Presentation on theme: "Chemical Equilibrium Chapter 13. Chemical Equilibrium The state where the concentrations of all reactants and products remain constant with time. On the."— Presentation transcript:
Chemical Equilibrium The state where the concentrations of all reactants and products remain constant with time. On the molecular level, there is frantic activity. Equilibrium is not static, but is a highly dynamic situation.
Equilibrium & Kinetics At equilibrium, R f = R r but k f is not necessarily equal to k r. k f [Reactant] = k r [Product] If equilibrium lies far to the right, [Products] will be large and k r will be small while[Reactants] will be small and k f will be large.
Reactions That Appear to Run to Completion 1.Formation of a precipitate. 2. Formation of a gas. 3. Formation of a molecular substance such as water. These reactions appear to run to completion, but actually the equilibrium lies very far to the right. All reactions in closed vessels reach equilibrium.
Molecular representation of the reaction 2NO 2(g) ----> N 2 O 4(g). c) & d) represent equilibrium.
Concentration profile for the Haber Process which begins with only H 2(g) & N 2(g).
Shifting Gaseous Equilibrium If the size of a container is changed, the concentration of the gases change. A smaller container shifts the equilibrium to the right -- N 2(g) + 3H 2(g) ---> 2NH 3(g). Four gaseous molecules produce two gaseous molecules. A larger container shifts to the left -- two gaseous molecules produce four gaseous molecules.
The Law of Mass Action For jA + kB lC + mD The law of mass action (Cato Guldberg & Peter Waage) is represented by the equilibrium expression:
Notes on Equilibrium Expressions (EE) -The Equilibrium Expression for a reaction is the reciprocal of that for the reaction written in reverse. -When the equation for a reaction is multiplied by n, EE new = (EE original ) n -The units for K depend on the reaction being considered.
Equilibrium Expressions 1/2N 2(g) + 3/2H 2(g) ---> NH 3(g) K ´ ´ = K 1/2
Equilibrium Position For a given reaction at a given temperature, there is only one equilibrium constant (K), but there are an infinite number of equilibrium positions. Where the equilibrium position lies is determined by the initial concentrations of the reactants and products. The initial concentrations do not affect the equilibrium constant.
Three equilibrium positions but only one equilibrium constant (K).
KpKp K in terms of equilibrium partial pressures: Pressure and concentration are directly proportional when the temperature is held constant.
K versus K p For jA + kB lC + mD K p = K(RT) n n = sum of coefficients of gaseous products minus sum of coefficients of gaseous reactants.
K versus K p When n = 0 and K p = K(RT) o, then K p = K K and K p are equal numerically but do not have the same units. Under what situations would n = 0 ?
Heterogeneous Equilibria... are equilibria that involve more than one phase. CaCO 3 (s) CaO(s) + CO 2 (g) K = [CO 2 ] The position of a heterogeneous equilibrium does not depend on the amounts of pure solids or liquids present.
The position of the equilibrium CaCO 3(s) ---> CaO (s) + CO 2(g) does not depend upon the amounts of solid CaCO 3 or CaO.
Magnitude of K A K value much larger than 1 means that the equilibrium system contains mostly products -- equilibrium lies far to the right. A very small K value means the system contains mostly reactants -- equilibrium lies far to the left. The size of K and the time required to reach equilibrium are not directly related!
A physical analogy illustrating the difference between thermodynamic and chemical stability. The magnitude of K depends on E, but reaction rate depends on E a.
Reaction Quotient... helps to determine the direction of the move toward equilibrium. The law of mass action is applied with initial concentrations.
Reaction Quotient (continued) H 2 (g) + F 2 (g) 2HF(g)
Q versus K Q = K System is at equilibrium. Q > K System will shift to the left. Q < K System will shift to the right. See Sample Exercise 13.7 on page 628.
Solving Equilibrium Problems 1.Balance the equation. 2.Write the equilibrium expression. 3.List the initial concentrations. 4.Calculate Q and determine the shift to equilibrium.
Solving Equilibrium Problems (continued ) 5.Define equilibrium concentrations. 6.Substitute equilibrium concentrations into equilibrium expression and solve. 7.Check calculated concentrations by calculating K.
ICE See Sample Exercise 13.11 on pages 633-634. H 2(g) + F 2(g) 2HF (g) [H 2 ] o = [F 2 ] o = [HF] o = 3.000 mol/1.500 L = 2.000M 1.000 Q < K equilibrium shifts right.
ICE H 2(g) + F 2(g) 2HF (g) [H 2 ] [F 2 ] [HF] Initial (mol/L) 2.000 2.000 2.000 Change (mol/L) - x - x + 2x Equil. (mol/L) 2.000 - x 2.000 - x 2.000 + 2x
ICE Continued x = 1.528 [H 2 ] = [F 2 ] = 2.000 M - 1.528 M = 0.472 M [HF] = 2.000 M + 2(1.528 M) = 5.056 M Substitute into K expression to check validity!
ICE & the Quadratic Formula See Sample Exercise 13.12 on pages 637-638. Use the quadratic formula to solve for K. Obtain the quadratic formula program from your instructor.
Systems With Small K’s See example on pages 639-640. When K is small, the change (x) is small compared to the initial concentration and a simplification can be made for the calculation. This approximation must be checked to see if it is valid. When K is two powers of ten or more smaller than the initial concentration, the approximation should be OK.
Le Châtelier’s Principle... if a change is imposed on a system at equilibrium, the position of the equilibrium will shift in a direction that tends to reduce that change.
Le Chatelier’s Principle If a reactant or product is added to a system at equilibrium, the system will shift away from the added component. If a reactant or product is removed, the system will shift toward the removed component. See Sample Exercise 13.13 on pages 642-643.
Effects of Changes on the System 1.Concentration: The system will shift away from the added component. 2.Temperature: K will change depending upon the temperature (treat heat as a reactant or product).
Effects of Changes on the System (continued) 3.Pressure: a. Addition of inert gas does not affect the equilibrium position. b. Decreasing the volume shifts the equilibrium toward the side with fewer gaseous moles.
The system of N 2, H 2, and NH 3 are initially at equilibrium. When the volume is decreased, the system shifts to the right -- toward fewer molecules.