction of the reaction. [H 2 ] = 4.25 x 10 -3 M [I 2 ]= 3.97 x 10 -1 M [HI]= 9.83 x 10 -2 M
14 - 28 Reaction quotient example Q = = = 5.73 Since Q is < K c, the system is not in equilibrium and will proceed in the forward direction. [ HI ] 2 [ H 2 ] [ I 2 ] (9.83 x 10 -2 ) 2 (4.25 x 10 -3 )(3.97 x 10 -1 )
14 - 29 Calculating equilibrium concentrations If the stoichiometry and K c for a reaction is known, calculating the equilibrium concentrations of all species is possible. Commonly, the initial concentrations are known. One of the concentrations is expressed as the variable x. All others are then expressed in terms of x.
14 - 30 Equilibrium calculation example A sample of COCl 2 is allowed to decompose. The value of K c for the equilibrium COCl 2 (g) CO (g) + Cl 2 (g) is 2.2 x 10 -10 at 100 o C. If the initial concentration of COCl 2 is 0.095M, what will be the equilibrium concentrations for each of the species involved?
14 - 31 Equilibrium calculation example COCl 2 (g) CO (g) Cl 2 (g) Initial conc., M0.0950.0000.000 Change in conc.- X + X + X due to reaction Equilibrium Concentration, M(0.095 -X) X X K c == [ CO ] [ Cl 2 ] [ COCl2 ] X 2 (0.095 - X)
14 - 32 Equilibrium calculation example X 2 (0.095 - X) K c = 2.2 x 10 -10 = Rearrangement gives X 2 + 2.2 x 10 -10 X - 2.09 x 10 -11 = 0 This is a quadratic equation. Fortunately, there is a straightforward equation for their solution
14 - 33 Quadratic equations An equation of the form a X 2 + b X + c = 0 Can be solved by using the following x = Only the positive root is meaningful in equilibrium problems. -b + b 2 - 4ac 2a
14 - 34 Equilibrium calculation example -b + b 2 - 4ac 2a 2.2 x 10 -10 2.09 x 10 -11 X 2 + 2.2 x 10 -10 X - 2.09 x 10 -11 = 0 b c a b c X = - 2.2 x 10 -10 + [(2.2 x 10 -10 ) 2 - (4)(1)(- 2.09 x 10 -11 )] 1/2 2 X = 9.1 x 10 -6 M
14 - 35 Equilibrium calculation example Now that we know X, we can solve for the concentration of all of the species. COCl 2 = 0.095 - X = 0.095 M CO= X = 9.1 x 10 -6 M Cl 2 = X = 9.1 x 10 -6 M In this case, the change in the concentration of is COCl 2 negligible.
14 - 36 Summary of method of calculating equilibrium concentrations Write an equation for the equilibrium. Write an equilibrium constant expression. Express all unknown concentrations in terms of a single variable such as x. Substitute the equilibrium concentrations in terms of the single variable in the equilibrium constant expression. Solve for x. Use the value of x to calculate equilibrium concentrations.
14 - 37 Predicting shifts in equilibria Le Châtelier’s Principle When subjected to a stress, a system in equilibrium will act to relieve the stress. The position of a chemical equilibrium will shift in a direction to relieve a stress aA + bB cC + dDExample Adding A or B or removing C or D will shift equilibrium to the right.
14 - 38 Predicting shifts in equilibria Equilibrium concentrations are based on: –The specific equilibrium –The starting concentrations –Other factors such as: Temperature Pressure Reaction specific conditions Altering conditions will stress a system, resulting in an equilibrium shift.
14 - 39 Changes in concentration Changes in concentration do not change the value of the equilibrium constant at constant temperature. When a material is added to a system in equilibrium, the equilibrium will shift away from that side of the equation. When a material is removed from a system in equilibrium, the equilibrium will shift towards that sid of the equation.
14 - 40 Changes in concentration HI H2H2 I2I2 Log concentration Time Example. Example. I 2 is added to an equilibrium mixture. The system will adjust all of the concentrations to reestablish a new equilibrium with the same K c.
14 - 41 Changes in concentration HI H2H2 I2I2 Log concentration Time Example. Example. Some H 2 is removed. Again, the system adjusts all of the concentrations to reestablish a new equilibrium with the same K c.
14 - 42 Changes in pressure Changing the pressure does not change the value of the equilibrium constant at constant temperature. Solids and liquids are not effected by pressure changes. Changing pressure by introducing an inert gas will not shift an equilibrium. Pressure changes only effect gases that are a portion of an equilibrium.
14 - 43 Changes in pressure In general, increasing the pressure by decreasing volume shifts equilibria towards the side that has the smaller number of moles of gas. H 2 (g) + I 2 (g) 2HI (g) N 2 O 2 (g) 2NO 2 (g) Unaffected by pressure Increased pressure, shift to left
14 - 44 Changes in temperature Changes in temperature usually change the value of the equilibrium constant. K c can either increase or decrease with increasing temperature. The direction and degree of change is dependent on the specific reaction. T, o C K p 6492.7 x 10 0 7606.3 x 10 1 8718.2 x 10 2 9826.8 x 10 3 T, o C K p 6492.7 x 10 0 7606.3 x 10 1 8718.2 x 10 2 9826.8 x 10 3 T, o C K p 2279.0 x 10 -2 4278.1 x 10 -5 6271.3 x 10 -6 8279.7 x 10 -8 T, o C K p 2279.0 x 10 -2 4278.1 x 10 -5 6271.3 x 10 -6 8279.7 x 10 -8 CH 4 (g) + H 2 O (g) 2H 2 (g) + CO (g) N 2 (g) + 3H 2 (g) 2NH 3 (g)