46- تطبيقات حسابية على العلاقة بين Kp, Kc
ﺍﻟﻔﺼﻞ ﺍﻟﺴﺎﺩﺱ :ﺍﻹﺗﺰﺍﻥ ﺍﻟﻜﻴﻤﻴﺎﺋﻲ 1196 ﺇﻋﺪﺍﺩ ﺩ /ﻋﻤﺮ ﺑﻦ ﻋﺒﺪ ﺍ ﺍﳍﺰﺍﺯﻱ ﺍﻟﻌﻼﻗﺔ ﺑﲔ Kc ، Kp Relationship between Kp, Kc ﻟﺪﯾﻚ اﻟﺘﻔﺎﻋﻞ اﻟﻤﺘﺰن اﻟﺘﺎﻟﻲ: → cC + dD aA + bB ← وﺑﺪﻻﻟﺔ اﻟﺘﺮﻛﯿﺰ ﻓﺈن ﺛﺎﺑﺖ اﻹﺗﺰان : ][ E ] .[ D a b ][ A ] . [ B e d = Kc وﺑﻤﺎ أن اﻟﻌﻼﻗﺔ ﺑﯿﻦ اﻟﻀﻐﻂ واﻟﺘﺮﻛﯿﺰ اﻟﻤﻮﻻري ھﻲ : n -1 ) = P ( RT V ﻓﺈن ﺗﺮاﻛﯿﺰ اﻟﻤﻮاد ) (A, B, C, Dﺑﺪﻻﻟﺔ اﻟﻀﻐﻂ ھﻲ : PBb = PBb (RT)- b ( RT ) b = ]= PAa (RT)- a , [ B PAa b a ) ( RT PCc PDd d = ]= PCc (RT)- c , [ D = PDd (RT)- d ( RT ) c ( RT ) d = ][A a = ][ C c وﺑﺎﻟﺘﺎﻟﻲ ﻧﻌﻮض ﺑﮭﺬه اﻟﻘﯿﻢ ﻓﻲ اﻟﻤﻌﺎدﻟﺔ : ] [ E ] .[ D a b ][ A] .[ B d e = Kc 1196 PDF created with FinePrint pdfFactory Pro trial version http://www.pdffactory.com ﺍﻹﺗﺰﺍﻥ ﺍﻟﻜﻴﻤﻴﺎﺋﻲ:ﺍﻟﻔﺼﻞ ﺍﻟﺴﺎﺩﺱ 1197 ﺍﳍﺰﺍﺯﻱ ﻋﻤﺮ ﺑﻦ ﻋﺒﺪ ﺍ/ﺇﻋﺪﺍﺩ ﺩ : Kp وﺗﺼﺒﺢ [ E ] .[ D ] KC = a b [ A ] .[ B ] c -c d -d PC ) ( RT ) . ( PD ) ( RT ) ( KC = a b ( PA ) (RT) -a . ( PB ) (RT) -b c d - (c + d) PC ) . ( PD ) ( RT ) ( KC = × a b - (a + b) ( PA ) . ( PB ) ( RT ) -(c + d) + (a + b) K C = K P . ( RT ) (a + b) - (c + d) (a + b) - (c + d) K C = K P . ( RT ) ..............(divided by ( RT ) (a + b) - (c + d) KP. ( RT ) KC = (a + b) - (c + d) (a + b) - (c + d) (RT ) ( RT ) (c + d) - (a + b) K p = K C ( RT ) ∆n K p = K C ( RT ) e d g : ﺣﯿﺚ . ( اﻟﻔﺮق ﺑﯿﻦ ﻋﺪد ﻣﻮﻻت اﻟﻐﺎزات اﻟﻨﺎﺗﺠﺔ وﻋﺪد ﻣﻮﻻت اﻟﻐﺎزات اﻟﻤﺘﻔﺎﻋﻠﺔΔn (g) ) . ﺛﺎﺑﺖ ﻋﻨﺪ ﺛﺒﻮت درﺟﺔ اﻟﺤﺮارةKc و،( ﺛﺎﺑﺘﺔR ، Δn (g) ) وﻟﻨﻔﺲ اﻟﺘﻔﺎﻋﻞ ﻓﺈن : وﺑﺎﻟﺘﺎﻟﻲ ﻓﺈن K c (RT) Δn (g) = constant : وﺑﺎﻟﺘﺎﻟﻲ ﻓﺈن [P C ] [P D ] a b [P A ] [P B ] c d = c o n sta n t 1197 PDF created with FinePrint pdfFactory Pro trial version http://www.pdffactory.com ﺍﻟﻔﺼﻞ ﺍﻟﺴﺎﺩﺱ :ﺍﻹﺗﺰﺍﻥ ﺍﻟﻜﻴﻤﻴﺎﺋﻲ 1198 ﺇﻋﺪﺍﺩ ﺩ /ﻋﻤﺮ ﺑﻦ ﻋﺒﺪ ﺍ ﺍﳍﺰﺍﺯﻱ وھﺬا اﻟﻤﻘﺪار اﻟﺜﺎﺑﺖ ھﻮ ) (Kpﺣﯿﺚ ﯾﺪل اﻟﺮﻣﺰ ) (Pاﻟﺘﺤﺘﻲ ﻋﻠﻰ اﺳﺘﺨﺪام اﻟﻀﻐﻮط ﻟﻘﯿﺎس ﻗﯿﻤﺔ ھﺬا اﻟﺜﺎﺑﺖ. d b c . P D a . P B P C P A = P K ﺣﯿﺚ Kpﺛﺎﺑﺖ اﻹﺗﺰان اﻟﻜﯿﻤﯿﺎﺋﻲ ﺑﺪﻻﻟﺔ اﻟﻀﻐﻮط. واﻟﻌﻼﻗﺔ اﻟﺘﻲ ﺗﺮﺑﻂ ﺑﯿﻦ Kpو : Kc ∆n g ) ( RT Kp = Kc وﻋﻨﺪ ﺗﺴﺎوي ﻋﺪد ﻣﻮﻻت اﻟﻤﻮاد اﻟﻤﺘﻔﺎﻋﻠﺔ وﻋﺪد ﻣﻮﻻت اﻟﻤﻮاد اﻟﻨﺎﺗﺠﺔ ﻓﺈن : ) ( Δn (g) = 0وھﺬا ﯾﻌﻨﻲ أن : Kp = Kc ﻣﺜﺎﻝ )(١٥ إذا ﻋﻠﻤﺖ أن ) (Kc = 0.105ﻟﻠﺘﻔﺎﻋﻞ : )→ 2NH3 (g 3H 2 (g) + N 2 (g) ← ﻋﻨﺪ درﺟﺔ ﺣﺮارة )(427 ºC وإذا ﻋﻠﻤﺖ أن ) (R = 0.0821 atm L/K molﻓﺎﺣﺴﺐ ﻗﯿﻤﺔ Kpﻟﻠﻨﻈﺎم اﻟﻤﺘﻮازن؟ ﺍﳊﻞ ﻣﻦ اﻟﻌﻼﻗﺔ : 1198 PDF created with FinePrint pdfFactory Pro trial version http://www.pdffactory.com 1199 ﺍﻹﺗﺰﺍﻥ ﺍﻟﻜﻴﻤﻴﺎﺋﻲ:ﺍﻟﻔﺼﻞ ﺍﻟﺴﺎﺩﺱ ﺍﳍﺰﺍﺯﻱ ﻋﻤﺮ ﺑﻦ ﻋﺒﺪ ﺍ/ﺇﻋﺪﺍﺩ ﺩ K p = K c (RT) Δn Δn = n P - n R = 2 - 4 = - 2 → 2NH 3 (g) 3H 2 (g) + N 2 (g) ← nR = 4 nP = 2 K p = 0.105 × (0.0821 × 700) -2 Kp = 0.105 ( 0.0821 × 700 ) 2 K P = 3.179 × 10 -5 (١٦) ﻣﺜﺎﻝ واﻟﻀﻐﻮطKc ﺑﺪﻻﻟﺔ اﻟﺘﺮاﻛﯿﺰ اﻟﻤﻮﻻرﯾﺔ،أﻛﺘﺐ ﻣﻌﺎدﻟﺔ ﺛﺎﺑﺖ اﻹﺗﺰان ﻟﻠﺘﻔﺎﻋﻼت اﻟﺘﺎﻟﯿﺔ . ﻟﻜﻞ ﺗﻔﺎﻋﻞKp, Kc ﺛﻢ أﻛﺘﺐ اﻟﻌﻼﻗﺔ ﺑﯿﻦ،Kp اﻟﺠﺰﺋﯿﺔ → CH4 (g) + H2 O(g) 1......... CO(g) + 3H2 (g) ← → CO(g) + 3H2 (g) 2........ CH4 (g) + H2 O(g) ← → 2NH3 (g) 3........3H2 (g) + N2 (g) ← 1199 PDF created with FinePrint pdfFactory Pro trial version http://www.pdffactory.com ﺍﻹﺗﺰﺍﻥ ﺍﻟﻜﻴﻤﻴﺎﺋﻲ:ﺍﻟﻔﺼﻞ ﺍﻟﺴﺎﺩﺱ 1200 ﺍﳍﺰﺍﺯﻱ ﻋﻤﺮ ﺑﻦ ﻋﺒﺪ ﺍ/ﺇﻋﺪﺍﺩ ﺩ ﺍﳊﻞ 1......... Δ n = n P - n R = 2 - 4 = -2 → C H 4 (g) + H 2 O (g) C O (g) + 3H 2 (g) ← nR = 4 Kc = nP = 2 [C H 4 ]. [ H 2 O ] 3 [C O ]. [ H 2 ] K P = K c (R T ) = , Kp = (P ). (P ) ( P ). ( P ) CH 4 H 2O 3 CO H2 Kc -2 (R T ) 2 Δn = n P - n R = 4 - 2 = 2 → C O (g) + 3H 2 (g ) 2......... C H 4 (g) + H 2 O (g) ← nR = 2 nP = 4 [C O ]. [ H 2 ] [ C H 4 ]. [ H 2 O ] 3 Kc = K P = K c (R T ) , Kp = ( PC O ) . ( P H 2 ) 3 ( P ). ( P ) CH 4 H 2O 2 Δn = n P - n R = 2 - 4 = - 2 → 2N H 3 (g) 3........... 3H 2 (g) + N 2 (g) ← nR = 4 [NH 3 ] nP = 2 Kc = [H 2 ] .[ N 2 ] 3 K P = K c (R T ) = -2 (P ) (P ) .(P ) 2 2 , Kp = NH3 3 H2 N2 Kc (R T ) 2 1200 PDF created with FinePrint pdfFactory Pro trial version http://www.pdffactory.com ﺍﻹﺗﺰﺍﻥ ﺍﻟﻜﻴﻤﻴﺎﺋﻲ:ﺍﻟﻔﺼﻞ ﺍﻟﺴﺎﺩﺱ 1201 ﺍﳍﺰﺍﺯﻱ ﻋﻤﺮ ﺑﻦ ﻋﺒﺪ ﺍ/ﺇﻋﺪﺍﺩ ﺩ (١٧) ﻣﺜﺎﻝ : وﺟﺪ ﻓﻲ ﻧﻈﺎم ﻣﺘﻮازن ﻣﻦ اﻟﮭﯿﺪروﺟﯿﻦ واﻟﻨﯿﺘﺮوﺟﯿﻦ واﻟﻨﺸﺎدر → 2NH3 (g) N 2 (g) + 3H2 (g) ← : أن ﻛﻤﯿﺎت ھﺬه اﻟﻤﻮاد ھﻲ ([ H ] = 1.155 × 10 2 -2 mol/L, [ N 2 ] = 8.3 × 10 -3 mol/L , [NH 3 ] = 2.726 × 10 -5 mol/ L ) ﻓﺎﺣ ﺴﺐ ﻛ ﻼً ﻣ ﻦ،(1.26 atm) ( ﯾ ﺴﺎوي500 ºC) ﻓﺈذا ﻋﻠﻤﺖ أن ﺿﻐﻂ ھﺬا اﻟﻨﻈ ﺎم ﻋﻨ ﺪ Kp, Kc ﺍﳊﻞ Δn = n P - n R = 2 - 4 = -2 → 2NH 3 (g) N 2 g) + 3H 2 (g) ← nR = 4 nP = 2 [ NH3 ] 3 [ N 2 ] .[ H 2 ] 2 Kc = ( 2.726 × 10 ) = (1.155 × 10 ) × (8.3 × 10 ) -5 2 Kc -2 3 -3 K c = 0.0581 K p = K c (RT) -2 ⇒ K p = 0.0581( 0.0821 × 773 ) Kp = 0.0581 ( 0.0821 × 773) 2 -2 = 1.44 × 10-5 1201 PDF created with FinePrint pdfFactory Pro trial version http://www.pdffactory.com 1202 ﺍﻹﺗﺰﺍﻥ ﺍﻟﻜﻴﻤﻴﺎﺋﻲ:ﺍﻟﻔﺼﻞ ﺍﻟﺴﺎﺩﺱ ﺍﳍﺰﺍﺯﻱ ﻋﻤﺮ ﺑﻦ ﻋﺒﺪ ﺍ/ﺇﻋﺪﺍﺩ ﺩ (١٨) ﻣﺜﺎﻝ : ﻟﺪﯾﻚ اﻟﺘﻔﺎﻋﻞ اﻟﺘﺎﻟﻲ → CH3OH(g) 1..........................CO(g) + 2H2 (g) ← (K c = 1.4 × 10 7 ) ﺛﺎﺑﺖ اﺗﺰاﻧﮫ : ﻟﻠﺘﻔﺎﻋﻠﯿﻦ اﻟﺘﺎﻟﯿﯿﻦKp, Kc اﺣﺴﺐ ﻗﯿﻤﺔ، (25 ºC) ﻋﻨﺪ درﺟﺔ ﺣﺮارة → 2CH3OH(g) 2..........................2CO(g) + 4H 2 (g) ← → CO(g) + 2H 2 (g) 3..........................CH 3OH(g) ← ﺍﳊﻞ (2, 3) ﻟﻠﺘﻔﺎﻋﻠﯿﻦKc ﺣﺴﺎب ﻗﯿﻤﺔ ﺛﺎﺑﺖ اﻹﺗﺰان: ًأوﻻ nR = 6 nP = 2 → 2CH3 OH(g) 2..........................2CO(g) + 4H2 (g) ← nR = 1 nP = 3 → 3..........................CH3 OH(g) ← CO(g) + 2H2 (g) : (1, 2, 3) ﺑﻜﺘﺎﺑﺔ ﺛﺎﺑﺖ اﻹﺗﺰان ﻟﻠﻤﻌﺎدﻻت [CH3OH ] 2 [CO][ H 2 ] 2 CH 3OH ] [ Kc 2 = 2 4 [ CO] [ H 2 ] 2 CO ][ H 2 ] [ Kc3 = [ CH3OH ] Kc1 = 1202 PDF created with FinePrint pdfFactory Pro trial version http://www.pdffactory.com 1203 ﺍﻹﺗﺰﺍﻥ ﺍﻟﻜﻴﻤﻴﺎﺋﻲ:ﺍﻟﻔﺼﻞ ﺍﻟﺴﺎﺩﺱ ﺍﳍﺰﺍﺯﻱ ﻋﻤﺮ ﺑﻦ ﻋﺒﺪ ﺍ/ﺇﻋﺪﺍﺩ ﺩ : وﻧﻼﺣﻆ أن Kc2 = (Kc1 )2 Kc2 = (1.4 × 107 ) 2 Kc2 = 1.96 × 1014 : ﻛﺬﻟﻚ ﻧﻼﺣﻆ أن Kc3 = Kc3 = 1 Kc1 1 (1.4 × 107 ) Kc3 = 7.14 × 10- 8 : ﻟﻠﺘﻔﺎﻋﻼت اﻟﺜﻼﺛﺔKp ﺛﺎﻧﯿﺎً ﺣﺴﺎب ﻗﯿﻤﺔ : اﻟﺘﻔﺎﻋﻞ اﻷول nR = 3 P =1 n → CH 3OH(g) 1......CO(g) + 2H 2 (g) ← ∆n = 1 - 3 = -2 Kp1 = Kc1 (RT) ∆n Kp1 = 1.4 × 107 (0.082 × 298)-2 Kp1 = 1.4 × 107 ( 0.082 × 298 ) 2 Kp1 = 2.34 × 104 1203 PDF created with FinePrint pdfFactory Pro trial version http://www.pdffactory.com ﺍﻹﺗﺰﺍﻥ ﺍﻟﻜﻴﻤﻴﺎﺋﻲ:ﺍﻟﻔﺼﻞ ﺍﻟﺴﺎﺩﺱ 1204 ﺍﳍﺰﺍﺯﻱ ﻋﻤﺮ ﺑﻦ ﻋﺒﺪ ﺍ/ﺇﻋﺪﺍﺩ ﺩ : اﻟﺘﻔﺎﻋﻞ اﻟﺜﺎﻧﻲ nR = 6 nP = 2 → 2CH 3OH(g) 2.......2CO(g) + 4H 2 (g) ← ∆n = 2 - 6 = - 4 Kp 2 = Kc 2 (RT) ∆n Kp 2 = 1.96 × 1014 (0.082 × 298) - 4 Kp 2 = 1.96 × 1014 ( 0.082 × 298 ) 4 Kp 2 = 5.47 × 108 : اﻟﺘﻔﺎﻋﻞ اﻟﺜﺎﻟﺚ nR = 1 nP = 3 → CO(g) + 2H 2 (g) 3......CH 3OH(g) ← Δn = 3 - 1 = 2 Kp3 = Kc3 (RT)Δn Kp3 = 7.14 × 10-8 (0.082 × 298) 2 Kp3 = 4.3 × 10-5 : وﯾﻤﻜﻦ ﺗﻠﺨﯿﺺ اﻟﻨﺘﺎﺋﺞ اﻟﺴﺎﺑﻘﺔ ﻛﻤﺎ ﯾﻠﻲ (1) اﻟﺘﻔﺎﻋﻞ (2) اﻟﺘﻔﺎﻋﻞ (3) اﻟﺘﻔﺎﻋﻞ Kc1 Kp1 Kc2 Kp2 Kc3 Kp3 1.4 × 107 2.34 × 104 1.96 × 1014 5.47 × 108 7.14 × 10-8 4.3 × 10-5 1204 PDF created with FinePrint pdfFactory Pro trial version http://www.pdffactory.com 1205 ﺍﻹﺗﺰﺍﻥ ﺍﻟﻜﻴﻤﻴﺎﺋﻲ:ﺍﻟﻔﺼﻞ ﺍﻟﺴﺎﺩﺱ ﺍﳍﺰﺍﺯﻱ ﻋﻤﺮ ﺑﻦ ﻋﺒﺪ ﺍ/ﺇﻋﺪﺍﺩ ﺩ (١٩)ﻣﺜﺎﻝ Kp, Kc ﺛﻢ أﻛﺘﺐ اﻟﻌﻼﻗﺔ ﺑﯿﻦ،( ﻟﻠﺘﻔﺎﻋﻼت اﻟﺘﺎﻟﯿﺔKc) أﻛﺘﺐ ﻣﻌﺎدﻟﺔ ﺛﺎﺑﺖ اﻹﺗﺰان → CH 4 (g) + H 2 O(g) 1...................CO(g) + 3H 2 (g) ← → 2NH 3 (g) 2...................3H 2 (g) + N 2 (g) ← → Zn(s) + CO 2 (g) 3....................ZnO(s) + CO(g) ← → CaO(s) + CO 2 (g) 4...................CaCO 3 (s) ← → Ag + (aq) + Cl - (aq) 5..................AgCl(s) ← → I 2 (g) + 5CO 2 (g) 6...................5CO(g) + I 2O 5 (S) ← → CS 2 (g) + 4H 2 (g) 7....................2H 2S(g) + CH 4 (g) ← ﺍﳊﻞ : اﻟﻤﻌﺎدﻟﺔ اﻷوﻟﻰ nR = 4 nP = 2 → CH 4 (g) + H 2O(g) 1................... CO(g) + 3H 2 (g) ← ∆n = n P - n R = 2 - 4 = -2 Kc : اﻟﺘﻌﺒﯿﺮ ﻋﻦ ﺛﺎﺑﺖ اﻹﺗﺰان Kc = [CH 4 ].[ H 2O] 3 [CO].[ H 2 ] 1205 PDF created with FinePrint pdfFactory Pro trial version http://www.pdffactory.com 1206 ﺍﻹﺗﺰﺍﻥ ﺍﻟﻜﻴﻤﻴﺎﺋﻲ:ﺍﻟﻔﺼﻞ ﺍﻟﺴﺎﺩﺱ ﺍﳍﺰﺍﺯﻱ ﻋﻤﺮ ﺑﻦ ﻋﺒﺪ ﺍ/ﺇﻋﺪﺍﺩ ﺩ Kp, Kc : اﻟﻌﻼﻗﺔ ﺑﯿﻦ K p = K c (RT) Δn K p = K c (RT) -2 Kp = Kc (RT) 2 : اﻟﻤﻌﺎدﻟﺔ اﻟﺜﺎﻧﯿﺔ n =4 n =2 R P → 2NH (g) 2................... 3H 2 (g) + N 2 (g) ← 3 ∆n = n P - n R = 2 - 4 = -2 Kc : اﻟﺘﻌﺒﯿﺮ ﻋﻦ ﺛﺎﺑﺖ اﻹﺗﺰان [ NH3 ] 3 [ H 2 ] .[ N 2 ] 2 Kc = Kp, Kc : اﻟﻌﻼﻗﺔ ﺑﯿﻦ K p = K c (RT) Δn K p = K c (RT) -2 Kp = Kc (RT) 2 : اﻟﻤﻌﺎدﻟﺔ اﻟﺜﺎﻟﺜﺔ nR = 1 nP = 1 → Zn(s) + CO 2 (g) 3.................... ZnO(s) + CO(g) ← ∆n = n P - n R = 1 - 1 = 0 1206 PDF created with FinePrint pdfFactory Pro trial version http://www.pdffactory.com 1207 ﺍﻹﺗﺰﺍﻥ ﺍﻟﻜﻴﻤﻴﺎﺋﻲ:ﺍﻟﻔﺼﻞ ﺍﻟﺴﺎﺩﺱ ﺍﳍﺰﺍﺯﻱ ﻋﻤﺮ ﺑﻦ ﻋﺒﺪ ﺍ/ﺇﻋﺪﺍﺩ ﺩ Kc : اﻟﺘﻌﺒﯿﺮ ﻋﻦ ﺛﺎﺑﺖ اﻹﺗﺰان Kc = [CO 2 ] [CO ] Kp, Kc : اﻟﻌﻼﻗﺔ ﺑﯿﻦ K p = K c (RT) Δn K p = K c (RT) 0 Kp = Kc : اﻟﻤﻌﺎدﻟﺔ اﻟﺮاﺑﻌﺔ nR = 0 nP = 1 → CaO(s) + CO 2(g) 4................... CaCO3 (s) ← ∆n = n P - n R = 1 - 0 = 1 Kc : اﻟﺘﻌﺒﯿﺮ ﻋﻦ ﺛﺎﺑﺖ اﻹﺗﺰان K c = [ CO2 ] Kp, Kc : اﻟﻌﻼﻗﺔ ﺑﯿﻦ K p = K c (RT) Δn K p = K c (RT)1 K p = K c (RT) : اﻟﻤﻌﺎدﻟﺔ اﻟﺨﺎﻣﺴﺔ nR = 0 nP = 0 → Ag + (aq) + Cl - (aq) 5.................. AgCl(s) ← ∆n = 0 1207 PDF created with FinePrint pdfFactory Pro trial version http://www.pdffactory.com 1208 ﺍﻹﺗﺰﺍﻥ ﺍﻟﻜﻴﻤﻴﺎﺋﻲ:ﺍﻟﻔﺼﻞ ﺍﻟﺴﺎﺩﺱ ﺍﳍﺰﺍﺯﻱ ﻋﻤﺮ ﺑﻦ ﻋﺒﺪ ﺍ/ﺇﻋﺪﺍﺩ ﺩ Kc : اﻟﺘﻌﺒﯿﺮ ﻋﻦ ﺛﺎﺑﺖ اﻹﺗﺰان K c = Ag + . Cl- . ﻷﻧﮫ ﻻ ﯾﻮﺟﺪ ﻓﯿﮫ ﻏﺎزات ﻣﺘﻔﺎﻋﻠﺔ أو ﻧﺎﺗﺠﺔKp وﻻ ﯾﻌﺒﺮ ﻋﻦ ھﺬا اﻟﺘﻔﺎﻋﻞ ﺑـ : اﻟﻤﻌﺎدﻟﺔ اﻟﺴﺎدﺳﺔ nR = 5 nP = 6 → 6...................5CO(g) + I 2O 5(s) ← I 2(g) + 5CO 2(g ) ∆n = n P - n R = 6 - 5 = 1 Kc : اﻟﺘﻌﺒﯿﺮ ﻋﻦ ﺛﺎﺑﺖ اﻹﺗﺰان [ I2 ].[CO 2 ] 5 [CO] 5 Kc = Kp, Kc : اﻟﻌﻼﻗﺔ ﺑﯿﻦ K p = K c (RT) Δn K p = K c (RT)1 K p = K c RT : اﻟﻤﻌﺎدﻟﺔ اﻟﺴﺎﺑﻌﺔ nR = 3 nP = 5 →CS 2(g) + 4H 2(g ) 7....................2H 2S(g) + CH 4(g) ← Δn = n P - n R = 5 - 3 = 2 Kc : اﻟﺘﻌﺒﯿﺮ ﻋﻦ ﺛﺎﺑﺖ اﻹﺗﺰان 4 Kc C S 2 . H 2 = 2 H 2 S . C H 4 1208 PDF created with FinePrint pdfFactory Pro trial version http://www.pdffactory.com ﺍﻟﻔﺼﻞ ﺍﻟﺴﺎﺩﺱ :ﺍﻹﺗﺰﺍﻥ ﺍﻟﻜﻴﻤﻴﺎﺋﻲ 1209 ﺇﻋﺪﺍﺩ ﺩ /ﻋﻤﺮ ﺑﻦ ﻋﺒﺪ ﺍ ﺍﳍﺰﺍﺯﻱ اﻟﻌﻼﻗﺔ ﺑﯿﻦ Kp, Kc : K p = K c (RT)Δn K p = K c (RT)2 ﻣﺜﺎﻝ )(٢٠ ﻟﺪﯾﻚ اﻟﺘﻔﺎﻋﻞ اﻟﻤﺘﺰن اﻟﺘﺎﻟﻲ ﻋﻨﺪ درﺟﺔ ﺣﺮارة ):(500 ºC K c = 6 × 10-12 وإذا ﻛﺎﻧﺖ اﻟﺘﺮاﻛﯿﺰ ﻋﻨﺪ اﻹﺗﺰان ھﻲ : )→ 2NH 3 (g 3H 2 (g) + N 2 (g) ← H 2 = 0.25 mol/L, NH 3 = 0.05 mol/L أ( اﺣﺴﺐ ﺗﺮﻛﯿﺰ ] [N2ﻋﻨﺪ اﻹﺗﺰان ب( اﺣﺴﺐ ﻗﯿﻤﺔ Kpﻟﮭﺬا اﻟﺘﻔﺎﻋﻞ ﻋﻨﺪ ﻧﻔﺲ درﺟﺔ اﻟﺤﺮارة. ﺍﳊﻞ أ( ﯾﻤﻜﻦ ﺣﺴﺎب ﺗﺮﻛﯿﺰ ] [N2ﻣﻦ ﻗﯿﻤﺔ Kcﻛﻤﺎ ﯾﻠﻲ : ] [ NH3 = Kc 3 ] [ H 2 ] .[ N 2 2 ] NH3 [ = ] [ N2 3 [H2 ] . Kc 2 (0.05) 2 = ] [ N2 ) (0.25)3 .(6 × 10-12 [ N 2 ] = 2.67 × 1010 mol/L 1209 PDF created with FinePrint pdfFactory Pro trial version http://www.pdffactory.com ﺍﻹﺗﺰﺍﻥ ﺍﻟﻜﻴﻤﻴﺎﺋﻲ:ﺍﻟﻔﺼﻞ ﺍﻟﺴﺎﺩﺱ 1210 ﺍﳍﺰﺍﺯﻱ ﻋﻤﺮ ﺑﻦ ﻋﺒﺪ ﺍ/ﺇﻋﺪﺍﺩ ﺩ : Kp ب( ﺣﺴﺎب ﻗﯿﻤﺔ nR = 4 nP = 2 → 3H 2 (g) + N 2 (g) ← 2NH3 (g) K c = 6 × 10-12 ∆n = 2 - 4 = -2 K p = K c (RT)Δn K p = ( 6 × 10-12 ) (0.082 × 773) -2 6 × 10-12 Kp = (0.082 × 773) 2 K p = 1.49 × 10-15 (٢١) ﻣﺜﺎﻝ → PCl3 (g) + Cl 2 (g) : ﻟﺪﯾﻚ اﻟﺘﻔﺎﻋﻞ اﻟﻤﺘﺰن اﻟﺘﺎﻟﻲ PCl5 (g) ← : ( ﻓﺈن ﻋﺪد اﻟﻤﻮﻻت10 L) ﻓﺈذا ﻋﻠﻤﺖ أﻧﮫ ﻋﻨﺪ اﻹﺗﺰان ﻓﻲ ﺣﺠﻢ ﻗﺪره (n PCl5 = 0.0080 mol, n PCl3 = 0.0114 mol, n Cl2 = 0.0114 mol ) . ﻟﮭﺬا اﻟﺘﻔﺎﻋﻞKc اﺣﺴﺐ ﻗﯿﻤﺔ ﺍﳊﻞ : ﻧﻮﺟﺪ اﻟﺘﺮاﻛﯿﺰ ﻛﻤﺎ ﯾﻠﻲKc ﻟﺤﺴﺎب ﻗﯿﻤﺔ 0.008 = 0.0008 10 0.0114 = 0.00114 [ PCl3 ] = 10 0.0114 = 0.00114 [Cl2 ] = 10 [ PCl5 ] = 1210 PDF created with FinePrint pdfFactory Pro trial version http://www.pdffactory.com ﺍﻟﻔﺼﻞ ﺍﻟﺴﺎﺩﺱ :ﺍﻹﺗﺰﺍﻥ ﺍﻟﻜﻴﻤﻴﺎﺋﻲ 1211 ﺇﻋﺪﺍﺩ ﺩ /ﻋﻤﺮ ﺑﻦ ﻋﺒﺪ ﺍ ﺍﳍﺰﺍﺯﻱ ﺛﻢ ﻧﺤﺴﺐ : Kc ] [ PCl3 ][ Cl2 ] [ PCl5 = Kc )(0.00114) × (0.00114 )(0.0008 = Kc K c = 1.62 × 10-3 ﻣﺜﺎﻝ )(٢٢ اﺣﺴﺐ ﻛﻼً ﻣﻦ Kc, Kpﻟﻠﺘﻔﺎﻋﻞ اﻟﻤﺘﺰن اﻟﺘﺎﻟﻲ : )→ 2NO2 (g N 2 O4 (g) ← ﻋﻨﺪ درﺟﺔ ﺣﺮارة ) ، (25 ºCﻋﻠﻤﺎً ﺑﺄن اﻟﻀﻐﻮط اﻟﺠﺰﺋﯿﺔ : )(P(N2O4 ) = 0.5625 atm, Pt = 0.844 atm ﺍﳊﻞ ﻧﺤﺴﺐ أوﻻً اﻟﻀﻐﻂ اﻟﺠﺰﺋﻲ ﻟـ NO2ﻷﻧﮫ ﻟﯿﺲ ﻣﻌﻄﻰ : ) Pt = P(N2O4 ) + P(NO2 ) 0.844 = 0.5625 + P(NO 2 P(NO2 ) = 0.844 - 0.5625 P(NO2 ) = 0.2815 atm 1211 PDF created with FinePrint pdfFactory Pro trial version http://www.pdffactory.com ﺍﻟﻔﺼﻞ ﺍﻟﺴﺎﺩﺱ :ﺍﻹﺗﺰﺍﻥ ﺍﻟﻜﻴﻤﻴﺎﺋﻲ 1212 ﺇﻋﺪﺍﺩ ﺩ /ﻋﻤﺮ ﺑﻦ ﻋﺒﺪ ﺍ ﺍﳍﺰﺍﺯﻱ وﺑﺎﻟﺘﺎﻟﻲ ﻟﺤﺴﺎب : Kp 2 ) ) P(NO2 ( ) P(N2O4 = Kp (0.2815) 2 = Kp )(0.5625 K p = 0.141 وﻟﺤﺴﺎب Kcﻧﺘﺒﻊ اﻟﻌﻼﻗﺔ : K p = K c (RT)Δn Kp = Kc (RT)Δn )(0.141 = Kc )(0.082 × 298 K c = 5.8 × 10-3 ﻣﺜﺎﻝ )(٢٣ ﻟﺪﯾﻚ اﻟﺘﻔﺎﻋﻞ اﻟﻤﺘﺰن اﻟﺘﺎﻟﻲ ﻋﻨﺪ درﺟﺔ ﺣﺮارة ): (450 ºC K p = 0.66 )→ N 2O 4 (g 2NO 2 (g) ← إذا ﻋﻠﻤﺖ أن اﻟﻀﻐﻂ اﻟﺠﺰﺋﻲ ﻟـ N2O4ھﻮ ) ،(0.13 atmاﺣﺴﺐ Kc ﺍﳊﻞ ﻧﺤﺴﺐ أوﻻً اﻟﻀﻐﻂ اﻟﺠﺰﺋﻲ ﻟـ NO2ﻷﻧﮫ ﻟﯿﺲ ﻣﻌﻄﻰ : 1212 PDF created with FinePrint pdfFactory Pro trial version http://www.pdffactory.com 1213 ﺍﻹﺗﺰﺍﻥ ﺍﻟﻜﻴﻤﻴﺎﺋﻲ:ﺍﻟﻔﺼﻞ ﺍﻟﺴﺎﺩﺱ ﺍﳍﺰﺍﺯﻱ ﻋﻤﺮ ﺑﻦ ﻋﺒﺪ ﺍ/ﺇﻋﺪﺍﺩ ﺩ Kp = (PN2O4 ) (PNO2 ) 2 (0.13) (PNO2 ) 2 0.66 = (P ) 2 (NO2 ) = 0.13 0.66 ( P ) = 0.197 ( P ) = 0.44 atm 2 (NO2 ) (NO2 ) : ﻧﺘﺒﻊ اﻟﻌﻼﻗﺔKc وﻟﺤﺴﺎب K p = K c (RT)Δn Kc = Kp (RT)Δn (0.141) Kc = (0.082 × 298)-1 K c = (0.141) (0.082 × 298) K c = 39.13 (٢٤) ﻣﺜﺎﻝ → 2SO3 (g) 2SO 2 (g) + O 2 (g) ← : ﻟﺪﯾﻚ اﻟﺘﻔﺎﻋﻞ اﻟﺘﺎﻟﻲ : وﻋﻨﺪ اﻹﺗﺰان ﻛﺎﻧﺖ اﻟﻀﻐﻮط اﻟﺠﺰﺋﯿﺔ ﻛﻤﺎ ﯾﻠﻲ (PSO2 = 0.3 atm, PO2 = 0.35 atm, Ptotal = 1.35 atm) .(1100 K) ( ﻋﻨﺪ درﺟﺔ ﺣﺮارةKp, Kc) ﻓﺎﺣﺴﺐ ﻗﯿﻤﺔ 1213 PDF created with FinePrint pdfFactory Pro trial version http://www.pdffactory.com 1214 ﺍﻹﺗﺰﺍﻥ ﺍﻟﻜﻴﻤﻴﺎﺋﻲ:ﺍﻟﻔﺼﻞ ﺍﻟﺴﺎﺩﺱ ﺍﳍﺰﺍﺯﻱ ﻋﻤﺮ ﺑﻦ ﻋﺒﺪ ﺍ/ﺇﻋﺪﺍﺩ ﺩ ﺍﳊﻞ : SO3 ( ﻧﺤﺴﺐ اﻟﻀﻐﻂ اﻟﺠﺰﺋﻲ ﻟﻐﺎزPtotal) ﻣﻦ ﻣﺠﻤﻮع اﻟﻀﻐﻂ اﻟﻜﻠﻲ Ptotal = PSO2 + PO2 + PSO3 PSO3 = Ptotal - (PSO2 + PO2 ) PSO3 = 1.35 - (0.3 + 0.35) PSO3 = 0.7 atm : ﻛﻤﺎ ﯾﻠﻲKp ﻧﺤﺴﺐ ﻗﯿﻤﺔ Kp = (PSO3 )2 (PSO2 )2 (PO2 ) ( 0.7 ) Kp = 2 (0.3) 2 (0.35) K p = 15.6 : ﻧﺘﺒﻊ اﻟﻌﻼﻗﺔKc وﻹﯾﺠﺎد ﻗﯿﻤﺔ K p = K c (RT) ∆n Kc = Kp (RT) ∆n 15.6 Kc = (0.082 × 1100) -1 K c = 15.6 × (0.082 × 1100) K c = 1407.12 (٢٥) ﻣﺜﺎﻝ : ﻟﺪﯾﻚ اﻟﺘﻔﺎﻋﻞ اﻟﺘﺎﻟﻲ اﻟﻤﺘﺰن → H 2 O(g) H 2 O(L) ← PH 2O = 0.0131 atm Kp, Kc اﺣﺴﺐ ﻗﯿﻤﺔ،25 ºC ﻋﻨﺪ درﺟﺔ ﺣﺮارة 1214 PDF created with FinePrint pdfFactory Pro trial version http://www.pdffactory.com ﺍﻟﻔﺼﻞ ﺍﻟﺴﺎﺩﺱ :ﺍﻹﺗﺰﺍﻥ ﺍﻟﻜﻴﻤﻴﺎﺋﻲ 1215 ﺇﻋﺪﺍﺩ ﺩ /ﻋﻤﺮ ﺑﻦ ﻋﺒﺪ ﺍ ﺍﳍﺰﺍﺯﻱ ﺍﳊﻞ nR = 0 P=1 n → H 2 O(L) ← ) H 2 O(g ∆n = 1 - 0 = 1 ﺣﺴﺎب ﻗﯿﻤﺔ : Kp K p = PH O = 0.0131 2 وﻟﺤﺴﺎب Kcﻧﺘﺒﻊ اﻟﻌﻼﻗﺔ : K P = K c (RT)∆n KP (RT)∆n 0.0131 = Kc (0.0821 × 298)1 = Kc K c = 5.35 × 10-4 ﻣﺜﺎﻝ )(٢٦ ﻟﺪﯾﻚ اﻟﺘﻔﺎﻋﻞ اﻟﺘﺎﻟﻲ Kc = 2.6 × 108 : )→ 2H2S(g 2H2 (g) + 2S(g) ← ﻋﻨﺪ درﺟﺔ ﺣﺮارة ) ، (552 ºCوﻛﺎﻧﺖ اﻟﺘﺮاﻛﯿﺰ ﻋﻨﺪ اﻹﺗﺰان ][H2] = 0.002 M, [S = 0.001 M اﺣﺴﺐ ﺗﺮﻛﯿﺰ ][H2S ﺍﳊﻞ ﻣﻦ ﻗﺎﻧﻮن ﺛﺎﺑﺖ اﻹﺗﺰان ﯾﻤﻜﻦ ﺣﺴﺎب ﺗﺮﻛﯿﺰ ] [H2Sﻛﻤﺎ ﯾﻠﻲ : 1215 PDF created with FinePrint pdfFactory Pro trial version http://www.pdffactory.com ﺍﻹﺗﺰﺍﻥ ﺍﻟﻜﻴﻤﻴﺎﺋﻲ:ﺍﻟﻔﺼﻞ ﺍﻟﺴﺎﺩﺱ 1216 ﺍﳍﺰﺍﺯﻱ ﻋﻤﺮ ﺑﻦ ﻋﺒﺪ ﺍ/ﺇﻋﺪﺍﺩ ﺩ [ H 2S] 2 2 [ H 2 ] .[S] 2 Kc = [ H 2S] 2 2.6 × 10 = 8 (0.002) 2 × (0.001)2 [ H 2S] = (2.6 × 108 ) × (0.002)2 × (0.001)2 [ H 2S] = (2.6 × 108 ) × (0.002)2 × (0.001)2 [ H 2S] = 1.04 × 10-3 [ H 2S] = 0.032 mol/L 2 (٢٧) ﻣﺜﺎﻝ : ﻟﺪﯾﻚ اﻟﺘﻔﺎﻋﻞ اﻟﺘﺎﻟﻲ → CO(g) + 2H 2 (g) CH3OH(g) ← K p = 1.14 × 103 at (275 ο C) .(275 ºC) ﻟﻠﺘﻮازن ﻋﻨﺪ درﺟﺔ ﺣﺮارةKc اﺣﺴﺐ ﻗﯿﻤﺔ ﺍﳊﻞ : ﻣﻦ اﻟﻌﻼﻗﺔKc ﯾﻤﻜﻦ ﺣﺴﺎب ﻗﯿﻤﺔ nR = 1 nP = 3 → CH 3OH(g) ← CO(g) + 2H 2 (g) ∆n = 3 - 1 = 2 K p = K c (RT) ∆n Kc = Kc = Kp (RT) 2 1.14 × 103 ( 0.082 × 548 ) 2 K c = 0.56 1216 PDF created with FinePrint pdfFactory Pro trial version http://www.pdffactory.com ﺍﻟﻔﺼﻞ ﺍﻟﺴﺎﺩﺱ :ﺍﻹﺗﺰﺍﻥ ﺍﻟﻜﻴﻤﻴﺎﺋﻲ 1217 ﺇﻋﺪﺍﺩ ﺩ /ﻋﻤﺮ ﺑﻦ ﻋﺒﺪ ﺍ ﺍﳍﺰﺍﺯﻱ ﻣﺜﺎﻝ )(٢٨ ﺳﺨﻨﺖ ﻛﺮﺑﻮﻧﺎت اﻟﻜﺎﻟﺴﯿﻮم ﻋﻨﺪ ) (800 ºCﻓﻲ إﻧﺎء ﻣﻐﻠﻖ : )→ CaO(s) + CO2 (g CaCO3 (s) ← ووﺟﺪ أن اﻟﻀﻐﻂ داﺧﻞ اﻟﺤﯿﺰ ﻗﺪ ﺛﺒﺖ ) ،(0.236 atmﻓﻜﻢ ﺗﺮﻛﯿﺰ ﺛﺎﻧﻲ أﻛﺴﯿﺪ اﻟﻜﺮﺑﻮن CO2 ﺍﳊﻞ ﺑﺘﻄﺒﯿﻖ اﻟﻌﻼﻗﺔ : ) Kp = P(CO2 K p = 0.236 وﺑﺘﻄﺒﯿﻖ اﻟﻌﻼﻗﺔ : K p = K c (RT)Δn 0.236 = K c (0.082 × 1073)1 0.236 )(0.082 × 1073 = Kc K c = 2.68 × 10-3 وﻟﺤﺴﺎب ﺗﺮﻛﯿﺰ ] [CO2ﻧﻄﺒﻖ اﻟﻌﻼﻗﺔ : )→ CaO(s) + CO 2 (g CaCO3 (s) ← ] K c = [ CO 2 ⇒ [ CO 2 ] = 2.68 × 10 -3 mol/L 1217 PDF created with FinePrint pdfFactory Pro trial version http://www.pdffactory.com ﺍﻟﻔﺼﻞ ﺍﻟﺴﺎﺩﺱ :ﺍﻹﺗﺰﺍﻥ ﺍﻟﻜﻴﻤﻴﺎﺋﻲ 1218 ﺇﻋﺪﺍﺩ ﺩ /ﻋﻤﺮ ﺑﻦ ﻋﺒﺪ ﺍ ﺍﳍﺰﺍﺯﻱ ﻣﺜﺎﻝ )(٢٩ ﻟﺪﯾﻚ اﻟﺘﻔﺎﻋﻞ اﻟﺘﺎﻟﻲ : )K c = 0.105 at (427 οC )→ 2NH 3 (g 3H 2 (g) + N 2 (g) ← اﺣﺴﺐ ﻗﯿﻤﺔ ) (Kpﻟﻠﻨﻈﺎم اﻟﻤﺘﺰن ،ﻋﻠﻤﺎً ﺑﺄن )(R = 0.082 L.atm/K. mol ﺍﳊﻞ nR = 4 nP = 2 → 3 H 2 (g ) + N 2 (g ) ← ) 2 N H 3 (g ∆ n = n P - n R = 2 - 4 = -2 ﺑﺘﻄﺒﯿﻖ اﻟﻌﻼﻗﺔ : K p = K c (RT) Δn K p = 0.105 × (0.082 × 700) -2 0.105 (0.082 × 700) 2 0.105 = Kp 3294.76 K p = 3.2 × 10-5 = Kp ﻣﺜﺎﻝ )(٣٠ ﻟﺪﯾﻚ اﻟﺘﻔﺎﻋﻞ اﻟﺘﺎﻟﻲ : )→ 2CO(g CO2 (g) + C(s) ← وﻋﻨﺪ درﺟﺔ ﺣﺮارة ) (760 ºCﻛﺎن اﻟﻀﻐﻂ اﻟﻜﻠﻲ ) ،(Pt = 2.048 atmﻓﺈذا ﻋﻠﻤﺖ أن اﻟﻀﻐﻂ اﻟﺠﺰﺋﻲ ﻷول أﻛﺴﯿﺪ اﻟﻜﺮﺑﻮن ھﻮ )(PCO = 1.7 atm اﺣﺴﺐ ﻛﻼً ﻣﻦ Kp, Kc 1218 PDF created with FinePrint pdfFactory Pro trial version http://www.pdffactory.com ﺍﻟﻔﺼﻞ ﺍﻟﺴﺎﺩﺱ :ﺍﻹﺗﺰﺍﻥ ﺍﻟﻜﻴﻤﻴﺎﺋﻲ 1219 ﺇﻋﺪﺍﺩ ﺩ /ﻋﻤﺮ ﺑﻦ ﻋﺒﺪ ﺍ ﺍﳍﺰﺍﺯﻱ ﺍﳊﻞ ﻧﺤﺴﺐ ﻣﻦ اﻟﻤﻌﻄﯿﺎت اﻟﻀﻐﻂ اﻟﺠﺰﺋﻲ ﻟﺜﺎﻧﻲ أﻛﺴﯿﺪ اﻟﻜﺮﺑﻮن ) (CO2ﻛﻤﺎ ﯾﻠﻲ : Ptotal = PCO2 + PCO PCO2 = Ptotal - PCO PCO2 = 2.048 - 1.7 PCO2 = 0.348 atm وﺑﺎﻟﺘﺎﻟﻲ ﻓﺈن : Kp (P(CO) ) 2 ) ) (P(CO2 = Kp (1.7) 2 = Kp )(0.348 K p = 8.30 وﻟﺤﺴﺎب ﻗﯿﻤﺔ Kcﻧﺘﺒﻊ اﻟﻌﻼﻗﺔ : K p = K c (RT)Δn )8.3 = K c (0.082 × 1033 8.3 )(0.082 × 1033 K c = 0.098 = Kc ﻣﺜﺎﻝ )(٣١ ﺳﺨﻨﺖ ﻛﻤﯿﺔ ﻣﻦ ﻛﻠﻮﯾﺪ اﻷﻣﻮﻧﯿﻮم اﻟﺼﻠﺐ ﻓﻲ إﻧﺎء ﺣﺠﻤﮫ ) (5 Lاﻟﻰ ) (500 ºCوﻋﻨﺪ اﻹﺗﺰان اﻟﺘﺎﻟﻲ : )→ NH3 (g) + HCl(g NH 4Cl(s) ← 1219 PDF created with FinePrint pdfFactory Pro trial version http://www.pdffactory.com ﺍﻟﻔﺼﻞ ﺍﻟﺴﺎﺩﺱ :ﺍﻹﺗﺰﺍﻥ ﺍﻟﻜﻴﻤﻴﺎﺋﻲ 1220 ﺇﻋﺪﺍﺩ ﺩ /ﻋﻤﺮ ﺑﻦ ﻋﺒﺪ ﺍ ﺍﳍﺰﺍﺯﻱ وﺟﺪ أن ﻛﻤﯿﺔ اﻷﻣﻮﻧﯿﺎ ) (NH3ﺗﺴﺎوي ) ،(2 molاﺣﺴﺐ ﻗﯿﻤﺔ ﺛﺎﺑﺖ اﻹﺗﺰان )(Kc ﻟﻠﺘﻔﺎﻋﻞ. ﺍﳊﻞ ﻣﻦ اﻟﻤﻌﺎدﻟﺔ → NH3 (g) + HCl(g) : NH 4Cl(s) ← ﻓﺈن ﺗﺮﻛﯿﺰ ] [NH3اﻟﻤﺘﻜﻮن ﯾﺴﺎوي ﺗﺮﻛﯿﺰ ] ،[HClﻷن ﻛﻼھﻤﺎ ﯾﺸﻜﻼن ﻓﻲ اﻟﻤﻌﺎدﻟﺔ ﻧﻔﺲ ﻋﺪد اﻟﻤﻮﻻت : )→ NH 3 (g) + HCl(g NH 4 Cl(s) ← 1 mol 1 mol وﺑﻤﺎ أن اﻟﻨﺎﺗﺞ ﻣﻦ اﻟﻨﺸﺎدر ھﻮ ) ،(2 molﻓﺈن اﻟﻨﺎﺗﺞ ﻣﻦ ) (HClھﻮ ):(2 mol 2 = 0.4 mol/L 5 = ][ NH3 ] = [ HCl وﯾﻤﻜﻦ ﺣﺴﺎب Kcﻟﻠﺘﻔﺎﻋﻞ اﻟﺴﺎﺑﻖ ﻛﻤﺎ ﯾﻠﻲ : ] K c = [ NH 3 ] .[ HCl K c = 0.4 × 0.4 = 0.16 ﻣﺜﺎﻝ )(٣٢ إذا ﻛﺎن ﺧﻠﯿﻂ ﻣﻦ ﻏﺎزي NO2 ،N2O4ﻓﻲ ﺣﺎﻟﺔ اﺗﺰان ﻋﻨﺪ درﺟﺔ ﺣﺮارة ): (25 ºC )→ 2NO 2 (g N 2 O 4 (g) ← وﻛﺎن اﻟﻀﻐﻂ اﻟﻜﻠﻲ ﻋﻨﺪ اﻹﺗﺰان ) ،(0.844 atmواﻟﻀﻐﻂ اﻟﺠﺰﺋﻲ ﻟﻐﺎز N2O4ھﻮ ) ، (0.5625 atmاﺣﺴﺐ ﻗﯿﻤﺔ Kp 1220 PDF created with FinePrint pdfFactory Pro trial version http://www.pdffactory.com ﺍﻟﻔﺼﻞ ﺍﻟﺴﺎﺩﺱ :ﺍﻹﺗﺰﺍﻥ ﺍﻟﻜﻴﻤﻴﺎﺋﻲ 1221 ﺇﻋﺪﺍﺩ ﺩ /ﻋﻤﺮ ﺑﻦ ﻋﺒﺪ ﺍ ﺍﳍﺰﺍﺯﻱ ﺍﳊﻞ ﻟﺤﺴﺎب ﻗﯿﻤﺔ Kpﻻ ﺑﺪ ﻣﻦ ﻣﻌﺮﻓﺔ اﻟﻀﻐﻂ اﻟﺠﺰﺋﻲ ﻟـ NO2وذﻟﻚ ﻣﻦ اﻟﻌﻼﻗﺔ : ) Ptotal = P( N2O4 ) + P( NO2 ) 0.844 = 0.5625 + P(NO2 P( NO2 ) = 0.844 - 0.5625 P( NO2 ) = 0.2815 atm وﺑﺎﻟﺘﺎﻟﻲ ﻓﺈن ﻗﯿﻤﺔ : Kp ) ( ) (P 2 ) P(NO2 = Kp ) (N 2 O4 )( 0.2815 )( 0.5625 2 = Kp K p = 0.141 ﻣﺜﺎﻝ )(٣٣ إذا ﻛﺎن اﻟﻀﻐﻂ اﻟﻜﻠﻲ ﻟﻠﺘﻔﺎﻋﻞ اﻟﺘﺎﻟﻲ ﻋﻨﺪ اﻹﺗﺰان ھﻮ ) (2.04 atmﻋﻨﺪ ): (760 ºC )→ 2CO(g CO 2 (g) + C(s) ← اﺣﺴﺐ Kpﻟﻠﺘﻔﺎﻋﻞ إذا ﻛﺎن اﻟﻀﻐﻂ اﻟﺠﺰﺋﻲ ﻷول أﻛﺴﯿﺪ اﻟﻜﺮﺑﻮن ﻋﻨﺪ اﻹﺗﺰان (1.79 ).atm ﺍﳊﻞ ﻓﻲ ھﺬه اﻟﻤﺴﺄﻟﺔ ﻋﺮف اﻟﻀﻐﻂ اﻟﺠﺰﺋﻲ ﻷﺣﺪ اﻟﻐﺎزﯾﻦ واﻟﻀﻐﻂ اﻟﻜﻠﻲ ﻟﻠﻐﺎزﯾﻦ وﺑﺎﻟﺘﺎﻟﻲ ﻟﺤﺴﺎب اﻟﻀﻐﻂ اﻟﺠﺰﺋﻲ ﻟﻐﺎز ﺛﺎﻧﻲ أﻛﺴﯿﺪ اﻟﻜﺮﺑﻮن ﻧﺘﺒﻊ اﻟﻌﻼﻗﺔ : 1221 PDF created with FinePrint pdfFactory Pro trial version http://www.pdffactory.com ﺍﻟﻔﺼﻞ ﺍﻟﺴﺎﺩﺱ :ﺍﻹﺗﺰﺍﻥ ﺍﻟﻜﻴﻤﻴﺎﺋﻲ 1222 ﺇﻋﺪﺍﺩ ﺩ /ﻋﻤﺮ ﺑﻦ ﻋﺒﺪ ﺍ ﺍﳍﺰﺍﺯﻱ )Pt = P(CO2 ) + P(CO )P(CO2 ) = Pt - P(CO P(CO2 ) = 2.04 - 1.79 ⇒ P(CO2 ) = 0.25 atm وﺑﺎﻟﺘﺎﻟﻲ ﻟﺤﺴﺎب : Kp 2 ) (P )(CO ) P(CO 2 = Kp (1.79) 2 = Kp )(0.25 K p = 12.82 ﻣﺜﺎﻝ )(٣٤ ﻓﻲ ﻧﻈﺎم ﻣﻦ ﻣﺘﻮازن اﻟﮭﯿﺪروﺟﯿﻦ واﻟﻨﯿﺘﺮوﺟﯿﻦ واﻟﻨﺸﺎدر : )→ 2NH3 (g 3H 2 (g) + N 2 (g) ← وﺟﺪ أن ﻛﻤﯿﺎت اﻟﻤﻮاد ﻓﯿﮫ ھﻲ : n H 2 = 1.155 × 10 -2 mol n N 2 = 8.3 × 10 -3 mol n NH = 2.726 × 10 -5 mol 3 ﻓﺈذا ﻋﻠﻤﺖ أن ﺿﻐﻂ ھﺬا اﻟﻨﻈﺎم ﯾﺴﺎوي ) (1.26 atmﻋﻨﺪ ) (500 ºCﻓﻲ ﺣﺠﻢ )، (1 L ﻓﺎﺣﺴﺐ ﻛﻼً ﻣﻦ Kp, Kc 1222 PDF created with FinePrint pdfFactory Pro trial version http://www.pdffactory.com ﺍﻟﻔﺼﻞ ﺍﻟﺴﺎﺩﺱ :ﺍﻹﺗﺰﺍﻥ ﺍﻟﻜﻴﻤﻴﺎﺋﻲ 1223 ﺇﻋﺪﺍﺩ ﺩ /ﻋﻤﺮ ﺑﻦ ﻋﺒﺪ ﺍ ﺍﳍﺰﺍﺯﻱ ﺍﳊﻞ ﺗﺮاﻛﯿﺰ ھﺬه اﻟﻤﻮاد ﻋﻨﺪ اﻹﺗﺰان ھﻲ ﻧﻔﺴﮭﺎ ﻋﺪد اﻟﻤﻮﻻت ﺑﺴﺒﺐ أن ﺣﺠﻢ اﻟﻮﻋﺎء ﯾﺴﺎوي ﻟﺘﺮ واﺣﺪ ،وﺑﺎﻟﺘﺎﻟﻲ ﻓﺈن : [ H 2 ] = 1.155 × 10-2 M -3 [ N 2 ] = 8.3 × 10 M -5 [ NH3 ] = 2.726 × 10 M وﻟﺤﺴﺎب : Kc ] [ NH3 3 ] [ H2 ] [ N2 2 = Kc ) ( 2.726 × 10 = ) (1.155 × 10 ) × ( 8.3 × 10 -5 2 -2 3 -3 Kc K c = 0.058 وﻧﺤﺴﺐ Kpﻣﻦ اﻟﻌﻼﻗﺔ : K p = K c (RT) Δn K p = 0.058(0.082 × 773) -2 0.058 2 ) × 773 ( 0.082 = Kp 0.058 4017.78 K p = 1.44 × 10 -5 = Kp 1223 PDF created with FinePrint pdfFactory Pro trial version http://www.pdffactory.com
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