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Unit 5 - Chpt 13 & 17 - Equilibrium and Thermochemistry Part II Equilibrium basics Equilibrium Expressions with pressures Heterogeneous Equilibrium & Applications.

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Presentation on theme: "Unit 5 - Chpt 13 & 17 - Equilibrium and Thermochemistry Part II Equilibrium basics Equilibrium Expressions with pressures Heterogeneous Equilibrium & Applications."— Presentation transcript:

1 Unit 5 - Chpt 13 & 17 - Equilibrium and Thermochemistry Part II Equilibrium basics Equilibrium Expressions with pressures Heterogeneous Equilibrium & Applications Le Chateliers Principle Thermo - Entropy and Free Energy HW set1: Chpt 13 - pg , #10-14, 21ac, 22ac, 23ac, 24ac, 28, Due Thurs. Jan 17 HW set2: Chpt 13 #40, 43, 48, 52, 57, 63, 64 Due Wed Jan 23

2 Chemical Equilibrium The state where the concentrations of all reactants and products remain constant with time. On the molecular level, there is enormous activity. Equilibrium is not static, but is a highly dynamic situation. Macroscopically static Microscopically dynamic

3 Concentration with time Changes in Concentration N 2 (g) + 3H 2 (g) 2NH 3 (g) Concentrations reach levels where the rate of the forward reaction equals the rate of the reverse reaction.

4 Rates with time The Changes with Time in the Rates of Forward and Reverse Reactions

5 Concept check equilibrium Consider an equilibrium mixture in a closed vessel reacting according to the equation: H 2 O(g) + CO(g) H 2 (g) + CO 2 (g) You add more H 2 O(g) to the flask. How does the concentration of each chemical compare to its original concentration after equilibrium is reestablished? Justify your answer.

6 jA + kB lC + mD A, B, C, and D = chemical species. Square brackets = concentrations of species at equilibrium. j, k, l, and m = coefficients in the balanced equation. K = equilibrium constant (given without units). j l k m [B][A] [D] [C] K = Consider the following reaction at equilibrium:

7 Equilibrium Expression Math Equilibrium expression for a reaction is the reciprocal of that for the reaction written in reverse. When balanced equation for a reaction is multiplied by a factor of n, the equilibrium expression for the new reaction is the original expression raised to the nth power; thus K new = (K original ) n. K values are usually written without units.

8 K constant - more stuff K always has the same value at a given temperature regardless of the amounts of reactants or products that are present initially. For a reaction, at a given temperature, there are many equilibrium positions but only one value for K. Equilibrium position is a set of equilibrium concentrations. K involves concentrations. K p involves pressures.

9 Equilibrium Expressions using Pressures N 2 (g) + 3H 2 (g) 2NH 3 (g)

10 Example problem N 2 (g) + 3H 2 (g) 2NH 3 (g) Equilibrium pressures at a certain temperature:

11 Example continued

12 Relationship between K p & K K p = K(RT) Δn Δn = sum of the coefficients of the gaseous products minus the sum of the coefficients of the gaseous reactants. R = L·atm/mol·K T = temperature (in kelvin) PV = nRT ; C in conc =n/V (molar volume) ; C = P/RT K expression plug in P/RT for each, derived pg 602-3

13 Calc K from K p (last problem) N 2 (g) + 3H 2 (g) 2NH 3 (g) Using the value of K p (3.9 × 10 4 ) from the previous example, calculate the value of K at 35°C.

14 Homogeneous Equilibria Homogeneous equilibria – involve the same phase: N 2 (g) + 3H 2 (g) 2NH 3 (g) HCN(aq) H + (aq) + CN - (aq)

15 Heterogeneous equilibria Heterogeneous equilibria – involve more than one phase: 2KClO 3 (s) 2KCl(s) + 3O 2 (g) 2H 2 O(l) 2H 2 (g) + O 2 (g)

16 Heterogeneous equilibrium (cont) The position of a heterogeneous equilibrium does not depend on the amounts of pure solids or liquids present. The concentrations of pure liquids and solids are constant. 2KClO 3 (s) 2KCl(s) + 3O 2 (g)

17 Applications of Equilibrium Extent of a reaction A value of K much larger than 1 means that at equilibrium the reaction system will consist of mostly products – the equilibrium lies to the right. Reaction goes essentially to completion. A very small value of K means that the system at equilibrium will consist of mostly reactants – the equilibrium position is far to the left. Reaction does not occur to any significant extent.

18 Reaction Quotient - Q Apply the law of mass action using initial concentrations instead of equilibrium concentrations in K expression. Q = K; The system is at equilibrium. No shift will occur. Q > K; The system shifts to the left. Consuming products and forming reactants, until equilibrium is achieved. Q < K; The system shifts to the right. Consuming reactants and forming products, to attain equilibrium.

19 K vs Q graphic

20 Exercise - Applications ICE table ( Initial, Change, Equilibrium) Consider the reaction represented by the equation: Fe 3+ (aq) + SCN - (aq) FeSCN 2+ (aq) Trial #1: 6.00 M Fe 3+ (aq) and 10.0 M SCN - (aq) are mixed at a certain temperature and at equilibrium the concentration of FeSCN 2+ (aq) is 4.00 M. What is the value for the equilibrium constant for this reaction?

21 Trial 1 (cont) - ICE table Fe 3+ (aq) + SCN – (aq) FeSCN 2+ (aq) Initial Change – 4.00 – Equilibrium K = 0.333

22 Example 2 - trial 2 Consider the reaction represented by the equation: Fe 3+ (aq) + SCN - (aq) FeSCN 2+ (aq) Trial #2: Initial: 10.0 M Fe 3+ (aq) and 8.00 M SCN (aq) (same temperature as Trial #1) use -x and +x for changes, then solve quadratic equation. Equilibrium: ? M FeSCN 2+ (aq) 5.00 M FeSCN 2+

23 Example 3 - trial 3 Consider the reaction represented by the equation: Fe 3+ (aq) + SCN - (aq) FeSCN 2+ (aq) Trial #3: Initial: 6.00 M Fe 3+ (aq) and 6.00 M SCN (aq) ; use -x and +x for changes, then solve quadratic equation. Equilibrium: ? M FeSCN 2+ (aq) 3.00 M FeSCN 2+

24 Solving Equilibrium Problems 1) Write the balanced equation for the reaction. 2) Write the equilibrium expression using the law of mass action. 3) List the initial concentrations. 4) Calculate Q, and determine the direction of the shift to equilibrium.

25 Solving Equilibrium Problems 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 they give the correct value of K.

26 More practice (answers last slide) Consider the reaction represented by the equation: Fe 3+ (aq) + SCN - (aq) FeSCN 2+ (aq) Fe 3+ SCN - FeSCN 2+ Trial #19.00 M5.00 M1.00 M Trial #23.00 M2.00 M5.00 M Trial #32.00 M9.00 M6.00 M Find the equilibrium concentrations for all species.

27 Example - ICE ignore x A 1.00 mol sample of N 2 O 4 (g) is placed in a 10.0 L vessel and allowed to reach equilibrium according to the equation: N 2 O 4 (g) 2NO 2 (g) K = 4.00 x Calculate the equilibrium concentrations of: N 2 O 4 (g) and NO 2 (g). Concentration of N 2 O 4 = M Concentration of NO 2 = 6.32 x M

28 Le Chateliers 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.

29 Effects of changes on a system 1. Concentration: The system will shift away from the added component. If a component is removed, the opposite effect occurs. 2.Temperature: K will change depending upon the temperature (endothermic – energy is a reactant; exothermic – energy is a product).

30 Effects of changes on a system 3. Pressure: a) The system will shift away from the added gaseous component. If a component is removed, the opposite effect occurs. b) Addition of inert gas does not affect the equilibrium position. c) Decreasing the volume shifts the equilibrium toward the side with fewer moles of gas.

31 Change in volume effect on equilibrium ??

32 More practice answers Trial #1: [Fe 3+ ] = 6.00 M; [SCN - ] = 2.00 M; [FeSCN 2+ ] = 4.00 M Trial #2: [Fe 3+ ] = 4.00 M; [SCN - ] = 3.00 M; [FeSCN 2+ ] = 4.00 M Trial #3: [Fe 3+ ] = 2.00 M; [SCN - ] = 9.00 M; [FeSCN 2+ ] = 6.00 M


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