Chemstry 13 NT Be true to your work, your word, and your frend. Henry Davd Thoreau 1 Chem 13 NT Chemcal Equlbrum Module Usng the Equlbrum Constant Interpretng the Equlbrum Constant Predctng the Drecton of a Reacton Oscllatng patterns formed by a reacton far from equlbrum Revew Equlbrum and the equlbrum constant, K c. Obtanng equlbrum constants for reactons. The equlbrum constant, K p. Equlbrum constants for sums of reactons. Heterogeneous equlbrum. 3 1
4 Usng the Equlbrum Constant In the last module, we looked at how a chemcal reacton reaches equlbrum and how ths equlbrum s characterzed by an equlbrum constant. In ths module, we wll look at how we can use the equlbrum constant. 5 Usng the Equlbrum Constant In the last module, we looked at how a chemcal reacton reaches equlbrum and how ths equlbrum s characterzed by an equlbrum constant. We ll look at the followng uses. 1. Qualtatvely nterpretng the equlbrum constant. 6
Usng the Equlbrum Constant In the last module, we looked at how a chemcal reacton reaches equlbrum and how ths equlbrum s characterzed by an equlbrum constant. We ll look at the followng uses.. Predctng the drecton of a reacton. 7 Usng the Equlbrum Constant In the last module, we looked at how a chemcal reacton reaches equlbrum and how ths equlbrum s characterzed by an equlbrum constant. We ll look at the followng uses. 3. Calculatng equlbrum concentratons. 8 Qualtatvely Interpretng the Equlbrum Constant As stated n the last module, f the equlbrum constant s large, you know mmedately that the products are favored at equlbrum. For example, consder the Haber process N (g) + 3H(g) NH 3 (g) At 5 o C the equlbrum constant equals 4.1 x 10 8. 9 3
Qualtatvely Interpretng the Equlbrum Constant As stated n the last module, f the equlbrum constant s large, you know mmedately that the products are favored at equlbrum. For example, consder the Haber process N (g) + 3H(g) NH 3 (g) In other words, at ths temperature, the reacton favors the formaton of ammona at equlbrum. 10 Qualtatvely Interpretng the Equlbrum Constant As stated n the last module, f the equlbrum constant s small, you know mmedately that the reactants are favored at equlbrum. On the other hand, consder the reacton of ntrogen and oxygen to gve ntrc oxde, NO. N (g) + O (g) NO(g) At 5 o C the equlbrum constant equals 4.6 x 10-31. 11 Qualtatvely Interpretng the Equlbrum Constant As stated n the last module, f the equlbrum constant s small, you know mmedately that the reactants are favored at equlbrum. On the other hand, consder the reacton of ntrogen and oxygen to gve ntrc oxde, NO. N (g) + O (g) NO(g) In other words, ths reacton occurs to a very lmted extent. 1 4
Predctng the Drecton of Reacton How could one predct the drecton n whch a reacton at non-equlbrum condtons wll shft to re-establsh equlbrum? To answer ths queston, you substtute the current concentratons nto the reacton quotent expresson and compare t to K c. The reacton quotent, Q c, s an expresson that has the same form as the equlbrumconstant expresson but whose concentratons are not necessarly at equlbrum. 13 Predctng the Drecton of Reacton For the general reacton aa + bb cc + dd the Q c expresson would be: Q c c a d b [C] [D] = [A] [B] where the subscrpt sgnfes ntal or current concentratons. 14 Predctng the Drecton of Reacton For the general reacton aa + bb cc + dd the Q c expresson would be: Q c c a d b [C] [D] = [A] [B] If Q c > K c, the reacton wll shft left toward reactants. 15 5
Predctng the Drecton of Reacton For the general reacton aa + bb cc + dd the Q c expresson would be: Q c c a d b [C] [D] = [A] [B] If Q c < K c, the reacton wll shft rght toward products. 16 Predctng the Drecton of Reacton For the general reacton aa + bb cc + dd the Q c expresson would be: Q c c a d b [C] [D] = [A] [B] If Q c = K c, the reacton s at equlbrum and wll not shft. 17 A Problem To Consder Consder the followng equlbrum. N (g) + 3H(g) NH 3 (g) A 50.0 L vessel contans 1.00 mol N, 3.00 mol H, and 0.500 mol NH 3. In whch drecton (toward reactants or toward products) wll the system shft n order to reestablsh equlbrum at 400 o C? The K c for the reacton at 400 o C s 0.500. 18 6
A Problem To Consder Frst, calculate concentratons from moles of substances. N (g) + 3H(g) NH 3 (g) 1.00 mol 50.0 L 3.00 mol 50.0 L 0.500 mol 50.0 L 19 A Problem To Consder Frst, calculate concentratons from moles of substances. N (g) + 3H(g) NH 3 (g) 0.000 M 0.0600 M 0.0100 M The Q c expresson for the system would be: [NH Q c = [N ][H 3] 3 ] 0 A Problem To Consder Frst, calculate concentratons from moles of substances. N (g) + 3H(g) NH 3 (g) 0.000 M 0.0600 M 0.0100 M Substtutng these concentratons nto the reacton quotent gves: (0.0100) Qc = = 3.1 3 (0.000)(0.0600) 1 7
A Problem To Consder Frst, calculate concentratons from moles of substances. N (g) + 3H(g) NH 3 (g) 0.000 M 0.0600 M 0.0100 M Because Q c = 3.1 s greater than K c = 0.500, the reacton wll go to the left (toward reactants) as t approaches equlbrum. Once you have determned the equlbrum constant for a reacton, you can use t to calculate the concentratons of substances n the equlbrum mxture. 3 For example, consder the followng equlbrum. CO(g) + 3 H (g) CH 4(g) + HO(g) Suppose a gaseous mxture contaned 0.30 mol CO, 0.10 mol H., 0.00 mol H O and an unknown amount of CH 4 per lter. What s the concentraton of the CH 4 n ths mxture? The equlbrum constant K c equals 3.9. 4 8
Frst, calculate concentratons from moles of substances. CO(g) + 3 H (g) CH 4(g) + HO(g) 0.30 mol 1.0 L 0.10 mol 1.0 L?? 0.00 mol 1.0 L 5 Frst, calculate concentratons from moles of substances. CO(g) + 3 H (g) CH 4(g) + HO(g) 0.30 M 0.10 M?? 0.00 M The equlbrum-constant expresson s: [CH4][H O] K c = [CO][H 3 ] 6 Frst, calculate concentratons from moles of substances. CO(g) + 3 H (g) CH 4(g) + HO(g) 0.30 M 0.10 M?? 0.00 M Substtutng the known concentratons and the value of K c gves: [CH 4 ]( 0.00M) 3.9 = 3 (0.30M)( 0.10M) 7 9
Frst, calculate concentratons from moles of substances. CO(g) + 3 H (g) CH 4(g) + HO(g) 0.30 M 0.10 M?? 0.00 M You can now solve for [CH 4 ]. (3.9)(0.30M)(0.10M) [CH ] = (0.00M) 3 4 = 0.059 The concentraton of CH 4 n the mxture s 0.059 mol/l. 8 Suppose we begn a reacton wth known amounts of startng materals and want to calculate the quanttes at equlbrum? 9 Consder the followng equlbrum. CO(g) + H O(g) CO (g) + H Suppose you start wth 1.000 mol each of carbon monoxde and water n a 50.0 L contaner. Calculate the molartes of each substance n the equlbrum mxture at 1000 o C. The K c for the reacton s 0.58 at 1000 o C. (g) 30 10
Frst, calculate the ntal molartes of CO and H O. CO(g) + H O(g) CO (g) + H(g) 1.000 mol 50.0 L 1.000 mol 50.0 L 31 Frst, calculate the ntal molartes of CO and H O. CO(g) + H O(g) CO (g) + H(g) 0.000 M 0.000 M 0 M 0 M The startng concentratons of the products are 0. We must now set up a table of concentratons (startng, change, and equlbrum expressons n x). 3 Let x be the moles per lter of product formed. Startng Equlbrum CO (g) + H O(g) CO (g) + H (g) 0.000 0.000 0.000 0.000 The equlbrum-constant expresson s: [CO ][H ] K c = [CO ][H O ] 0 0 +x +x x x 33 11
Solvng for x. Startng Equlbrum CO (g) + H O(g) CO (g) + H (g) 0.000 0.000 0.000 0.000 0 0 +x +x x x Substtutng the values for equlbrum concentratons, we get: (x)(x) 0.58 = (0.000 - x)(0.000 - x) 34 Solvng for x. Startng Equlbrum Or: CO (g) + H O(g) CO (g) + H (g) 0.000 0.000 0.000 0.000 x 0.58 = (0.000 - x) 0 0 +x +x x x 35 Solvng for x. Startng Equlbrum CO (g) + H O(g) CO (g) + H (g) 0.000 0.000 0.000 0.000 0 0 +x +x x x Takng the square root of both sdes we get: 0.58 = x (0.000 - x) 36 1
Solvng for x. Startng Equlbrum CO (g) + H O(g) CO (g) + H (g) 0.000 0.000 Takng the square root of both sdes we get: 0.76 = 0.000 0.000 x (0.000- x) 0 0 +x +x x x 37 Solvng for x. Startng Equlbrum CO (g) + H O(g) CO (g) + H (g) 0.000 0.000 0.000 0.000 Rearrangng to solve for x gves: 0.000 0.76 x = = 0.0086 1.76 0 0 +x +x x x 38 Solvng for equlbrum concentratons. Startng Equlbrum CO (g) + H O(g) CO (g) + H (g) 0.000 0.000 0.000 0.000 If you substtute for x n the last lne of the table you obtan the followng equlbrum concentratons. 0.0114 M CO 0.0086 M CO 0.0114 M H O 0.0086 M H 0 0 +x +x x x 39 13
The precedng example llustrates the three steps n solvng for equlbrum concentratons. 1. Set up a table of concentratons (startng, change, and equlbrum expressons n x).. Substtute the expressons n x for the equlbrum concentratons nto the equlbrum-constant equaton. 3. Solve the equlbrum-constant equaton for the the values of the equlbrum concentratons. 40 In some cases t s necessary to solve a quadratc equaton to obtan equlbrum concentratons. Remember, that for the general quadratc equaton: ax + bx+ c = 0 the roots are defned as: - b x = b a - 4ac 41 In some cases t s necessary to solve a quadratc equaton to obtan equlbrum concentratons. Remember, that for the general quadratc equaton: ax + bx+ c = 0 The next example llustrates how to solve such an equaton. 4 14
Consder the followng equlbrum. H (g) + I(g) HI(g) Suppose 1.00 mol H and.00 mol I are placed n a 1.00-L vessel. How many moles per lter of each substance are n the gaseous mxture when t comes to equlbrum at 458 o C? The K c at ths temperature s 49.7. 43 The concentratons of substances are as follows. Startng Equlbrum H (g) + I(g) 1.00 1.00.00.00 The equlbrum-constant expresson s: [HI ] Kc = [H][I] HI(g) 0 +x x 44 The concentratons of substances are as follows. Startng Equlbrum H (g) + I(g) 1.00 1.00.00.00 HI(g) 0 +x x Substtutng our equlbrum concentraton expressons gves: (x) 49.7 = (1.00 - x)(.00 - x) 45 15
Solvng for x. H (g) + I(g) Startng 1.00.00 Equlbrum 1.00.00 HI(g) 0 +x x Because the rght sde of ths equaton s not a perfect square, you must solve the quadratc equaton. (x) 49.7 = (1.00 - x)(.00 - x) 46 Solvng for x. H (g) + I(g) Startng 1.00.00 Equlbrum 1.00.00 HI(g) 0 +x x The equaton rearranges to gve: 0.90x - 3.00x +.00 = 0 47 Solvng for x. H (g) + I(g) Startng 1.00.00 Equlbrum 1.00.00 HI(g) 0 +x x The two possble solutons to the quadratc equaton are: x =.33 and x = 0.93 48 16
Solvng for x. H (g) + I(g) Startng 1.00.00 Equlbrum 1.00.00 HI(g) 0 +x x However, x =.33 gves a negatve value to 1.00 (the equlbrum concentraton of H ), whch s not possble. only x = 0.93 remans 49 Solvng for equlbrum concentratons. Startng Equlbrum H (g) + I(g) 1.00 1.00.00.00 If you substtute 0.93 for x n the last lne of the table you obtan the followng equlbrum concentratons. 0.07 M H 1.07 M I 1.86 M HI HI(g) 0 +x x 50 Operatonal Sklls Usng the reacton quotent, Q c Obtanng one equlbrum concentraton gven the others. Solvng equlbrum problems Tme for a few revew questons. 51 17
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