EQUILIBRIUM. Opposing reactions proceed at equal rates Concs. of reactants & products do not change over time

Transcription

1 EQUILIBRIUM Opposing reactions proceed at equal rates Concs. of reactants & products do not change over time Examples: vapor pressure above liquid saturated solution Now: equilibrium of chemical reactions Reactions we discussed so far could be described without equilibrium. i.e. CH O 2 CO H 2 O But this is not true for many reactions.

2 VAPOR PRESSURE an equilibrium process time rate of evaporation rate rate of condensation dynamic equilibrium (rates equal) time DYNAMIC EQUILIBRIUM rate forward = rate backward no net change although change is occuring

3 HABER PROCESS an equilibrium N 2 (g) + 3 H 2 (g) 2 NH 3 (g) Initial state: reactants only Initial state: products only Final state: ratio of products to reactants is the same for both sets of conditions

4 EQUILIBRIUM Equilibrium reached when rate of forward reaction equals rate of reverse reaction Example: N 2 O 4 (g) 2 NO 2 (g) forward N 2 O 4 NO 2 + NO 2 Rate f = k f [N 2 O 4 ] reverse NO 2 + NO 2 N 2 O 4 Rate r = k r [NO 2 ] 2 Equilibrium reached when Rate f = Rate r or k f [N 2 O 4 ] = k r [NO 2 ] 2 Reorganizing: 2 [ NO ] 2 [ NO] 2 4 k f = = k r equil. constant

5 EQUILIBRIUM CONSTANT aa + bb K c = [C]c [D] d [A] a [B] b K p = P C c P D d P Aa P B b cc + dd law of mass action K p = K c (RT) Δn Δn = change in number of moles in going from reactants to products K has no units [ ] means mol/l P means atm

6 EQUILIBRIUM CO + 3 H 2 CH 4 + H 2 O 1200 K start: 1 mol CO 3 mol H 2 in 10.0 L volume forward & reverse reactions both proceed 3 moles 2 1 CO H 2 CH 4 or H 2 O time At equilibrium: mol CO mol H mol H 2 O & CH 4 K c = [CH 4 ] [H 2 O] [CO] [H 2 ] 3

7 Calculate K C and K P CO + 3 H 2 CH 4 + H 2 O T= 1200K K c = [CH 4 ] [H 2 O] (0.0387)(0.0387) = = 3.93 [CO] [H 2 ] 3 (0.0613)(0.1839) 3 P CH4 P H2O K p = P CO P 3 H2 K p = [CH 4]RT [H 2 O]RT [CO]RT ([H 2 ]RT) 3 PV = nrt P = nrt V = n V RT K p = [CH 4] [H 2 O] (RT) 2 [CO] [H 2 ] 3 (RT) 4 = K c (RT) 2 K p = 3.93 = 4.05 x 10 4 {(0.0821)(1200)} 2 K p = K c (RT) Δn Δn = (2) (4) = 2 products reactants

8 DIRECTION OF EQUILIBRIUM 2 NO(g) N 2 (g) + O 2 (g) K c = [N 2 ][O 2 ] [NO] 2 N 2 (g) + O 2 (g) K c = 2 NO(g) [NO] 2 [N 2 ][O 2 ] K c expressions are reciprocals Same equilibrium -- can approach from either direction

9 EQUILIBIA FOR RELATED REACTIONS OLD EXAM QUESTION At a certain T, K c for the reaction H 2 + I 2 2 HI is 16. At the same T, what is K c for the reaction 1 1 HI 2 H I 2 A. 1/16 B. 4 C. 1/4 D. 16 E. None of these

10 MORE EQUILIBRIUM PROBLEMS K C for the reaction 2 H 2 O 2 (g) 2 H 2 O(g) + O 2 (g) is given by HO 2 O2 A. D. HO HO 2 O2 B. 2 E. HO C. [ ] [ ] [ 2 2] 2 [ ] [ ] [ 2 2] [ HO 2 ] [ O2] [ HO] [ HO 2 2] 2 [ HO 2 ] [ O2] [ HO 2 2] [ HO] [ O] 2 2

11 MORE EQUILIBRIUM PROBLEMS For which of the following reactions does K C = K P? A. I and II I. CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O(g) II. CO(g) + H 2 O(g) CO 2 (g) + H 2 (g) III. C 2 H 4 (g) + 3 O 2 (g) 2 CO 2 (g) + 2 H 2 O(g) B. I and III C. II and III D. I, II, and III E. none of the above

12 GENERAL APPROACH TO CALCULATE AN EQUILIBRIUM CONSTANT 1. Write the balanced equation 2. Write the general form for K 3. Set up a data table - initial conditions - changes in concentrations - equilibrium concentrations (may need algebraic unknown) 4. Substitute into K expression and solve

13 EQUM CONSTANT FROM KNOWN PRESS. A mixture of H 2 and N 2 is added to a reaction vessel at 472 o C and allowed to reach equilibrium. The equilibrium mixture of gasses was analyzed and found to contain 7.38 atm H 2, 2.46 atm N 2, and atm NH 3. From these data, calculate the equilibrium constants K P and K C for the reaction N 2 (g) + 3 H 2 (g) 2 NH 3 (g)

14 CALCULATION OF EQUM CONSTANT 2 HI(g) H 2 (g) + I 2 (g) 2.0 L flask with 0.20 mol HI At equm: [HI] = M Problem Statement [HI] [H 2 ] [I 2 ] Initial Change 2x +x +x Equm x x x [HI] = x = M so x = [H 2 ] = [I 2 ] = M K c = [H 2 ] [I 2 ] [HI] 2 = (0.011)(0.011) (0.078) 2 = 0.020

15 CALCULATION OF EQUM CONSTANT N 2 O 4 (g) 2 NO 2 (g) 1.0 L flask with mol N 2 O 4 At equm: [N 2 O 4 ] = M Problem Statement [N 2 O 4 ] [NO 2 ] Initial Change x +2x Equm x +2x [N 2 O 4 ] = x = M so x = K c [ NO2 ] [ NO] 2 4 ( x) ( x) ( ) ( 0.120) = = 2 = = 0.213

16 EQUILIBRIUM PROBLEM # mol SO 2 (g) and 1.00 mol O 2 (g) are added to a 1.00 L vessel and react until equilibrium is reached. At equilibrium, the vessel has mol SO 3 (g). KC = 238 for the reaction at 25 C. What are the equilibrium concentrations of SO 2 (g) and O 2 (g)? What is K P for the reaction?

17 EQUILIBRIUM PROBLEM #2 Enough glacial acetic acid (pure acetic acid) is added to pure water to make a M solution. Some of the acid dissociates according to the reaction HC 2 H 3 O 2 (aq) H + (aq) + C 2 H 3 O 2 (aq) At equilibrium, the concentration of H + is found to be 1.88 * 10 3 M. Calculate K C for the reaction. Assume that the H + conc. in pure water is zero.

18 HETEROGENEOUS EQUILIBRIUM Heterogeneous equm: reactants and products in more than one phase 3 Fe(s) + 4 H 2 O(g) Fe 3 O 4 (s) + 4 H 2 (g) What is K c?? K c ' = [Fe 3 O 4 ][H 2 ] 4 [Fe] 3 [H 2 O] 4 What is [Fe]? [Fe 3 O 4 ]? They are constants Put them into the equm constant [Fe] 3 [Fe 3 O 4 ] Correct K c is: K c ' = K c = [H 2 ] 4 [H 2 O] 4 [H 2 ] 4 [H 2 O] 4 Leave solids and pure liquids out of equm constant expressions

19 HETEROGENEOUS EQUILIBRIUM EXAMPLE A 1.0 L container holds 224 g of Fe and 5.00 mol H 2 O(l). It is heated to 1000 K and reaches equm g Fe are left unreacted. What is K c? 3 Fe(s) + 4 H 2 O(g) Fe 3 O 4 (s) + 4 H 2 (g) Init Final 1.0??? Change Equm [H 2 ] 4 K c = = [H 2 O] 4 (4.0) 4 (1.0) 4 = 256

20 HETEROGEN. EQUILIBRIUM EXAMPLE #2 Consider the reaction CO(g) + NiO(s) CO 2 (g) + Ni(s) 140 g of CO(g) and 523 g of NiO(s) are added to a 1.00 L reaction vessel at 500 o C and allowed to reach equilibrium. Analysis of the reaction mixtures shows that the concentration of CO(g) at equilibrium is 0.24 M. What mass of NiO(s) remains at equilibrium? What is the value of K C?

21 USING AN EQUILIBRIUM CONSTANT (1) Interpret value does equm favor reactants or products? (2) Predict direction of reaction from given conditions will move toward equlibrium Calculate Q, compare to K (3) Calculate equm concentrations

22 VALUE OF K c Reaction goes almost completely to products 1 K c [Pr] [Rc] K c big little so K large Reaction doesn t go almost no products 2 K c [Pr] [Rc] K c little big so K small Both reactants and products present 3 K c [Pr] [Rc] [Pr] [Rc] K < K < 10

23 Example of small K PbI 2 (s) Pb +2 (aq) + 2 I (aq) yellow colorless colorless Reactants dominate K = 7.1 x 10 9 Examples of large K Cu +2 (aq) + 4 NH 3 (aq) pale blue colorless purple K = 5.0 x Cu(NH 3 ) 4 +2 (aq) Ni +2 (aq) + 6 NH 3 (aq) Products dominate K = 5.5 x 10 8 Ni(NH 3 ) 6 +2 (aq) pale green colorless dark blue

24 PREDICTING THE DIRECTION OF A REACTION Reaction Quotient Q aa + bb cc + dd Q= [C]c [D] d [A] a [B] b The concentrations used in Q are NOT equilibrium concentrations When Q = K c system is at equm When Q < K c reaction needs to move to right to produce more product When Q > K c reaction needs to move to left to produce more reactant

25 CO(g) + H 2 O(g) CO 2 (g) + H 2 (g) K c = [CO 2 ] [H 2 ] [CO] [H 2 O] = 4.0 Initial conditions: [CO] = [H 2 O] = [CO 2 ] = [H 2 ] = 0.02 M Is the system at equilibrium? If not, which direction will it go? Q = (0.02)(0.02) (0.02)(0.02) = 1 System is not at equm Q < K c For Q K, need larger numerator & smaller denominator CO(g) + H 2 O(g) CO 2 (g) + H 2 (g) reaction will go to right

26 PRODEDURE TO DETERMINE EQUILIBRIUM CONCS write the balanced equation for reaction write the equm constant expression list the initial conditions calculate Q and determine the direction of shift to reach equm define the change needed to reach equm for each species define the equm concs for each species (initial change) substitute equm concs into equm expression solve for unknown check equm concs found by substituting them into the equm expression

27 CALCULATION OF EQUILIBRIUM CONCS 2 IBr(g) Br 2 (g) + I 2 (g) [Br 2 ] [I 2 ] K c = = 2.5 x 10 3 [IBr] 2 Problem Statement 0.05 mol of each gas is put in a 1.0 L vessel. What are the equm concs of all three gases?

28 2 IBr(g) Br 2 (g) + I 2 (g) [IBr] [Br 2 ] [I 2 ] Initial Change 2x x x Equm x 0.05 x 0.05 x (0.05 x)(0.05 x) ( x) 2 = 2.5 x 10 3 Take square root (0.05 x) ( x) = 0.05 x = [I 2 ] = [Br 2 ] = = M [IBr] = (0.043) = M Check the answer: (0.007)(0.007) (0.136) 2 = 2.5 x 10 3

29 CALCULATION OF EQUILIBRIUM CONCS H 2 (g) + F 2 (g) K c = [HF] 2 [H 2 ] [F 2 ] 2 HF(g) = 1.15 x mol of each gas was added to a 1.5 L vessel. What are the equilibrium concs of all species?

30 H 2 (g) + F 2 (g) 2 HF(g) Initial concs: [H 2 ] = [F 2 ] = [HF] = 2.0 M Q = (HF) 2 (H 2 )(F 2 ) = (2.0) 2 (2.0)(2.0) = 1.0 Q < K c H 2 (g) + F 2 (g) 2 HF(g) initial change x x + 2x equm 2.0 x 2.0 x x 1.15 x 10 2 = ( x) 2 (2.0 x) (2.0 x) x = 1.53 [H 2 ] = [F 2 ] = 2.0 x = 0.47 M [HF] = x = 5.06 M Check the answer: (5.06) 2 (0.47) (0.47) = 1.15 x 102

31 CALCULATION OF EQUILIBRIUM CONCS #3 A 0.50 L reaction vessel is filled with mole of CO(g) and H 2 O(l) and is then filled with 1.00 mole of CO 2 (g) and H 2 (g). The vessel is then heated to 1200 K and the mixture allowed to reach equilibrium. K C = 4.0 for the reaction. Calculate the equilibrium concentrations of all species. Calculate partial pressure of H 2.

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33 PRACTISE THE SET-UP Consider the reaction 2 NH 3 (g) N 2 (g) + 3 H 2 (g). Suppose that 3 moles of pure NH3 were placed in a 1.0 L reaction vessel and allowed to reach equilibrium. If 3X represents the concentration of H 2 (g) at equilibrium, which of the following represents the concentration of NH3 at equilibrium in moles per liter? A. 3 X B. 3 2X C X D. X E. 2X Write the expression that you would use to solve for the equilibrium concentrations if you knew K C. You do not need to solve this expression.

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35 OLD EXAM QUESTION Consider this equilibrium H 2 (g) + Br 2 (g) 2 HBr(g) K p = at 500 K 1.00 atm of HBr is placed in a 5.00 L vessel at 500 K and allowed to reach equilibrium. What is the partial pressure of HBr at equilibrium? A atm B atm C atm D atm E atm

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