SESSION 2023/2024 TOPIC 6: CHEMICAL EQUILIBRIUM CHEMISTRY SK015 113 TUTORIAL 6.0 6.1: Dynamic Equilibrium & 6.2: Equilibrium Constant 1. a) Define The chemical reaction takes place in both forward and reverses directions. b) characteristic of the system at dynamic equilibrium The rate of the forward reaction is equal to the rate of the reverse reaction. The concentration of reactants and products remains unchanged at equilibrium. c) after t1 from the graph: The concentration of N2O4 and NO2 are constant Rate decomposition of N2O4 is the same as the rate formation of NO2 2. Expression a) H2CO3(l) + H2O(l) ⇋ HCO3 - (aq) + H3O+ (aq) = [3 −] [3 +] not applicable : heterogeneous b) 2NaCl(aq) + CaCO3(s) ⇋ Na2CO3(s) + CaCl2(s) = 1 [] 2 not applicable : heterogeneous c) Hb(CO)4(aq) + 4O2(g) ⇋ Hb(O2)4(aq) + 4CO(g) = [(2)4][] 4 [()4][2 ] 4 = () 4 (2 ) 4 : heterogeneous d) C2H2(g) + 2Br2(g) ⇋ C2H2Br4(g) = [224 ] [22] [2] 2 = (224 ) (22 ) (2 ) 2 : homogeneous e) Br2(l) ⇋ Br2(g) = [2 ] = 2 : heterogeneous f) 3Fe(s) + 4H2O (l) ⇋ Fe3O4(s) + 4H2(g) = [2] 4 = (2 ) 4 : heterogeneous 3. value of Kc Kc = 8 c 1 1 K 3.6 10 = = 5.3×10−5 4. Kp ’ and Kc ’ Kp ’ = 25 K 3.0 10 p = = 5.5×1012
SESSION 2023/2024 TOPIC 6: CHEMICAL EQUILIBRIUM CHEMISTRY SK015 114 Kp ’ = Kc ’ (RT)-1/2 = ' RT K c Kc ’ = Kp ’ RT = 5.5×1012 (0.08206 298 )( ) = 2.7×1013 5. Kc = [ ] 2 [2][2] 3 = (0.412) 2 (1.15)(1.35) 3 = 0.06 6. Kp = 2 2 2 ( ) ( ) NOCl NO Cl P P P = (1.20) 2 (0.05) 2(0.3) = 1.90 x 103 7. At equilibrium the concentration of I2 is 6.61 × 10−4M so that 1.0×10−3−x=6.61×10−4 x=1.0×10−3− 6.61×10−4 =3.39×10−4M Finally, substitute the equilibrium concentrations into the K expression and solve: Kc = [I3 −] [I2 ][I −] = (3.39 x 10−4 ) (6.61×10-4)(6.61×10 -4) =776
SESSION 2023/2024 TOPIC 6: CHEMICAL EQUILIBRIUM CHEMISTRY SK015 115 8. Kc = [2] [] 4.90 x 102 = (0.1) 2 [] X = 2.0×0−4M 9. KC = [PCl3 ][Cl2 ] [PCl5 ] 0.0211 = (x)(x) (2.00−x) x2 + 0.0211x − 0.0422 = 0 x = 0.195 M [PCl5] = 2.00 − x = 2.00 − 0.195 = 1.80 M [PC3] = [Cl2] = x = 0.195 M 10. 2NOCl(g) ⇌ 2NO(g) + Cl2(g) Species 2NOCl(g) 2NO(g) Cl2(g) Pi 0.50 - - P∆ -2x +2x +x P⇋ 0.5-2x 2x x Kp = () 22 () 2 () 2 = (2) 2 () (0.5−2) 2 = 4.0 x 10-4 4x2 -2x + 0.25 = 1.6 x 10-3 x 3 4x3 - 1.6 x 10-3 x 2 +8.0 x 10-4x – 1 x 10-4 = 0 X= 0.0271 atm = 0.4458 atm PNO = 0.0542 atm PCl2 = 0.0271 atm Species PCl5 (g) PCl3 (g) Cl2 (g) [ ]i 2.00 - - [ ]c -x +x +x [ ]e 2.00-x x x
SESSION 2023/2024 TOPIC 6: CHEMICAL EQUILIBRIUM CHEMISTRY SK015 116 11. Kc = [22 ] [2 ][2 ] = 0.047 2 ( 0.103 2 ) ( 0.303 2 ) = 3.012 Kp = Kc (RT)-1 = (3.012)(0.08206)(298.15)-1 = 0.123 12. ( ) ( ) x 0.3914 M 8.3 1 2x x NO Kc 2 2 2 = = − = = N2O4 [N2O4] = 0.3914 M, [NO2] = 0.2172 M 13. Kp Species NH4SH(s) NH3(g) H2S(g) Pi - P∆ - +x +x P⇋ - x x PT = + NH H S 3 2 P P 0.307 atm = x + x x = 0.1535 atm Kp = 3 2 ( )( ) P P NH H S = (0.1535)(0.1535) = 0.02356 14. a) equilibrium constant Kc and Kp. Species 2NOBr(g) 2NO(g) Br2(g) [ ]i 1 - - [ ]c -2x +2x +x [ ]e 1-2x 2x x =0.045 x = 0.045 M = [Br2] , [NO] = 0.09M, [NOBr] = 0.91M Kc = 2 2 2 NO Br NOBr = (0.09) 2(0.045) (0.91) 2 = 4.4016×10−4 Kp = Kc (RT)1 SO2(g) + Cl2(g) ⇋ SO2Cl2(g) n 0.150 0.350 - n∆ -x -x +x n⇋ 0.150-x 0.350 - x X= 0.047 [ ]⇋ 0.150−x 2 = 0.103 2 0.350 − x 2 = 0.303 2 2 = 0.047 2 Species 2NO2(g) N2O4(g) [ ]i 1 - [ ]c -2x +x [ ]e 1-2x x
SESSION 2023/2024 TOPIC 6: CHEMICAL EQUILIBRIUM CHEMISTRY SK015 117 = (4.4016×10−4 )(0.08206)(350.15) = 1.26 ×10−2 b) degree of dissociation of NOBr. = amount dissociate initial amount = (0.91) (1) = 0.91 15. Kc & Kp Species I2(g) 2I(g) [ ]i 0.1125 - [ ]∆ -x +2x [ ]⇋ 0.1125 –x 2x α = amount dissociate initial amount 0.0126 = 0.1125 x = 1.4175×10−3M At equilibrium; [I2] = 0.1111 M [I] = 2.835×10−3 M Kc = [] 2 [2 ] = (2.835 10 -3 ) 2 0.1111 = 7.2342×10−5 Kp = Kc (RT)1 = (7.2342×10−5 )(0.08206)(587) = 3.4847 x 10−3 16. Direction Qc = [] 2 [] 2[2 ] = (6.8) 2 (4.0×10 −2) 2(1.66×10 −2) = 1.74×106 Qc > Kc ● More product in the system. ● Position of equilibrium shift to the left. 17. Direction, Kc = 1.2 Qc = [NH3] 2 [2][2] 3 = (0.25) 2 (0.35)(0.64) 3 = 0.6811 Qc < Kc; ● more reactant in the system, ● thus, system will move forward to achieve the new equilibrium. ● Therefore, [NH3] will increase, [N2] and [H2] will decrease. 18. a) Kc Kc = [][2] [2 ][2 ] = (0.025)(0.03) (0.066)(0.035) = 0.3247 b) Concentration Species CO2 H2 CO H2O
SESSION 2023/2024 TOPIC 6: CHEMICAL EQUILIBRIUM CHEMISTRY SK015 118 [ ]i 0.066 M 0.5 M 0.025 M 0.03 M [ ] ⇋ (0.066-x) M (0.5-x) M (0.025+x) M (0.03 + x) M Kc = [][2] [2 ][2 ] 0.3247 = (0.025+)(0.03+) (0.066−)(0.5−) 2+0.05+0.0006 0.033−0.566+ 2 = 0.3247 x1 = 0.0378 M (choosen) x2 = -0.3945 Therefore; [CO2] = 0.0282 M [CO] = 0.0628 M [H2] = 0.4622 M [H2O] = 0.0678 M 6.3 Le Chatelier’s Principle 1. a) the temperature When T increased, the equilibrium will shift forward in order to lower temperature by absorbed the heat supply and produced more product. Thus, Kp will increased. When T decreased, the equilibrium will shift backward in order to increase temperature by releasing heat and produced more reactant. Thus, Kp will decreased. b) the pressure Value of Kp remain unchanged. 2. a) Calculate concentration of H2O and H2 at equilibrium. Species Fe2O3(s) 3H2(g) 2Fe(s) 3H2O(g) [ ]i - 0.067 M - 0 M [ ]c - -3x M - +3x M [ ]⇋ - (0.067-3x) M - 3x M Kc = [2] 3 [2 ] 3 0.064 = (3) 3 (0.067−3) 3 x = 6.381 x 10-3 M [H2] = 4.79×10−2 M [H2O] = 1.91×10−2 M b) Kp
SESSION 2023/2024 TOPIC 6: CHEMICAL EQUILIBRIUM CHEMISTRY SK015 119 Kp = Kc (RT)0 = 0.064 c) direction of the reaction i. Shift to the right. ii. unchanged iii. unchanged 3. Exothermic or endothermic From table, when T increased, Kp is decreased. When T increased, the equilibrium was shifted backward in order to overcome the high temperature and increase the amount of reactant. Thus backward reaction is endothermic while forward reaction is exothermic. 4. increasing the product ● Removed N2O4. ● Add NO2. ● Decrease Volume. ● Decrease Temperature.
SESSION 2023/2024 TOPIC 6: CHEMICAL EQUILIBRIUM CHEMISTRY SK015 120 MEKA 6 1. Expression a) 3 2 2 2 3 c N H [NH ] K = , 3 N H 2 NH p (P )(P ) (P ) K 2 2 3 = , homogeneous b) HCl H O Cl K 3 c + − = , No Kp , heterogenous c) Kc = CO2 , P CO2 K =P heterogenous d) [NH ] [CO ] 1 K 2 2 3 c = , (P ) (P ) 1 K 3 CO2 2 NH p = heterogenous 2. Relationship a) K1 = 2 3 3 2 2 NH N H K2 = 3 1/2 3/2 2 2 NH N H K2 = K1 b) K1 = 2 2 2 HI H I K2 = 2 2 2 H I HI K2 = 1 1 K 3. Kc Kc = 2 2 2 [HI] [H ][I ] = ( ) ( )( ) 2 0.770 0.302 0.508 = 3.86 4. a) Derive From equation; Kc = 2 2 CO [CO] Kp = ( ) CO2 2 CO P P From, ideal gas equation, PV = nRT, v n where V nRT P = = P = RT Therefore CO 2 2 P CO RT = P CO RT CO =
SESSION 2023/2024 TOPIC 6: CHEMICAL EQUILIBRIUM CHEMISTRY SK015 121 Thus, Kp = ( ) 2 2 CO RT CO RT = 2 1 2 2 [CO ] RT [CO] − KP = KC(RT)−1 b) Calculate KP Kp = (7.14×10─2) (0.08206×973.15)─ 1 = 8.94×10─4 5. a) Concentration at equilibrium Assume volume of the container is constant. 2HI(g) H2(g) + I2(g) [ ]I, M 2.0 - - [ ]c, M -2x + x + x [ ]e, M 2.0-2x x x Given, % dissociation of HI = 20% % dissociation = dissociated initial HI x100% HI 20% = 2x x100% 2.0 x = 0.20 M Therefore concentration of HI, H2 and I2 after equilibrium; [HI] = 2.0-2x [H2] = [I2] = 0.20 M = 2.0-2(0.20) = 1.6 M b) Equilibrium constant, Kc Kc = 2 2 2 [H ][I ] [HI] = ( )( ) ( ) 2 0.20 0.20 1.6 = 0.016 6. a) Concentration at equilibrium Assume volume of the container is constant. N2O4(g) 2NO2(g) [ ]I, M 0.0645 - [ ]c, M -x +2x [ ]e, M 0.0645-x 2x Kc = 2 2 2 4 NO N O 4.61×10−3 = ( ) ( ) 2 2x 0.0645 x − x = 8.0642×10−3 M
SESSION 2023/2024 TOPIC 6: CHEMICAL EQUILIBRIUM CHEMISTRY SK015 122 Therefore concentration of N2O4 and NO2 [N2O4] = 0.05644 M [NO2] = 0.01613 M b) Degree of dissociation α = 2 4 dissociated 2 4 initial N O N O = 3 8.0642 x10 0.0645 − = 0.125 7. a) Calculate Qp QP = ( )( ) ( ) Br H 2 2 2 HBr P P P = ( )( ) ( ) 2 0.010 0.010 0.20 = 2.5×10−3 b) direction ● Qp >Kp ● the system is not at equilibrium. ● The net reaction will proceed from right to the left until the equilibrium is reestablished. 8. shift in equilibrium a) The equilibrium will shift forward to increase the concentration of H2 gas. b) The equilibrium will shift backward to increase the temperature. c) Does not change the system, its only increase the rate of reaction. d) The equilibrium will shift backward to increase the concentration C2H6. e) When V increases, the P will be decreased. The equilibrium will shift forward to raise the pressure up. f) The equilibrium will shift forward to produce more gaseous particles
SESSION 2023/2024 TOPIC 6: CHEMICAL EQUILIBRIUM CHEMISTRY SK015 123 KUMBE 6 1. a) Partial pressure of each gas n SO2Cl2 = 0.0235 mol P SO2Cl2 = ( )( )( ) ( ) 1 1 0.0235 mol 0.08206 atm L mol K 373.15 K 1 L − − =0.720 atm P, atm SO2Cl2(g) SO2(g) + Cl2(g) PI 0.72 - - Pc -x +x +x Pe 0.72-x x x PT = 0.72 –x + x + x = 1.3 x = 0.58 atm P SO2Cl2 = 0.72-0.58 = 0.140 atm P SO2 = P Cl2 = 0.580 atm b) Calculate Kc Kp = ( )( ) ( ) 2 2 2 2 SO Cl SO Cl P P P = (0.58) 0.58 ( ) (0.14) = 2.40 Kp = Kc (RT) Kc = ( ) K p RT = 2.40 (0.08206)(373.15) =0.0780 2. Calculate KP % NH3 = 15 % % N2 = 21.25 % % H2 = 63.75 % P NH3 = 150 kPa P N2 = 212.5 kPa P H2 =637.5 kPa Kp = ( ) ( )( ) 3 2 2 2 NH 3 N H P P P = ( ) ( )( ) 2 3 150 212.5 637.5 = 4.09×10−7 3. a) Direction [PCl3] = 3.49 x 10-4 M [Cl2] = 0.0114 M [PCl5] = 3.93 x 10-4 M Qc = 5 3 2 PCl PCl Cl = ( ) ( )( ) 4 4 3.93 10 0.0114 3.49 10 − − = 98.78 Qc > Kc ● More product in the system. ● Position of equilibrium shift to the left.
SESSION 2023/2024 TOPIC 6: CHEMICAL EQUILIBRIUM CHEMISTRY SK015 124 b) Concentration of each species at equilibrium M PCl3(g) + Cl2(g) PCl5(g) [ ]i 3.49 x 10-4 0.0114 3.93 x 10-4 [ ]c +x +x -x [ ]e 3.49 x 10-4 + x 0.0114 + x 3.93 x 10-4 - x KC = 5 3 2 PCl PCl Cl 2.63 = ( ) ( )( ) 4 4 3.93 10 0.0114 3.49 10 x x x − − − + + x = 3.71×10−4 M [PCl3] = 7.2×10−4 M [Cl2] = 0.012 M [PCl5] = 2.2×10−5 M 4. a) Calculate Kp and Kc Kp = ( ) CO2 P = 0.220 Kp = Kc(RT) Kc = Kp RT = ( ) ( ) 0.220 0.08206 (798.15) = 3.36×10−3 b) shift in equilibrium position i. Does not change. ii. Shift to the left. iii. Shift to the right. iv. Does not change. v. Shift to the right. 5. a) Calculate Kc Kp = Kc(RT)−2 Kc = KP(RT)2 = 1.50×10−5 (0.08206×773.15)2 = 0.0604 b) Calculate Kc ’ Kc ’ = c 1 K = 1 0.0604 = 4.07 c) i. Direction Qc = 2 3 3 2 2 NH N H = ( ) ( )( ) 2 3 1.6 0.2 0.6 = 59.3 Qc < Kc ● More reactant in the system. ● Reaction will shift forward. ii. Effect of increasing volume • When V increase, the P will be decreased. The equilibrium will shift backward to raise the pressure up.