Chapter 5: Chemial Equilibrium ahoot!. At eq, the rate of the forward reation is the rate of the reverse reation. equal to, slower than, faster than, the reverse of. Selet the statement that BEST desribes a system at equilibrium. The reation stops sine all reatants have beome produts. The reation stops sine all of the atalyst has been used. As soon produt forms other moleules form reatants. The reation stops beause equilibrium amounts havebeen reahed. 3. The eq expression for 4A + 3B < > C + D is. = [C][D]/[A][B ], = [A][B ]/[C][D], = [C] [D]/[A] 4 [B ] 3, = [A] 4 [B ] 3 /[C] [D] 4. The reverse eq expression for the 4A + 3B < > C + D is. = [C][D]/[A][B ], = [A][B ]/[C][D], = [C] [D]/[A] 4 [B ] 3, = [A] 4 [B ] 3 /[C] [D] 5. Equilibrium onstants typially have units of. M, M, M, None of the above 6. p = when. the reation is at eq, the reation is exothermi, all of the gases present are at the same pressure, n gas_produts = n gas_reatants 7. What is the p of Cl (g) + CO (g) < > COCl (g) at 35C given is 5.0?.0, 0.0, 0.0, 0.00 8. What is the for 3O < > O 3 if the original was 5 for 4O 3 < > 6O? 5, 0., 0 5 9. If the value of the is large, then at equilibrium mostly will be present. reatants, produts, atalysts, water 0. HA < > H + + A [HA] =.65 0 M, & [H + ] = [A ] = 5.44 0 4 M at eq. =..79 0,.79 0 3,.79 0 4,.79 0 5. The reation quotient Q is usually represented by. [reatants]/[produts], [produts]/[reatants], [reatants] x [produts], [reatants] + [produts]. Co(s) + H + (aq) < > Co + (aq) + H (g) Whih Q is orret? Q = [Co][H + ] /[Co + ][H ], Q = [Co + ][H ]/[H + ], Q = [Co + ][H ]/[Co][H + ], Q = [H + ] /[Co + ][H ] 3. Whih of the following is true if reation quotient Q is at eq? Q >, Q <, Q =, Q = 4. If =.39 and Q = 5. whih diretion do we go in to reah equilibrium? reverse, forward, remain the same, annot be predited. 5. Q for heterogeneous equilibria do not inlude onentrations of. pure liquids, pure solids, both, neither 6. CO + H < > CO + H O If H is added, [CO] eq will. inrease, derease, remain unhanged, 7. CO + H < > CO + H O If all speies are gases & the ontainer is ompressed, [CO] eq will. inrease, derease, remain unhanged, 8. CO + H < > CO + H O If H O is added, [CO] eq will. inrease, derease, remain unhanged, 9. CO + H < > CO + H O If CO is removed, [CO] eq will. inrease, derease, remain unhanged, 0. CO + H < > CO + H O Adding a atalyst will ause [CO] eq to. inrease, derease, remain unhanged,. CO + H < > CO + H O If this is an endothermi reation and T is inreased, [CO] eq will. inrease, derease, remain unhanged,. CO + H < > CO + H O If this is an endothermi reation and T is inreased, the value of will. inrease, derease, remain unhanged, 3. CO + H < > CO + H O Adding a Ne to this reation will ause the will ause [CO] eq to. inrease, derease, remain unhanged, 4. Ni(CO) 4(g) < > Ni (s) + 4 CO (g) Adding nikel to this reation will ause the equilibrium to. Shift toward produts, shift toward reatants, remain unhanged, hange based on the amount added Numeri Example: A reation vessel ontains an eq mix of the following: P SO = 0.008 atm, P O = 0.003 atm, and P SO3 = 0.066 atm. What is the eq onstant for the following reation? SO( g) O( g) SO3( g)
PSO 0.066 SO O atm 3 4 p.66 0 P P (0.008 atm) (0.003 atm) Another Example: What is the p of the reation below at 35C given = 5.0? Cl CO COCl ( g) ( g) ( g) [ COCl] L units are [ Cl ][ CO] mol n L Latm p ( RT) 5.0 0.0806 (35 73.5) mol mol p 0.0 Appliation of Mass Ation Rules I: Calulate the eq onst for D A Bgiven the info below. AB C 3.3 C D 0.04 we need to look from the reverse diretion for both of these reations C AB 0.303 3.3 D C 4.4 0.04 D AB 0.3034.4 7.39 Appliation of Mass Ation Rules II: If the eq. onstant at a given temperature is.4 x 0 3 for SO O SO what is the eq. onstant for the reations below? ( g) ( g) 3( g) a.) SO O SO ( g) ( g) 3( g) b.) SO3( g) SO( g) O( g) = (.4 x 0 ) 0.049-3 -3 - = (.4 x 0 ) 47.) SO3( g) SO( g) O ( g) = (.4 x 0 ) 0.4 Calulating Equilibrium Constant when all eq onentrations are known: What is the for SO( g) O( g) SO3( g) if their equilibrium onentrations are [SO ]=0.5 M, [O ]=0.68, [SO 3 ]=.5. Write down the equilibrium expression symbolially: 3 SO SO O. Plug the give equilibrium values into the expression and solve: SO O.5 0.5 0.68 3.5 0 SO Calulating Eq Constant with both initial and some eq onentrations known: we use an "ICE" table, "I"nitial, "C"hange, "E"q & stoihiometry -- Ex: What is the for SO( g) O( g) SO3( g) if the initial onentrations of reatants were [SO ]=0.50M and [O ]=0.680M and the equilibrium onentration of the produt is [SO 3 ]=0.050M?. Setup the ICE table SO O SO 3 Initial 0.5 0.68 0.0 Change x x +x Eq?? 0.05-3
. Use the given equilibrium onentration to identify x x 0.050 x 0.05 3. Plug x into the table to get the missing equilibrium onentrations SO O SO 3 Initial 0.50 0.680 0.00 Change (0.05) -0.05 +x Eq 0.00 0.655 0.05 4. Write down the equilibrium expression symbolially and plug in values SO O 0.05 0.0 0.655 3 3.8 0 SO Calulating Equilibrium onstant when the initial onentrations and %dissoiation is known: -- Ex: What is the for SO( g) O( g) SO3( g) if 0.500M of both reatants will be 0.5% dissoiated in order to reah equilibrium?. Setup the ICE table SO O SO 3 Initial 0.500 0.500 0.0 Change x x +x Eq???. Use the given %dissoiation to identify x % dissoiation x initial _ onentration x 0.5 0.5 0.005 00 00 3. Plug x into the table to get the missing equilibrium onentrations SO O SO 3 Initial 0.500 0.500 0.00 Change (0.005) -0.005 +x Eq 0. 4950 0.4975 0.0050 4. Write down the equilibrium expression symbolially and plug in values SO O 0.0050 0.4950 0.4975 3 4.05 0 SO Example with Q: Given the data below is the reation in equilibrium and if not in whih diretion will need to go in order to reah eq? A B =, [A] = 0.0 M, [B] =.0 M [ B].0 Q 0 < = therefore it will go in the forward diretion [ A] 0. Calulating Equilibrium Conentrations using "ICE" table, "I"nitial, "C"hange, "E"q ICE Example I: The value of = 0.0900 at 98 for the reation below, determine the eq onentrations if initially [H O] = 0.0043 M and [Cl O] = 0.0044 M. H O( g ) Cl O( g) HOCl( g). Write down the ICE table H O Cl O HOCl Initial 0.0043 0.0044 0.0 Change x x +x Eq 0.0043 x 0.0044 x +x. Write down the equilibrium expression symbolially [ HOCl] [ H O] Cl O 3. Fill in the expression and get the quadrati equation 3
x 4x 0.0900 5 0.0043 x 0.0044 x.90940 8.740 xx 5 0.0900 (.90940 8.740 xx ) 4x 6 4.7850 7.8660 x0.0900x 4x 4 6 0 3.900x 7.8660 x.7850 4. Use the quadrati equation to find x (hoose the positive value) Reall the quadrati equation: b b 4a for ax bx 0 x a in our ase, a = 3.900, b = 7.8660, =.7850 4 4 6 4 4 6 7.8660 7.8660 43.900 -.7850 x 3.900 4 7 5 7.8660 6.870.6880 x 7.8 4 7.8660 0.00544 4 x x 5.700 7. 8 or 4 x 7.70 5. Plug the x value into the E row of the table and find the onentrations: x must be greater than zero therefore x = 5.70x0 4 Using this value we an determine what the onentrations are at eq: [H O] = 0.0043 x = 0.0043 0.00057 = 0.00375 M [Cl O] = 0.0044 x = 0.00385 M [HOCl] = x = 0.004 M ICE Example II: The value of for the thermal deomposition of hydrogen sulfide HS H S ( g ) ( g) ( g) is. x 0 6 at 400. A sample of gas in whih [H S] = 0.600M is heated to 400 in a sealed vessel. After hemial eq has been ahieved, what is the value of [H S]? Assume no H and S was present in the original sample.. Write down the ICE table H S H S Initial 0.600 0.0 0.0 Change x +x +x Eq 0.600 x +x +x. Write down the equilibrium expression symbolially: [ H][ S] HS 3. Fill in the expression x x 3 4x 0.600 x 0.600 x 6.0 4. Here we annot use quadrati so instead we must use an assumption to find x: assume 0.600 >> x0.600 x 0.600 3 3 4x 6 4x.0 x 0.00583M 0.600 x 0.36 5. Verify assumption: 0.00583 00%.94% 5% valid 0.600 6. Use the E row to find the equilibrium onentrations:
Using x = 0.00734 M we an determine what the onentrations are at eq: [H S] = 0.600 x = 0.600 *0.00583 = 0.588 M [H ] = x = 0.07 M and [S ] = 0.00583 M ICE Example III: What are the eq onentrations of eah of the speies in the following reation, given the = 5. at 700 and the initial onentration of all speies is 0.050 M? CO( g ) H O( g) CO( g) H( g) CO H O CO H Initial 0.050 0.050 0.050 0.050 Change x x +x +x Eq 0.050 x 0.050 x 0.050+x 0.050+x [ CO][ H] (0.050 x)(0.050 x) (0.050 x) 5. [ CO][ HO] (0.050 x)(0.050 x) (0.050 x) (0.050 x) (0.050 x) 5..58 (0.050 x) (0.050 x).58(0.050 x) 0.050 x 0.9.58x 0.050 x x 0.093 M using this value we determine the onentrations: [CO] = [H O] = (0.050 0.093)M = 0.03 M [CO ] = [H ] = (0.050+0.093)M = 0.069 M LCP Example I: For eah senario predit the diretion the reation goes to attain eq: CO(g) H(g) CH3 OH(g) a.) CO is added. reation goes toward produt (forward) b.) CH 3 OH is added. reation goes toward reatants (reverse).) Pressure is redued. n reatants = 3, n produts = reation goes toward reatants (reverse) d.) Volume is inreased. reation goes toward reatant (forward) LCP Example II: In what diretion will the eq shift when eah of the following hanges are made to the system at eq? NO 4 NO H 58.0kJ g g (a) NO 4g is added. reation goes toward produt (b) NO g is removed. reation goes toward produt () the total pressure is inreased by adding N g. reation remains unhanged sine the partial pressures of the reating speies is onstant at onstant volume (d) the volume is dereased. reation goes toward reatant sine n reatants =, n produts = (e) the temperature is dereased. reation goes toward reatant sine the proess is endothermi or NO 58.0 kj NO 4 g g 5