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WHEN A REACTION IS AT EQUILIBRIUM, BOTH THE FORWARD AND REVERSE REACTIONS ARE OCCURING AT THE SAME RATE.

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Presentation on theme: "WHEN A REACTION IS AT EQUILIBRIUM, BOTH THE FORWARD AND REVERSE REACTIONS ARE OCCURING AT THE SAME RATE."— Presentation transcript:

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2 WHEN A REACTION IS AT EQUILIBRIUM, BOTH THE FORWARD AND REVERSE REACTIONS ARE OCCURING AT THE SAME RATE.

3 CONCENTRATION REMAINS CONSTANT AT EQUILIBRIUM …..BECAUSE REACTANTS ARE BEING USED UP TO PRODUCE PRODUCTS AT THE SAME RATE THAT PRODUCTS ARE BEING USED UP TO PRODUCE THE REACTANTS.

4 K c or K eq or K P are EQUILIBRIUM CONSTANTS. THEY SHOW A RELATIONSHIP BETWEEN REACTANTS AND PRODUCTS IN A REACTION. CONCENTRATIONS ARE GIVEN IN MOLARITY [ ]. FOR THE REACTION: aA + bB  cC + dD K eq = [C] c [D] d NOTE, PRODUCTS OVER [A] a [B] bREACTANTS.

5 N 2 O 4 (g)  2 NO 2 (g) When the reactant, N 2 O 4 is put in an evacuated container at 100 o C, it decomposes to NO 2. In the beginning only NO 2 is formed, but as soon as it forms, it begins going back to forming N 2 O 4. Eventually, the rate of the forward and reverse reactions are equal. Thus, the reaction has reached equilibrium. The following data and diagram depict this reaction at equilibrium.

6 Time (s) Conc. N 2 O 4 (M) Conc. NO 2 (M) N2O4N2O4 NO 2 Conc(M) Time (s)

7 Regardless of quantities of reactants or products started with Regardless of pressure Regardless of volume THE RATIO OF PRODUCTS TO REACTANTS WILL BE A CONSTANT AT EQUILIBRIUM AT ANY GIVEN TEMPERATURE. EX. [NO 2 ] 2 [N 2 O 4 ] Exp 1.[NO 2 ] 2 [N 2 O 4 ] = (0.120) 2 / = 0.36 Exp 2.[NO 2 ] 2 [N 2 O 4 ] = (0.072) 2 / = 0.37 Exp. 3.[NO 2 ] 2 [N 2 O 4 ] = (0.160) 2 / = 0.36 But if temp. is increased to 150 o C the K eq = 3.2

8 AN EQUILIBRIUM EXPRESSION IS ASSOCIATED WITH A REACTION N 2 O 4 (g)  2 NO 2 (g) Keq = [NO 2 ] 2 / [N 2 O 4 ] = 0.36 ½ N 2 O 4 (g)  NO 2 (g) Keq = [NO 2 ] / [N 2 O 4 ] 1/2 = (0.36) 1/2 = NO 2 (g)  N 2 O 4 (g) Keq = [N 2 O 4 ] / [NO 2 ] 2 = 1 / 0.36 = 2.8

9 PURE LIQUIDS AND SOLIDS ARE NOT INCLUDED IN THE EQUILIBRIUM EXPRESSION CO 2(g) + H 2(g)  CO (g) + H 2 O ( l ) K eq = [CO] [H 2 ][CO 2 ] I 2(s)  2 I (g) K eq = [I ] 2

10 CuO (s) + H 2(g)  Cu (s) + H 2 O (g) K eq = [H 2 O] [H 2 ] NOTICE THAT THE SOLIDS ARE NOT PRESENT IN THE EQUILIBRIUM EXPRESSION.

11 CALCULATE K eq FOR THE FOLLOWING REACTION: NH 4 Cl (s)  NH 3(g) + HCl (g) 2 moles of NH 3 and 2 moles of HCl and 1 mole of NH 4 Cl are in 5.0 L at equilibrium. K eq = [NH 3 ][HCl] [HCl] = 2MOLES 5.0 L = 0.4M [NH 3 ] = 2MOLES 5.0 L = 0.4M K eq = 0.4 x 0.4 = 0.16

12 FOR THE FOLLOWING REACTION: 2HI (g)  H 2(g) + I 2(g) STARTING WITH 0.100M HI, AND THEN AT EQUILIBRIUM, [H 2 ] = M. CALCULATE [I 2 ], [HI], AND K eq. SOLUTION: AT THE START, H 2 AND I 2 ARE ZERO AND ARE ALSO A 1:1 RATIO. THEY WILL HAVE THE SAME MOLAR CONC. 2 [HI] = [H 2 ] THAT IS, 2 [HI]ARE REQUIRED TO MAKE ONE [H 2 ]. SO WHATEVER THE [H 2 ] IS, THE [HI] IS DECREASING BY DOUBLE THAT AMOUNT. THEREFORE, [H 2 ] x 2 = 0.020M DECREASE IN [HI], THEREFORE, [HI] – = 0.080M [HI] AT EQUILIBRIUM. CONTINUED…….

13 K eq = [I 2 ][H 2 ] [HI] 2 = (0.010) (0.010) (0.080) 2 K eq = Why do we care about the equilibrium constant? IT SHOWS TO WHAT EXTENT A REACTION WILL PROCEED. IT SHOWS THE DOMINATING DIRECTION IN WHICH THE REACTION WILL GO TO REACH EQUILIBRIUM. IT SHOWS THE CONCENTRATIONS OF SPECIES PRESENT AT EQUILIBRIUM.

14 FOR N 2 O 4 (g)  2NO 2(g) K eq = 0.36 CALCULATE THE QUOTIENT OF PRODUCTS TO REACTANTS (Q) AND DETERMINE THE DIRECTION THE REACTION WILL SHIFT TO REACH EQUILIBRIUM FOR THE FOLLOWING CONDITIONS: a)0.20 MOLE / 4.0 L N 2 O 4 = 0.05 M b)0.20 MOLE / 4.0 L N 2 O 4 AND 0.20 MOLE / 4.0 L NO 2 Solution for part a: Q = 0 2 = 0, Q < K THEREFORE THE REACTION 0.05 WILL SHIFT TO THE RIGHT.

15 b) 0.20 MOLE / 4.0 L N 2 O 4 = 0.05 M 0.20 MOLE / 4.0 L NO 2 = 0.05 M Q = = 0.05 < 0.36 therefore the reaction 0.05 proceeds to the right

16 WE HAVE USED K eq, THE EQUILIBRIUM CONSTANT, WHICH SHOWS THE RATIO OF THE PRODUCTS TO THE REACTANTS AT EQUILIBRIUM. WE KNOW THAT K c REPRESENTS MOLAR CONCENTRATIONS OF SPECIES AT EQUILIBRIUM AND K p REPRESENTS THE RATIO OF PARTIAL PRESSURES OF GASES AT EQUILIBRIUM. THE REACTION QUOTIENT, Q, IS LIKE K BUT IT IS NOT NECESSARILY AT EQUILIBRIUM. COMPARING IT TO K IS A WAY IN WHICH WE MAY DETERMINE THE DIRECTION IN WHICH THE REACTION WILL PROCEED IN ORDER TO RE-ESTABLISH OR REACH EQUILIBRIUM.

17 RELATIONSHIPS OF K, Q, AND  G o Q < K……………………. Q > K……………………. Q = K…………………….  G o 0, K>1  G o > 0, LnK < 0, K<1  G o = 0, LnK = 0, K=1 REACTION SHIFTS RIGHT REACTION SHIFTS LEFT REACTION AT EQUILIBRIUM EQUILIBRIUM MIXTURE IS MOSTLY PRODUCTS EQUILIBRIUM MIXTURE IS MOSTLY REACTANTS EQUILIBRIUM MIXTURE HAS COMPARABLE AMOUNTS OF REACTANTS AND PRODUCTS

18 EXAMPLE: At the start of a reaction there are the following species in a 3.50 L reaction vessel at 430 o C: mole H 2, mole I 2, mole HI. K c = Determine if the system is at equilibrium and if not, determine the direction in which it will proceed. H 2(g) + I 2(g)  2 HI (g) [H 2 ] = mol / 3.50 L = M [I 2 ] = mol / 3.50 L = M [HI] = mol / L = M Q = = 19.5 Q < K therefore the reaction shifts x RIGHT

19 GIVEN THE REACTION; N 2 (g) + O 2 (g)  2 NO (g) Keq = [NO] 2 = 1 x at 25 o C [N 2 ] [O 2 ] IN ORDER FOR Keq TO BE THAT SMALL, THE NUMERATOR MUST BE SMALL COMPARED TO THE DENOMINATOR. THIS MEANS THAT [NO] << [N 2 ] [O 2 ]. THEREFORE, THIS Keq SHOWS VERY LITTLE PRODUCTION OF PRODUCTS (THE NO). IT IS NOT A FEASABLE REACTION. WE CAN CALCULATE THE [NO], GIVEN [N 2 ] = M AND [O 2 ] = M Keq = 1 x 10 –30 = [NO] 2 = 4 x = [NO] 2 (0.010)(0.040) 2 x = [NO]

20 Example: FOR THE FOLLOWING REACTION, WHERE Keq = 0.64 AT 900K, AND TO START THE REACTION, CO 2 AND H 2 ARE BOTH 0.10 M, WHAT ARE THE EQUILIBRIUM CONCENTRATIONS OF ALL SPECIES? CO 2 (g) + H 2 (g)  CO (g) + H 2 O (g) CHOOSE A SPECIES TO CALL ‘X’ AND DETERMINE INITIAL AND FINAL CONCENTRATIONS. CO 2 H 2 COH 2 O INITIAL CHANGE-X-X+X+X FINAL0.10-X0.10-XXX continued…………..

21 0.64 = x 2 (0.64) 1/2 = x (0.10 – x) – x X = M [CO 2 ] = – = = [H 2 ] [CO] = = [H 2 O] USING THE SAME EQUATION AND Keq, BUT WITH CO 2 = M AND H 2 = M, CALCULATE THE EQUILIBRIUM CONCENTRATIONS = x 2 = x 2 (0.200 – x) (0.100 – x) ( – 0.200x – 0.100x + x 2 ) Using the quadratic equation, [ x ]= M = [CO] = [H 2 O] 0.10 – = = [CO 2 ], – = = [H 2 ]

22 Example: N 2 O 4 (g)  2NO 2 (g) K c = 0.36 at 100. o C And starting concentration of N 2 O 4 = M What are the equilibrium concentrations of [NO 2 ] and [N 2 O 4 ] ? N 2 O 4 NO 2 INITIAL CHANGE-X+2X FINAL0.100 – X2X K c = (2X) 2 = 4X X0.100 – X 4X 2 – 0.35 X = 0, USING THE QUADRATIC EQUATION; X = 0.060, 2X = M = [NO 2 ], – = = [N 2 O 4 ]

23 THE EFFECT OF CHANGES IN CONDITIONS ON A SYSTEM AT EQUILIBRIUM WHEN CONDITIONS ARE CHANGED ON A SYSTEM AT EQUILIBRIUM, THE EQUILIBRIUM IS DISRUPTED AND THE CONCENTRATIONS MAY CHANGE. SOME THINGS THAT CAN DISRUPT EQUILIBRIUM: 1.ADDING OR REMOVING REACTANT OR PRODUCT. 2.CHANGING THE VOLUME OF THE SYSTEM 3.CHANGING THE TEMPERATURE To determine the direction the equilibrium will shift, we apply LeChatelier’s principle. We calculate Kc to determine the concentrations, as before.

24 Example: GIVEN THAT Kc = at 520 o C AND CONCENTRATION OF [HI] = M AND [ I 2 ] = [H 2 ] = M At EQUILIBRIUM, WHAT WOULD THE NEW CONCENTRATIONS BE WHEN EQUILIBRIUM IS RESTORED, IF THE [HI] IS TEMPRORARILY RAISED TO M? INITIAL CONCENTRATIONS ARE THOSE IMMEDIATELY FOLLOWING THE DISRUPTION IN EQUILIBRIUM: 2HI  H 2 + I 2 [HI] [I 2 ][H 2 ] INITIAL0.096 M M0.010 M CHANGE-2X +X +X FINAL0.096 – 2X X X

25 K c = = ( X)( X) = ( X) 2 (0.096 – 2X) 2 (0.096 – 2X) 2 (0.016) 1/2 = X – 2X = X ALGEBRA TIME  – 2X X = M [H 2 ] = = M [I 2 ] = = M [HI] = – 2(0.0017) = M NOTE: THIS VALUE IS BETWEEN THE STARTING CONC., AND HIGH VALUE OF THIS MAKES SENSE!

26 ALWAYS CHECK THAT ANSWERS MAKE SENSE WHEN YOU COMPLETE A PROBLEM! EQUILIBRIUM OF A SYSTEM CAN ONLY BE DISRUPTED BY ADDING OR REMOVING SPECIES IF THAT SPECIES APPEARS IN THE EQUILIBRIUM EXPRESSION. REMEMBER, SOLIDS AND PURE LIQUIDS ARE NOT INCLUDED IN EQUILIBRIUM EXPRESSIONS. example CaCO 3 (s)  CaO (s) + CO 2(g) K c = [CO 2 ] THEREFORE, ADDING AND REMOVING CaCO 3(s) or CaO (s) DOES NOT AFFECT EQUILIBRIUM.

27 VOLUME CHANGES AND THE EFFECT ON EQUILIBRIUM N 2 O 4 (g)  2NO 2 (g) NOTE THAT THERE ARE TWO MOLES OF GAS ON THE RIGHT AND ONE MOLE OF GAS ON THE LEFT. LETS DETERMINE WHAT WILL HAPPEN IF THE VOLUME OF THE CONTAINER IS DECREASED, (PRESSURE INCREASED), OR THE VOLUME OF THE CONTAINER IS INCREASED, (PRESSURE DECREASED). WE WILL USE LeCHATELIER’S PRINCIPLE TO DETERMINE THE SHIFT. WE TAKE INTO CONSIDERATION THE NUMBER OF GAS PARTICLES ON THE LEFT COMPARED WITH THE NUMBER ON THE RIGHT.

28 N 2 O 4 (g)  2NO 2 (g) IF THE PRESSURE IS INCREASED, (VOLUME DECREASED), THERE WILL BE AN INCREASE IN GAS PARTICLES PER UNIT VOLUME. THEREFORE THE REACTION WILL SHIFT IN THE DIRECTION TO DECREASE GAS PARTICLES. SINCE THE RIGHT HAS TWICE AS MANY PARTICLES AS THE LEFT, THE REACTION WILL SHIFT LEFT (TOWARD THE LESSER MOLES OF GAS. THUS, NO 2 WILL FORM MORE N 2 O 4. THE OPPOSITE WILL OCCUR IF THE PRESSURE IS DECREASED AND THE VOLUME INCREASED.

29 AFFECTS OF PRESSURE ON EQUILIBRIUM   = N 2 O 4  = NO 2                         2NO 2 N 2 O 4 THESE CHANGES IN CONCENTRATION CAN BE CALCULATED AS BEFORE. 

30 Example Predict the direction of the shift in the reaction: a.When the volume is increased b.When the pressure is increased C (s) + H 2 O (g)  CO (g) + H 2(g) SO 2(g) + ½ O 2(g)  SO 3(g) 1a. 2 MOLES OF GAS ON THE RIGHT AND ONE ON THE LEFT, THEREFORE, SHIFTS RIGHT. 1b. AN INCREASE IN PRESSURE RESULTS IN A SHIFT TO THE LEFT, TOWARD THE LESSER MOLES OF GAS. 2a. A DECREASE IN PRESSURE, SO A SHIFT TO THE LEFT 2b. AN INCREASE IN PRESSURE CAUSES A RIGHT SHIFT.

31 PRESSURE INCREASE = SHIFT TOWARD LESSER MOLES OF GAS. (TO DECREASE THE NUMBER OF PARTICLES) PRESSURE DECREASE = SHIFT TOWARD GREATER MOLES OF GAS. (INCREASES THE NUMBER OF PARTICLES)

32 WE CAN INCREASED PRESSURE TO A SYSTEM AT EQUILIBRIUM BY ADDING ANOTHER GAS WHILE AT CONSTANT VOLUME. IF WE ADD A GAS, AT CONSTANT VOLUME, THAT IS UNREACTIVE TO OUR SYSTEM AT EQUILIBRIUM, THE PRESSURE WILL INCREASE, BUT THERE WILL BE NO SHIFT IN EQUILIBRIUM. WE CAN INCREASE PRESSURE TO A SYSTEM AT EQUILIBRIUM BY ADDING ANOTHER GAS WHILE AT CONSTANT VOLUME.

33 HOW WOULD YOU INCREASE THE YIELD OF NO 2 IN THE FOLLOWING REACTION: NO (g) + ½ O 2(g)  NO 2(g) a.INCREASE PRESSURE BY COMPRESSION? b.INCREASE VOLUME? c.ADD AN INERT GAS? THE ANSWER IS ‘a’. AN INCREASE IN PRESSURE WILL RESULT IN A SHIFT TOWARD THE LESSER MOLES OF GAS.

34 THE EFFECTS OF TEMPERATURE CHANGE ON A SYSTEM AT EQUILIBRIUM. ACCORDING TO LeCHATELIER’S PRINCIPLE, IF TEMPERATURE IS INCREASED ON A SYSTEM AT EQUILIBRIUM, THE REACTION WILL SHIFT IN A DIRECTION TO COUNTERACT AND THUS ABSORB THE HEAT.

35 IF THE REACTION IS ENDOTHERMIC, THAT IS, ONE THAT ABSORBS HEAT, INCREASING TEMPERATURE WILL CAUSE A SHIFT FARTHER TO THE RIGHT. IF THE REACTION IS EXOTHERMIC, THAT IS, ONE THAT EXPELLS HEAT, AN INCREASE IN TEMPERATURE WILL CAUSE THE REACTION TO SHIFT IN THE DIRECTION THAT ABSORBS HEAT, WHICH IS A SHIFT TO THE LEFT.

36 N 2 O 4(g)  2NO 2(g) WHAT EFFECT WILL AN INCREASE IN TEMPERATURE HAVE ON THIS SYSTEM?  H = +58.2KJ THE REACTION WILL SHIFT RIGHT TO ABSORB THE EXCESS HEAT. N 2 O 4(g)  2NO 2(g)  H = -58.2KJ HOW WOULD A TEMPERATURE INCREASE AFFECT THIS SYSTEM AT EQUILIBRIUM? SINCE THIS REACTION IS EXOTHERMIC, A TEMPERATURE INCREASE WILL CAUSE A SHIFT TO THE LEFT.

37 IF THE REACTION IS ENDOTHERMIC, Kc BECOMES LARGER WITH A TEMPERATURE INCREASE. (A SHIFT TO THE RIGHT CAUSES AN INCREASE IN THE PRODUCTION OF PRODUCTS. IF THE REACTION IS EXOTHERMIC, THE REACTION WILL SHIFT LEFT WITH A TEMPERATURE INCREASE, THEREBY DECREASING THE PRODUCTS AND SO Kc ALSO DECREASES.

38 Example I 2(g)  2I (g)  H = KJ If this system is at equilibrium at 1000 o C, what directional shifts would occur when: a.I atoms are added b.The system is compressed c.The temperature is increased d.Which of these would affect Kc if any, and what would be the affect?

39 a.If I atoms were added the reaction would shift left to use up the excess I atoms. b.If the system was compressed the reaction would shift left toward the lesser moles of gas. c.If the temperature was increased the reaction would shift in the direction that absorbs heat, which in this case is right because it is an endothermic reaction. d.The temperature increase would cause an increase in Kc because the rightward shift increases the production of products.

40 The Relationship of Kc and Kp Kc refers to solutions with concentration expressed in Molarity. Kp refers to gases with concentration expressed as partial pressures. For example: Kp = (P NO 2 ) 2 P N 2 O 4 Terms for pure liquids or solids do not appear in the expressions for Kp or Kc. Kp and Kc have different numeric values.

41 For the reaction: N 2 O 4(g)  2NO 2(g) At 100 o CKc = 0.36 Kp = 11 At 150 o C Kc = 3.2 Kp = 110 For example;

42 TO RELATE Kp TO Kc WE HAVE TO HAVE A RELATIONSHIP FOR PARTIAL PRESSURE AND MOLARITY. MOLARITY = n / V AND FROM THE IDEAL GAS LAW, P A = n RT V THEREFORE, AT EQUILIBRIUM, P A =[A] x RT (because [A ] = n/V)

43 Example N 2 O 4(g)  2NO 2(g) Kp = (P NO 2 ) 2 P N 2 O 4 SO, Kp = [NO 2 ] 2 x (RT) 2 [N 2 O 4 ] x (RT) = [NO 2 ] 2 x RT [N 2 O 4 ] And since, Kc = [NO 2 ] 2 [N 2 O 4 ] THEN, Kp = Kc x RT FOLLOWING IS A GENERAL EQUATION WHICH IS VALID FOR ALL SYSTEMS:

44 Kp = Kc x (RT)  ng Where  ng = the change in moles of gas from products to reactants. (moles gaseous products – moles gaseous reactants).

45 Example N 2 O 4(g)  2NO 2(g  ng = 2 – 1 = 1 N 2(g) + 3 H 2(g)  2 NH 3(g)  ng = 2 – 4 = -2 At 300 o C Kc = 9.5 for the following reaction: N 2(g) + 3 H 2(g)  2 NH 3(g) Calculate Kp. T = 573 K,  ng = 2 – 4 = -2 Kp = Kc (RT) -2, Kp = 9.5 = 4.3 x 10 –3 ( x 573) 2

46 Example Calculate Kp at 520 o C for the following: 2HI (g)  H 2(g) + I 2(g)  ng = 2 – 2 = 0 T = 793 K Kc = Kp = (RT) 0 = Solution:

47 THE RELATIONSHIP OF FREE ENERGY (  G), AND K THE STANDARD, GIBBS FREE ENERGY IS REPRESENTED BY THE SYMBOL,  G 0. (TO BE COVERED IN DEPTH IN THE THERMODYNAMICS CHAPTER). THE EQUATIONS THAT RELATE FREE ENERGY TO K ARE,  G O = - RT(ln K) AND  G =  G O + RT(ln K)

48 THE STANDARD FREE ENERGY,  G O CAN BE CALCULATED IN MUCH THE SAME WAY AS  H, USING THERMODYNAMIC TABLES.  G o =   G O products -   G O reactants VALUES FOR THE STANDARD FREE ENERGY,  G O, ARE TAKEN TO BE AT 1 atm PRESSURE AND R= J/MOLE K, AND TEMP. IN KELVIN. A NEGATIVE VALUE FOR  G O INDICATES A SPONTANEOUS REACTION.

49 USING THE EQUATION,  G O = - RT (lnK), YOU CAN SEE THAT IF  G O IS NEGATIVE, THEN lnK IS POSITIVE AND K>1. THE REACTION PROCEEDS IN THE FORWARD DIRECTION. IF  G O IS POSITIVE, THEN lnK IS NEGATIVE AND K<1 SO THE REVERSE REACTION PROCEEDS SPONTANEOUSLY. IF  G O = 0, lnK = 0 AND K = 1. THE REACTION IS AT EQUILIBRIUM. THEREFORE A LARGE +K MEANS A FORWARD REACTION THAT WILL GO TO COMPLETION.

50 Example Calculate K at 25 o C given the following: a.  G O = KJ/mol, T = 298 K b.  G O = KJ/mol, T = 298 K a KJ/mol =( KJ/mol K) (298K) lnk 16.1 = ln k, e 16.1 = k, 1.03 x 10 7 = k Note the large + k, and the negative  G O b = ( 298K) ln k 9.7 x 10 –8 = k note the +  G O and the k<1

51 THE FOLLOWING ARE MORE EXAMPLES: CALC.  G o IF K = 1.0 x AT 100 o C  G o = j/mol K ( 373K)ln(1.0 x )  G o = -7.1 x 10 4 j = x 10 1 kj For CaCO 3(s)  CaO (s) + CO 2(g)  G o = 0 at 1110 K CALCULATE Kp AND Kc. 0 = -RT(lnK),0 = (1110) lnK 0 = lnK = 0,  K= 1 = Pco 2(g) = 1atm (1110) Continued…..

52 …..and since Kp = Kc (RT)  ng Kp = Kc = 1.1 x M = [CO 2 ] ( x 1110) 1

53 A 1L REACTION VESSEL IS CHARGED WITH mole H 2 AND mole I 2 AT 430 o C. CALC. THE CONCENTRATIONS OF H 2, I 2, AND HI AT EQUILIBRIUM. Kc = 54.3 H 2 (g) + I 2 (g)  2HI (g) Always make a table of before and after concentrations: H 2 I 2 HI Initial Change-X-X+2X Final X X2X

54 Kc = [HI] = (2X) 2 [H 2 ] [I 2 ] (0.500-X)(0.500-X)  (54.3) 1/2 = 2X  2X – X 7.37 (0.500) – 7.37X = 2X 3.68 = 9.37X.393 = X    –     –  

55 A 9.60 L REACTION VESSEL AT 430 o C is charged with 4.20 mole HI. CALCULATE ALL EQUILIBRIUM CONCENTRATIONS. Kc = 54.3 (Same equation as the last example) 4.20 mole HI / 9.60 L = M H 2 I 2 HI Initial Change+x+x-2x Finalxx0.438 – 2x 54.3 = [HI] 2 ans. HI=0.345 M [H 2 ] [I 2 ]I 2 =H 2 = M

56 From an earlier example we calculated a value for Q for the reaction of H 2 + I 2  2HI The species were in a 3.50 L vessel at 430 o C and Kc = The concentrations were as follows: H 2 = M, I 2 = M, HI = M We determined that at these conditions, the reaction would proceed right because Q { "@context": "http://schema.org", "@type": "ImageObject", "contentUrl": "http://images.slideplayer.com/4175678/13/slides/slide_55.jpg", "name": "From an earlier example we calculated a value for Q for the reaction of H 2 + I 2  2HI The species were in a 3.50 L vessel at 430 o C and Kc = 54.3.", "description": "The concentrations were as follows: H 2 = 0.00623 M, I 2 = 0.00414 M, HI = 0.0224 M We determined that at these conditions, the reaction would proceed right because Q

57 54.3 = ( x) ( –x)( – x) a= 50.3, b= 0.653, c = 8.98 x Using algebra and the quadratic equation the solution is : either x = M (more than we started with) or x = M

58 At 350 o C, Kc = 2.37 x 10 –3 for N 2(g) + 3H 2(g)  2 NH 3(g) The equilibrium concentrations were as follows: [N 2 ] = M, [H 2 ] = 8.80 M, [NH 3 ] = 1.05 M If the [NH 3 ] is quickly decreased to M by removing some of it, a) Predict the direction of shift. b) Prove this by calculating Qc and comparing with Kc. a)Shift right to produce more NH 3 b)(0.774) 2 = 1.29 x 10 –3 = Qc (0.683)(8.80) 3 Qc < Kc  shifts right


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