ANSWERS: UNIT 3 EQUILIBRIUM I

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1 ANSWERS: UNIT 3 EQUILIBRIUM I 1. a) [NaOH] = 4.00 g/40 g/mol / 0.10 L ii) [CaCl 2 ] = 16.0 g/111 g/mol / L = 1.00 M = M iii) [KOH] = 14.0 g/56 g/mol / L iv) [H 2 C 2 O 4 ] = 6.75 g/90 g/mol / L = 3.33 M = M 1. b) [ stock ] x # L (stock) = [ diluted ] x # L (diluted) dilution equation 0.56 M x 25 ml = [ diluted ] x 125 ml [ diluted ] = 0.11 M H 2 SO 4 1. c) [ stock ] x # L (stock) = [ diluted ] x # L (diluted) dilution equation 18.0 M x 25 ml = 1.50 M x # L (diluted) # L (diluted) = 300 ml 1. d) [ stock ] x # L (stock) = [ diluted ] x # L (diluted) dilution equation 1.50 M x 120 ml = 1.00 M x # L (diluted) # L (diluted) = 180mL so add 60 ml of water to make the dilution 2. [ mixture ] = total # moles / total # litres 0.5 M = (0.1 M * 1.0 L ) + ( 12 M * x L ) / (1.0 L + x L) 0.5 M = x / / (1.0 L + x L) x = x 11.5 x = 0.4 x = L or 34.8 ml 3. [ NaCl ] = [Na + ] = total # moles / total # litres [ KCl ] = [K + ] = total # moles / total # litres = (0.5 M * L ) / (0.075L) = (0.5 M * L)/ (0.075L) = 0.1 M = 0.4 M [Cl - ] = M = 0.5 M 4. Practice questions on working out equilibrium concentrations a) A + B AB b) A + B AB c) A 2 + B 2 2 AB d) A 2 + B 2 2AB [in] [in] [in] [in] [Eq] [Eq] [Eq] [Eq] e) A 2 + B 2 2 AB f) A 2 + B 2 2 AB g) 2A + B A 2 B h) 2A + B A 2 B [in] [in] [in] [in] [Eq] [Eq] [Eq] [Eq]

2 i) 2A + B 2 2 AB j) 2A + B 2 2 AB k) 2AB 3 A 2 + 3B 2 l) 2AB 3 A 2 + 3B 2 [in] [in] [in] [in] [Eq] [Eq] [Eq] [Eq] m) 2AB 3 A 2 + 3B 2 n) 2AB 3 A 2 + 3B 2 o) 2AB 3 A 2 + 3B 2 p) 2AB 3 A 2 + 3B 2 [in] [in] [in] [in] [Eq] [Eq] [Eq] [Eq] *LeChatelier's Principle. (14.8 pg ) 5a) Note that when 1 mole of reactant is consumed 2 moles of product are produced so products are forming at a rate 2x faster than reactants are disappearing. Both a & c are rejected because they are producing too much product. Rx e is rejected as it has the [reactants] = 0 ie not at equilibrium and rx b is an oscillatory rx not equilibrium. So the correct graph is d!! 5b) a) Adding any reactant shifts an equilibrium towards the products. (RIGHT) b) Adding any product shifts an equilibrium towards the reactants (LEFT ) c) Removing any product shifts an eq'm towards the products (RIGHT) d) A decrease in vol is the same as an increase in pressure so the eq'm will shift towards the side with the fewest moles of gaseous molecules - (LEFT) e) An increase in Temp pushes the eq'm towards the endothermic rx- (RIGHT) 5c) a) Adding a reactant pushes the equil towards products - [ CH 3 OH ] goes UP. b) Removing a reactant pushes the eq'm towards reactants - [CH 3 OH] goes DOWN c) Decreasing vol increases press so equil shifts right - [ CH 3 OH ] goes UP. d) Adding a catalyst has no effectcatalysts only decrease the time req'd to reach equil. e) Increasing Temp favours the endo thermic rx LEFT - [ CH 3 OH ] goes DOWN. 5d) a) Adding any reactant shifts an equilibrium towards the products. (RIGHT ) b) Removing any reactant shifts an equilibrium towards the reactants (LEFT ). c) Adding any product shifts an equilibrium towards the reactants (LEFT ) d) An increase in Temp pushes the eq'm towards the endothermic rx- (RIGHT) e) Adding an inert gas or non-reactant has no effect- * the volume of the can t change when this occurs f) A decrease in vol is the same as an increase in pressure so the eq'm will shift towards the side with the fewest moles of gaseous molecules - (LEFT)

3 5e) a) Removing reactants pushes equil towards the left - [ Cl 2 ] goes UP b) Adding any product pushes equil towards the left - [Cl 2 ] goes UP c) Raising temp favours the endothermic rx LEFT - [Cl 2 ] goes UP d) Decreasing vol = increasing press so eq'm moves RIGHT - [ Cl 2 ] goes DOWN 5f a) Adding any solid has no effect (has no effect ) b) Removing any solid has no effect ( has no effect ). c) Adding any product shifts an equilibrium towards the reactants (LEFT ) d) Increasing vol = decreasing press so eq'm moves to most gaseous moles (RIGHT) 6. SO 3 is a product & reaction is exothermic so a) Keep temperature down b) pressure high c) remove product as it forms and d) keep the concentration of reactants high. 7. H 2 O (l) H 2 O (g) ** note gas on right side and rx is endothermic Increasing the temp pushes equil towards the endothermic rx - RIGHT. But an increase in pressure pushes the equil LEFT away from the gas. So as water begins to boil the pressure goes up ( if in an enclosed area ) which causes it to stop boiling until a higher temp is reached which again causes the pressure to go up and stop boiling etc etc etc. ( a vicious circle ) This is the principle by which a pressure cooker works. Most food contains water and is thus limited by a max boiling pt of 100 C In a pressure cooker tho this temp can reach several hundred degrees so the food cooks much quicker. 8. 1) MnCl 2 (s) Mn 2+ + Cl - **** LEFT side favoured ( not soluble) 2) NaCl Na + + Cl - *** RIGHT SIDE favoured (dissolves well) Notice the common ion Cl - - Equil 1feels all this Cl - coming fron NaCl and shifts left to eliminate this stress and so would be much less soluble in NaCl sol'n 9a) The increasing tendency to go to completion a < c < b (smallest Kc largest Kc) b) K c = [ POCl 3 ] 2 ii) K c = [ SO 2 ] 2 [ O 2 ] iii) K c = [ NO] 2 [ H 2 O] 2 iv) K c = [ NO 2 ] 2 [H 2 O ] 8 [PCl 3 ] 2 [ O 2 ] [ SO 2 ] 2 [ N 2 H 4 ] [ O 2 ] 2 [ N 2 H 4 ] [ H 2 O c) i) K c = [CO] 2 ii) K c = [SO 2 ] [ H 2 O] iii) K c = [CO 2 ] [CH 4 ] iv) K c = [CO 2 ] [H 2 O] v) K c = [H 2 O] 5 [O 2 ] [ H 2 O] 2 [HF] 2

4 9d) The second equation is the reverse of the first! So the value of the new Kc will be the inverse of the first. K c = 1 x e) If you reverse the 2 nd equation ( take the inverse of the K c ) and double it (square the K c value) then add the 1 st and 2 nd reactions together you can obtain the desired reaction equation. Hence the K c for the 3 rd reaction will be the product of the 2 reaction K c values used to get it 2 CH 4 C 2 H 6 + H 2 K c = 9.5 x CH 3 OH + 2 H 2 2 CH H 2 O K c = 1.3 x CH 3 OH + H 2 C 2 H H 2 O K c = 1.2 x f) K c = [ HCl ] 2 ii) K c = [ HCl ] 2 K c for the 2 nd rx will be the square root of the 1 st [H 2 ] [ Cl 2 ] [H 2 ] ½ [ Cl 2 ] ½ 9g) K c = [H 2 ] [ Cl 2 ] This K c will be the inverse of the 1 st [ HCl ] a) K c = [ N 2 O ] [ H 2 O ] 2 b) if [ H 2 O ] is 3 X the [ N 2 O ] would be 1/9th ( since K = constant ) c) if [ N 2 O ] is 1/16 X the [ H 2 O ] would be 4 X 11a) K c = [ CH 3 OH ] [ H 2 ] 2 [ CO ] = ( )/ ( ) 2 ( ) = 3.97 x b) K c = [ C 2 H 5 OH ] [ C 2 H 4 ] [H 2 O] = ( 0.180)/ ( ) ( ) = c) K c = [ PCl 5 ] = 0.18 [ PCl 3 ] [Cl 2 ] = ( 0.005)/ ( ) ( ) = 4.96 * this does equal the Kc value so NOT at equilibrium Since the value is larger than the K c value the eq m must shift LEFT 11d) K c = [NO][ SO 3 ] = 85.0 [ SO 2 ] [NO 2 ] = ( )( / ( ) ( ) = 114 * this does equal the Kc value so NOT at equilibrium Since the value is larger than the K c value the eq m must shift LEFT

5 11e) K c = [ NH 3 ] 2 [ N 2 ] [ H 2 ] 3 **solving for [ H 2 ] 64 = ( 0.360) 2 / ( ) [ H 2 ] 3 [ H 2 ] 3 = [ H 2 ] = M 12. PCl 5 + H 2 2 HCl + PCl 3 K c = ( 3 mol/l 2 *1 mol/l) / 1 mol/l *1 mol/l = 9 b) for the reverse rx Ke = inverse of Ke (forward rx) = 1/9 13. A + 2 B 2 C [In] 3/3 5/3 0 ****** rem [ ] = mol /L [Eq] 1/3 1/3 4/3 K c = ( 4/3 ) 2 / ( 1/3 * (1/3 ) 2 ) = 48 14a) 2 HBr H 2 + Br 2 K c = [ H 2 ] [ Br 2 ] [In] [ HBr] 2-2x +x +x= [Eq] K c = * /( 0.309) 2 = b) CH 2 O H 2 + CO [In] x= [Eq] K c = 0.034*0.034 / = 1.8 x c). NO 2 + NO N 2 O + O 2 K c = (0.122) (0.315) [In] (0.118) ( 0.356) +x= = [Eq] d). 2 N 2 O + 3 O 2 4 NO 2 K c = (0.02) 4 [In] (0.010) 2 ( 0.041) 3-2x -3x +4x (x=0.005) = 23 [Eq]

6 15. Pb 2+ + Sn (s) Sn 2+ + Pb (s) Ke = 1.2 = [Sn 2+ ] / [ Pb 2+ ] [In] 0.5 constant 0 constant 1.2 = x / ( x ) -x +x x = x [Eq] x x 2.2 x = 0.6 So the equil'm [ Pb 2+ ] = M & [ Sn 2+ ] = M x = M 16a) 2 BrCl Br 2 + Cl 2 K c = [ Cl 2 ] [ Br 2 ] [In] [ BrCl] 2-2x +x +x [Eq] x x x = x * x /( x) 2 So the equil'm [ Cl 2 ] = [Br 2 ] = M = x / ( x) x = b) SO 3 + NO NO 2 + SO 2 K c = [ NO 2 ] [ SO 2 ] [In] [ SO 3 ] [NO] -x -x +x +x [Eq] x x x x = x * x /( x) 2 So the equil'm [ NO 2 ] = [SO 2 ] = M [ NO] = [SO 3 ] = M = x / ( x) x = c) CO + H 2 O CO 2 + H 2 K c = [ CO 2 ] [ H 2 ] [In] (Q= 1.0) [ CO] [H 2 O] +x +x -x -x [Eq] x x x x = (0.010-x) 2 /( x) 2 So the equil'm [ CO] = [H 2 O] = M [ CO 2 ] = [H 2 ] = 7.77 x10-3 M 0.632= (0.010-x) / ( x) x = 2.25 x 10-3

7 17. CO 2 + H 2 CO + H 2 O Kc = 0.02 = [CO][H 2 O] / [ CO 2 ][H 2 ] [In] = x 2 /(0.5 - x)(0.25-x) -x -x +x +x 0.02 = x 2 /( x + x 2 ) [Eq] x x x x x 2 = x x x x = 0 x = M So the equil'm [ CO 2 ] = 0.21 M & [ H 2 ] = 0.46 M & [ CO] = [H 2 O] = M graph [ ] 18a) SO 2 + NO 2 NO + SO 3 [In] (Q= 1.6) -x -x +x +x [Eq] x x x x So the equil'm [ SO 2 ] = M [ NO 2 ]= M [ NO] = M [SO 3 ] = M time Kc = 85 = [NO][SO 3 ] / [ SO 2 ][NO 2 ] 85 = (0.008+x)(0.012+x)/ (0.01-x) x) 85x x = x x + 9.6x x x = 0 x = M 18b) 2 BrCl Br 2 + Cl 2 K c = [ Cl 2 ] [ Br 2 ] [In] (Q= 0.375) [ BrCl] 2 +2x -x -x [Eq] x 0.03-x 0.02-x = (0.02-x) *( 0.03-x) /( x) x x = 0 x = So the equil'm [ Cl 2 ] = M, [Br 2 ] = M & [ BrCl] = M

8 19. SO 3 + NO NO 2 + SO 2 [Eq# 1] x -x +x +x [Eq# 2] X X X X Ke = 0.8 * * 0.4 = = (0.8 + X ) ( X ) ( X ) ( X ) 0.5 = ( X + X 2 ) ( X + X 2 ) X X 2 = X + X X X = 0 *** use the quadratic to solve for X X = 0.06 M or So the eq'm [ SO 3 ] = 0.94 M, [ NO ] = 0.34 M, [NO 2 ] = 0.86 M [ SO 2 ]= 0.18 M graph [ ] time

9 20 a). C + 2 H 2 CH 4 G = kj K p = e - G/ RT = e J/ *8.314*298 = e 20.5 = 8.0 x 10 8 ** Ke is large (rx goes to completion) so this method is worthwhile 20 b) ATP + H 2 O ADP + free phosphate G = - 33 kj at 37 C Ke = e - G/ RT = e J/ 8.314*310 = e 13 = 3.6 x c) CO + 2 H 2 CH 3 OH K p = 6.25 x 10 3 at 500 K G = - RT( ln K p ) = * 500 * ln 6.25 x 10 3 = * 500 * ( ) = 2.11 x 10 4 J = kj 20 d) CO + H 2 O HCHO 2 G = C K p = e - G/ RT K c = [ HCHO 2 ] = e J/ *8.314*673 [ CO] [ H 2 O] = e = 6.40 x 10-7 ** = ( 3.8x10-3 mol * R * 673 K) / 2.50 L ((0.04 * R * 673)/2.5 L) * ((0.022*R*673)/2.5 L) = 1.92x10-3 since K c > K p this rx is not at equilibrium at this temperature and must move to the left in order to reach equilibrium

10 21 2 NO + 2CO N 2 + CO 500 C H = ( 2(-394 )kj ) - ( 2(-110) + 2 ( 90.4) ) G = ( ) ( ) kj = = = kj = kj S = ( 2(213.6 ) ) - ( 2(197.9) + 2 ( 210.6) ) - G/ RT Ke = e = = e J/ 8.314*773 = J = e = 1.74 x CO + ½ O 2 CO 2 H = kj S = ( ) = J G = (573 * ) = kj - G/ RT Ke = e = e J/8.314*573 = e = 2.27 x b) T = H/ S = K = C ( Ke = 1 at this temp ) c) Rx should be spontaneous at all temp below C (it is spont at 300 C)