ExoLivre%20U1B .pdf



Nom original: ExoLivre%20U1B.pdfTitre: Microsoft Word - ExoLivre U1B.docAuteur: motec

Ce document au format PDF 1.4 a été généré par PScript5.dll Version 5.2 / Acrobat Distiller 7.0 (Windows), et a été envoyé sur fichier-pdf.fr le 15/11/2015 à 22:06, depuis l'adresse IP 41.109.x.x. La présente page de téléchargement du fichier a été vue 285 fois.
Taille du document: 365 Ko (6 pages).
Confidentialité: fichier public


Aperçu du document


‫اﻟﺘﻄﻮرات اﻟـﺮﺗــﻴﺒﺔ‬

‫اﻟﻜﺘﺎب اﻷول‬
‫اﻟﻮﺣﺪة ‪01‬‬

‫ﺗﻄﻮر آﻤﻴﺎت ﻣﺎدة اﻟﻤﺘﻔﺎﻋﻼت واﻟﻨﻮاﺗﺞ ﺧﻼل ﺗﺤﻮل آﻴﻤﻴــﺎﺋﻲ ﻓﻲ ﻣﺤﻠﻮل ﻣﺎﺋﻲ‬

‫ﺣﻠــﻮل ﺗﻤـــﺎرﻳﻦ اﻟﻜﺘﺎب اﻟﻤﺪرﺳﻲ‬

‫‪GUEZOURI Aek – Lycée Maraval - Oran‬‬

‫اﻟﺠﺰء اﻟﺜﺎﻧﻲ )ﺣﺴﺐ اﻟﻄﺒﻌﺔ اﻟﺠﺪﻳﺪة ﻟﻠﻜﺘﺎب اﻟﻤﺪرﺳﻲ اﻟﻤﻌﺘﻤﺪة ﻣﻦ ﻃﺮف اﻟﻤﻌﻬﺪ اﻟﻮﻃﻨﻲ ﻟﻠﺒﺤﺚ ﻓﻲ اﻟﺘﺮﺑﻴﺔ (‬
‫اﻟﺘﻤﺮﻳﻦ ‪13‬‬
‫‪ – 1‬ﺟﺪول اﻟﺘﻘﺪّم ‪:‬‬
‫‪I2 (aq) +‬‬

‫‪2‬‬

‫)‪2 I – (aq‬‬

‫=‬

‫‪2 H+ (aq) +‬‬

‫)‪H2O2 (aq‬‬

‫‪+‬‬

‫ﻣﻌﺎدﻟﺔ اﻟﺘﻔﺎﻋﻞ‬

‫)‪H2O (l‬‬
‫اﻟﺘﻘﺪم‬

‫ﺣﺎﻟﺔ اﻟﺠﻤﻠﺔ‬

‫زﻳﺎدة‬

‫‪0‬‬

‫)‪n (H+‬‬

‫)‪n (H2O2‬‬

‫‪0‬‬

‫اﻟﺤﺎﻟﺔ اﻻﺑﺘﺪاﺋﻴﺔ‬

‫زﻳﺎدة‬

‫‪x‬‬

‫‪n (H+) – 2 x‬‬

‫‪n (H2O2) - x‬‬

‫‪x‬‬

‫اﻟﺤﺎﻟﺔ اﻻﻧﺘﻘﺎﻟﻴﺔ‬

‫زﻳﺎدة‬

‫‪x max‬‬

‫‪n (H+) – 2 x max‬‬

‫‪n (H2O2) - x max‬‬

‫‪x max‬‬

‫اﻟﺤﺎﻟﺔ اﻟﻨﻬﺎﺋﻴﺔ‬

‫)‪(mol‬‬

‫آﻤﻴﺔ اﻟﻤـــﺎدة‬

‫‪ – 2‬ﻣﻦ اﻟﺠﺪول ﻟﺪﻳﻨﺎ ‪ ، n(I2) = x :‬وﻣﻦ ﺟﻬﺔ أﺧﺮى ﻟﺪﻳﻨﺎ‬

‫‪ ، n(I2) = [I2] V‬وﻣﻨﻪ ]‪x = 0,2 [I2‬‬

‫ﺑﻮاﺳﻄﺔ هﺬﻩ اﻟﻌﻼﻗﺔ اﻷﺧﻴﺮة ﻧﺤﺴﺐ ﻗﻴﻢ اﻟﺘﻘﺪم ﺑﺎﺳﺘﻌﻤﺎل اﻟﺘﺮاآﻴﺰ اﻟﻤﻮﻟﻴﺔ ﻟﺜﻨﺎﺋﻲ اﻟﻴﻮد اﻟﻤﺴﺠﻠﺔ ﻋﻠﻰ اﻟﺠﺪول ‪.‬‬
‫‪120‬‬

‫‪60‬‬

‫‪40‬‬

‫‪30‬‬

‫‪20‬‬

‫‪16‬‬

‫‪12‬‬

‫‪8‬‬

‫‪6‬‬

‫‪4‬‬

‫‪2‬‬

‫‪1‬‬

‫‪0‬‬

‫‪1,74‬‬

‫‪1,74‬‬

‫‪1,70‬‬

‫‪1,64‬‬

‫‪1,54‬‬

‫‪1,46‬‬

‫‪1,32‬‬

‫‪1,10‬‬

‫‪0,920‬‬

‫‪0,74‬‬

‫‪0,42‬‬

‫‪0,22‬‬

‫‪0‬‬

‫‪T10‬‬

‫اﻟﺒﻴﺎن ) ‪ : x = f (t‬اﻧﻈﺮ ﻟﻠﺸﻜﻞ ‪.‬‬
‫‪-3‬‬

‫أ( اﻟﺴﺮﻋﺔ اﻟﺤﺠﻤﻴﺔ ﻟﻠﺘﻔﺎﻋﻞ هﻲ ﺳﺮﻋﺔ اﻟﺘﻔﺎﻋﻞ ﻣﻦ أﺟﻞ ﺣﺠﻢ اﻟﻤﺰﻳﺞ اﻟﻤﺘﻔﺎﻋﻞ ‪.‬‬

‫‪1 dx‬‬
‫‪V dt‬‬

‫)‪t (mn‬‬
‫)‪x (mmol‬‬

‫) ‪x(mmol‬‬
‫‪T0‬‬

‫=‪v‬‬

‫اﻟﺴﺮﻋﺔ اﻟﺤﺠﻤﻴﺔ ﻟﻠﺘﻔﺎﻋﻞ ﻋﻨﺪ اﻟﻠﺤﻈﺔ ‪: t = 0‬‬
‫ﻧﺤﺴﺐ ﻣﻴﻞ اﻟﻤﻤﺎس ‪ T0‬وﻧﻘﺴﻢ اﻟﻨﺘﻴﺠﺔ ﻋﻠﻰ ﺣﺠﻢ اﻟﻤﺰﻳﺞ ‪. V‬‬
‫‪= 3,2 × 10 − 4 mol.mn −1‬‬

‫‪−3‬‬

‫‪⎛ dx ⎞ = 1,6 × 10‬‬
‫⎟ ⎜‬
‫‪5‬‬
‫‪⎝ dt ⎠ 0‬‬

‫‪1‬‬
‫ﻋﻨﺪ ‪× 3,2 × 10 − 4 = 1,6 × 10 −3 mol. L−1 .mn −1 : t = 10 mn‬‬
‫‪0,2‬‬

‫‪0,2‬‬

‫= ‪v0‬‬

‫)‪t(mn‬‬

‫‪40‬‬

‫‪−3‬‬
‫‪⎛ dx ⎞ = 0,8 × 10 = 5,3 × 10 −5 mol.mn −1‬‬
‫⎟ ⎜‬
‫‪15‬‬
‫‪⎝ dt ⎠10‬‬

‫‪1‬‬
‫‪× 5,3 × 10 −5 = 2,6 × 10 − 4 mol. L−1 .mn −1‬‬
‫‪0,2‬‬

‫= ‪v10‬‬

‫ب( ﻧﻼﺣﻆ ﻓﻲ اﻟﺠﺪول أن اﻟﺘﺮآﻴﺰ اﻟﻤﻮﻟﻲ ﻟﺜﻨﺎﺋﻲ اﻟﻴﻮد ﻳﺼﺒﺢ ﺛﺎﺑﺘﺎ اﺑﺘﺪاء ﻣﻦ ‪ ، t = 60 s‬وﺑﺎﻟﺘﺎﻟﻲ ‪ x‬آﺬﻟﻚ ‪.‬‬
‫‪1‬‬

‫‪0 5 10‬‬

‫ﻟﻮ رﺳﻤﻨﺎ اﻟﻤﻤﺎس ﻟﻠﺒﻴﺎن ) ‪ x = f (t‬ﻟﻜﺎن أﻓﻘﻴﺎ ‪ ،‬أي ﻣﻴﻠﻪ ﻣﻌﺪوم ‪ ،‬وﻣﻨﻪ ‪. v100 = 0‬‬
‫ﺟـ( ﻧﻼﺣﻆ أن ﺳﺮﻋﺔ اﻟﺘﻔﺎﻋﻞ ﺗﺘﻨﺎﻗﺺ ﺧﻼل اﻟﺰﻣﻦ ‪ ،‬واﻟﺴﺒﺐ هﻮ ﺗﻨﺎﻗﺺ ﺗﺮاآﻴﺰ اﻟﻤﺘﻔﺎﻋﻼت ‪.‬‬
‫اﻟﺘﻤﺮﻳﻦ ‪14‬‬
‫‪ – 1‬اﻟﺜﻨﺎﺋﻴﺘﺎن هﻤﺎ ‪:‬‬

‫)‪(aq‬‬

‫‪2+‬‬

‫‪/ Mn‬‬

‫)‪(aq‬‬

‫–‬

‫‪ MnO4‬و )‪CO2 (aq) / H2C2O4 (aq‬‬

‫اﻟﻤﻌﺎدﻟﺘﺎن اﻟﻨﺼﻔﻴﺘﺎن اﻹﻟﻜﺘﺮوﻧﻴﺘﺎن هﻤﺎ ‪:‬‬

‫) )‪2 × ( MnO4– (aq) + 5 e– + 8 H+(aq) = Mn2+(aq)+ 4 H2O(l‬‬
‫)–‪5 × (H2C2O4 (aq) = 2 CO2 (aq) + 2 H+ (aq) + 2 e‬‬
‫ﻣﻌﺎدﻟﺔ اﻷآﺴﺪة – ارﺟــﺎع ‪:‬‬

‫)‪2 MnO4 (aq) + 5 H2C2O4 (aq) + 6 H+ (aq) → 2 Mn2+ (aq) + 10 CO2 (aq) + 8 H2O (l‬‬
‫‪ – 2‬آﻤﻴﺔ ﻣﺎدة ﺷﺎردة اﻟﺒﺮﻣﻨﻐﻨﺎت ‪:‬‬

‫‪n (MnO4– ) = C1 V1 = 10 –3 × 0,05 = 5 × 10 –5 mol‬‬

‫آﻤﻴﺔ ﻣﺎدة ﺷﺎردة ﺣﻤﺾ اﻷآﺴﺎﻟﻴﻚ ‪n (H2C2O4 ) = C2 V2 = 10 –1 × 0,05 = 5 × 10 –3 mol :‬‬
‫‪ – 3‬ﻧﺤﺴﺐ آﻤﻴﺔ ﻣﺎدة ﺣﻤﺾ اﻷآﺴﺎﻟﻴﻚ اﻟﺘﻲ ﺗﻜﻔﻲ ﻟﺘﻔﺎﻋﻞ آﻞ آﻤﻴﺔ ﻣﺎدة اﻟﺒﺮﻣﻨﻐﻨﺎت اﻟﻤﻌﻄﺎة ‪:‬‬

‫)‪6 H+ (aq) = 2 Mn2+ (aq) + 10 CO2 (aq) + 8 H2O (l‬‬

‫‪5 H2C2O4 (aq) +‬‬

‫‪+‬‬

‫)–‪n (MnO4‬‬
‫–‬
‫‪ n (MnO4 ) - 2 x max‬ﻧﻬﺎﻳﺔ اﻟﺘﻔﺎﻋﻞ‬

‫)‪n (H2C2O4‬‬
‫‪n (H2C2O4) - 5 x max‬‬
‫ﻋﻨﺪ ﻧﻬﺎﻳﺔ اﻟﺘﻔﺎﻋﻞ ﻳﻜﻮن ﻟﺪﻳﻨﺎ ‪:‬‬

‫‪n (MnO4 –) - 2 x max = 0‬‬

‫)‪2 MnO4 (aq‬‬

‫‪t=0‬‬

‫)‪(1‬‬
‫)‪(2‬‬

‫‪n (H2C2O4) - 5 x max = 0‬‬
‫ﺑﺎﺳﺘﺨﺮاج ﻋﺒﺎرة ‪ x max‬ﻣﻦ )‪ (1‬وﺗﻌﻮﻳﻀﻬﺎ ﻓﻲ )‪ ، (2‬ﻧﺠﺪ ‪:‬‬

‫‪5‬‬
‫‪n (MnO4–) = 2,5 × 5 × 10 –5 = 12,5 × 10 –5 mol‬‬
‫‪2‬‬

‫= )‪، n (H2C2O4‬‬

‫وﻧﺤﻦ ﻟﺪﻳﻨﺎ آﻤﻴﺔ أآﺒﺮ ﻣﻦ هﺬﻩ )‪(5 × 10 –3 mol‬‬

‫‪E‬‬

‫)‪[Mn2+] (mol. L–1‬‬
‫‪•H‬‬

‫•‬

‫إذن ‪ ،‬ﻧﻌﻢ اﻟﻜﻤﻴﺔ آﺎﻓﻴﺔ ﻟﺰوال ﻟﻮن ﺑﺮﻣﻨﻐﻨﺎت اﻟﺒﻮﺗﺎﺳﻴﻮم ‪.‬‬

‫•‬

‫‪ - 4‬ﻧﺤﺴﺐ ﻣﻴﻞ آﻞ ﻣﻤﺎس ﻟﻠﺒﻴﺎن ‪ ،‬واﻟﺬي ﻳﻤﺜﻞ اﻟﺴﺮﻋﺔ‬

‫‪G‬‬
‫•‬

‫•‬

‫‪F‬‬

‫اﻟﺤﺠﻤﻴﺔ ﻟﺘﺸﻜﻞ ﺷﻮارد اﻟﻤﻨﻐﻨﻴﺰ ‪:‬‬
‫ﻓﻲ اﻟﻠﺤﻈﺔ ‪: t1 = 80 s‬‬

‫‪d [Mn 2+ ] AB 0,5 × 10 −4‬‬
‫=‬
‫=‬
‫‪= 3,38 × 10 −7‬‬
‫‪dt‬‬
‫‪CD‬‬
‫‪40 × 3,7‬‬
‫•‪K‬‬

‫‪v1 = 3,38 × 10 −7 mol .L−1 .s −1‬‬
‫ﻓﻲ اﻟﻠﺤﻈﺔ ‪: t2 = 200 s‬‬

‫‪d [Mn 2+ ] HI 3,75 × 10 −4‬‬
‫=‬
‫=‬
‫‪= 6,25 × 10 −6‬‬
‫‪dt‬‬
‫‪JI‬‬
‫‪60‬‬

‫•‬

‫‪v2 = 6,25 × 10 −6 mol .L−1 .s −1‬‬

‫•‬

‫‪A‬‬

‫‪B‬‬

‫)‪2 t (s‬‬

‫‪I‬‬

‫•‬

‫•‬

‫‪J‬‬

‫• ‪10 – 4‬‬
‫•‬

‫‪C‬‬

‫•‬

‫•‬

‫‪40‬‬

‫ﻓﻲ اﻟﻠﺤﻈﺔ ‪: t3 = 280 s‬‬
‫‪d [Mn 2+ ] EF 0,85 × 10 −4‬‬
‫=‬
‫=‬
‫‪= 4,72 × 10 −7‬‬
‫‪dt‬‬
‫‪GF‬‬
‫‪40 × 4,5‬‬

‫‪v3 = 4,72 × 10 −6 mol .L−1 .s −1‬‬
‫اﻻﺳﺘﻨﺘﺎج ‪ :‬ﻧﻼﺣﻆ أن ﺳﺮﻋﺔ ﺗﺸﻜﻞ ﺷﺎردة اﻟﻤﻨﻐﻨﻴﺰ ﺗﺰداد اﺑﺘﺪاء ﻣﻦ اﻟﻠﺤﻈﺔ ‪ ، t = 0‬ﺛﻢ ﺗﻤﺮ ﺑﻘﻴﻤﺔ ﻋﻈﻤﻰ ﺛﻢ ﺗﺘﻨﺎﻗﺺ ﺑﻌﺪ ذﻟﻚ ‪.‬‬
‫ﺗﻤﺮ ﺑﺎﻟﻘﻴﻤﺔ اﻟﻌﻈﻤﻰ ﻓﻲ ﻧﻘﻄﺔ اﻧﻌﻄﺎف اﻟﺒﻴﺎن )‪ ، . (K‬وهﺬﻩ اﻟﻘﻴﻤﺔ هﻲ ‪. v2‬‬
‫ﻣﻼﺣﻈﺔ ‪ :‬ﻟﻮ اﺳﺘﻌﻤﻠﻨﺎ ﺑﺪل ﺑﺮﻣﻨﻐﻨﺎت اﻟﺒﻮﺗﺎﺳﻴﻮم ﻣﺜﻼ ﺛﻨﺎﺋﻲ آﺮوﻣﺎت اﻟﺒﻮﺗﺎﺳﻴﻮم وﻣﺜﻠﻨﺎ اﻟﺒﻴﺎن )‪ [Cr3+] = f (t‬ﻟﻮﺟﺪﻧﺎ ﺑﻴﺎﻧﺎ ﺑﺎﻟﺸﻜﻞ اﻟﺘﺎﻟﻲ‬
‫إذن ﻣﺎ هﻮ اﻟﺴﺒﺐ ؟‬

‫]‪[Cr3+‬‬

‫ﻟﻤﻌﺮﻓﺔ اﻟﺴﺒﺐ ﻧﺠﺮي اﻟﺘﺠﺮﺑﺔ اﻟﺘﺎﻟﻴﺔ ‪ :‬ﻧﻜﻮّن ﻣﺰﻳﺠﻴﻦ ﻣﺘﻤﺎﺛﻠﻴﻦ ﻓﻲ اﻟﺘﺮاآﻴﺰ اﻟﻤﻮﻟﻴﺔ وﻓﻲ اﻟﺤﺠﻮم ﻣﻦ‬
‫ﺑﺮﻣﻨﻐﻨﺎت اﻟﺒﻮﺗﺎﺳﻴﻮم وﺣﻤﺾ اﻷآﺴﺎﻟﻴﻚ وﻧﻀﻴﻒ ﻷﺣﺪهﻤﺎ ﻓﻘﻂ ﺑﻌﺾ اﻟﻤﻠﻴﻤﺘﺮات اﻟﻤﻜﻌﺒﺔ ﻣﻦ ﻣﺤﻠﻮل‬
‫آﻠﻮر اﻟﻤﻨﻐﻨﻴﺰ ))‪ . ( Mn2+ (aq) , 2 Cl– (aq‬ﻧﻼﺣﻆ أن اﻟﻤﺰﻳﺞ اﻟﺬي أﺿﻔﻨﺎ ﻟﻪ آﻠﻮر اﻟﻤﻨﻐﻨﻴﺰ‬

‫‪t‬‬

‫ﻳﻜﻮن ﻓﻴﻪ اﻟﺘﻔﺎﻋﻞ أﺳﺮع ‪ ،‬ﻣﻌﻨﻰ هﺬا أن ﺷﻮارد اﻟﻤﻨﻐﻨﻴﺰ ﻣﺤﻔّﺰ ﻟﻬﺬا اﻟﺘﻔﺎﻋﻞ ‪.‬‬
‫إذن ﻣﺎذا ﻳﺤﺪث ﻟﻤﺎ ﻧﻤﺰج ﺑﺮﻣﻨﻐﻨﺎت اﻟﺒﻮﺗﺎﺳﻴﻮم وﺣﻤﺾ اﻷآﺴﺎﻟﻴﻚ ؟‬

‫ﺗﺴﻤﻰ هﺬﻩ اﻟﻈﺎهﺮة اﻟﺘﺤﻔﻴﺰ اﻟﺬاﺗﻲ ‪ ،‬أي أن أﺣﺪ ﻧﻮاﺗﺞ اﻟﺘﻔﺎﻋﻞ ﻳﻠﻌﺐ دور اﻟﻤﺤﻔﺰ آﺬﻟﻚ ‪ ،‬وﻓﻲ ﻣﺜﺎﻟﻨﺎ هﺬا ﺷﻮارد اﻟﻤﻨﻐﻨﻴﺰ ﺗﻠﻌﺐ‬
‫هﺬا اﻟﺪور ‪.‬‬
‫ﻓﻲ ﺑﺪاﻳﺔ اﻟﺘﻔﺎﻋﻞ ﻳﻜﻮن اﻟﺘﺮآﻴﺰ اﻟﻤﻮﻟﻲ ﻟﺸﻮارد اﻟﻤﻨﻐﻨﻴﺰ ﺿﻌﻴﻔﺎ ‪ ،‬ﻟﻬﺬا ﺗﻜﻮن ﺳﺮﻋﺔ ﺗﺸﻜﻞ اﻟﻤﻨﻐﻨﻴﺰ ﺿﻌﻴﻔﺔ )‪ 120‬ﺛﺎﻧﻴﺔ اﻷوﻟﻰ( ‪.‬‬
‫ﻋﻨﺪﻣﺎ ﻳﺘﺰاﻳﺪ اﻟﺘﺮآﻴﺰ اﻟﻤﻮﻟﻲ ﻟﺸﻮارد اﻟﻤﻨﻐﻨﻴﺰ ﻓﻲ اﻟﻤﺰﻳﺞ ﻳﺰداد اﻟﺘﺤﻔﻴﺰ ‪ ،‬وﺑﺎﻟﺘﺎﻟﻲ ﺗﺰداد ﺳﺮﻋﺔ ﺗﺸﻜﻞ اﻟﻤﻨﻐﻨﻴﺰ وﺗﻤﺮ ﺑﻘﻴﻤﺔ ﻋﻈﻤﻰ ‪ ،‬وذﻟﻚ‬
‫ﻋﻨﺪ اﻟﻠﺤﻈﺔ ‪. 200 s‬‬
‫ﺑﻌﺪ اﻟﻠﺤﻈﺔ ‪ 240 s‬ﺗﺘﻨﺎﻗﺺ اﻟﺴﺮﻋﺔ رﻏﻢ إزدﻳﺎد اﻟﺘﺮآﻴﺰ اﻟﻤﻮﻟﻲ ﻟﺸﻮارد اﻟﻤﻨﻐﻨﻴﺰ ‪ ،‬ﻷن اﻟﺘﺮاآﻴﺰ اﻟﻤﻮﻟﻴﺔ ﻟﻠﻤﺘﻔﺎﻋﻼت أﺻﺒﺤﺖ ﺿﻌﻴﻔﺔ وهﺬا‬
‫ﻳﺆﺛﺮ ﻋﻠﻰ ﺳﺮﻋﺔ ﺗﺸﻜﻞ اﻟﻤﻨﻐﻨﻴﺰ ﺳﻠﺒﺎ ‪.‬‬
‫اﻟﺘﻤﺮﻳﻦ ‪15‬‬
‫‪ – 1‬ﻣﻌﺎدﻟﺔ ﺗﻔﺎﻋﻞ اﻟﻤﻌﺎﻳﺮة ‪:‬‬
‫اﻟﺜﻨﺎﺋﻴﺘﺎن هﻤﺎ ‪ I2 / I – :‬و‬

‫–‪2‬‬

‫‪/ S2O3‬‬

‫–‪2‬‬

‫‪S4O6‬‬
‫– ‪I2 + 2 e – = 2 I‬‬

‫اﻟﻤﻌﺎدﻟﺘﺎن اﻟﻨﺼﻔﻴﺘﺎن اﻹﻟﻜﺘﺮوﻧﻴﺘﺎن هﻤﺎ ‪:‬‬

‫– ‪S4O62 – + 2 e‬‬
‫ﻣﻌﺎدﻟﺔ اﻷآﺴﺪة – إرﺟــﺎع ‪:‬‬
‫‪-2‬‬

‫–‪2 I‬‬

‫‪+‬‬

‫– ‪S4O62‬‬

‫‪0‬‬

‫‪0‬‬

‫‪2 xE‬‬

‫‪xE‬‬

‫= – ‪2 S2O32‬‬

‫– ‪I2 + 2 S2O32 – = S4O62 – + 2 I‬‬
‫→‬

‫– ‪2 S2O32‬‬

‫‪I2‬‬

‫‪+‬‬

‫)– ‪n (S2O32‬‬

‫)‪n (I2‬‬

‫‪n (S2O32 –) - 2 x E‬‬

‫‪n (I2) - x E‬‬

‫ﻋﻨﺪ اﻟﺘﻜﺎﻓﺆ ﻳﻜﻮن ﻟﺪﻳﻨﺎ ‪:‬‬

‫‪n (S2O32 –) – 2 x E = 0‬‬
‫‪n (I2) - x E = 0‬‬

‫)‪(1‬‬
‫)‪(2‬‬

‫‪3‬‬

‫‪t=0‬‬
‫اﻟﺘﻜﺎﻓﺆ‬

‫ﺑﺤﺬف ‪ x E‬ﺑﻴﻦ اﻟﻌﻼﻗﺘﻴﻦ )‪ (1‬و )‪ (2‬ﻧﺠﺪ ‪ ، n (I2) = 0,5 n (S2O32–) :‬وﺑﺎﻟﺘﺎﻟﻲ ‪:‬‬

‫'‪n (I2) = 0,5 C' V‬‬

‫‪ – 3‬اﻟﺮﺳﻢ اﻟﺒﻴـﺎﻧﻲ )‪n (I2) = f (t‬‬

‫)‪n(I2) ×10 –5(mol‬‬

‫‪A‬‬
‫•‬

‫•‬

‫•‪C‬‬

‫‪•B‬‬

‫‪1‬‬

‫)‪t (mn‬‬
‫‪-4‬‬

‫‪5‬‬

‫‪: t2 = 20 mn‬‬

‫أ( اﻟﺴﺮﻋﺔ اﻟﺤﺠﻤﻴﺔ اﻟﻤﺘﻮﺳﻄﺔ ﻟﺘﺸﻜﻞ ﺛﻨﺎﺋﻲ اﻟﻴﻮد ﺑﻴﻦ ‪ t1 = 10 mn‬و‬

‫ﻓﻲ اﻟﻠﺤﻈﺔ ‪ t1‬ﺗﺸﻜﻞ ‪ 3,4 × 10 – 5 mol‬ﻣﻦ ﺛﻨﺎﺋﻲ اﻟﻴﻮد ﻓﻲ ﺣﺠﻢ ﻗﺪرﻩ ‪ . 10 mL‬أﻣﺎ ﻓﻲ اﻟﻤﺰﻳﺞ اﻻﺑﺘﺪاﺋﻲ ‪ 100 mL‬ﺗﺸﻜﻠﺖ‬

‫اﻟﻘﻴﻤﺔ ‪n1 = 3,4 × 10 – 5 × 10 = 3,4 × 10 – 4 mol‬‬
‫ﻓﻲ اﻟﻠﺤﻈﺔ ‪ t2‬ﺗﺸﻜﻞ ‪ 5,2 × 10 – 5 mol‬ﻣﻦ ﺛﻨﺎﺋﻲ اﻟﻴﻮد ﻓﻲ ﺣﺠﻢ ﻗﺪرﻩ ‪ . 10 mL‬أﻣﺎ ﻓﻲ اﻟﻤﺰﻳﺞ اﻻﺑﺘﺪاﺋﻲ ‪ 100 mL‬ﺗﺸﻜﻠﺖ‬
‫اﻟﻘﻴﻤﺔ ‪) n2 = 5,2 × 10 – 5 × 10 = 5,2 × 10 – 4 mol‬ﻻ ﺗﻨﺲ أن إﺿﺎﻓﺔ اﻟﻤﺎء ﻻ ﻳﻐﻴّﺮ ﻋﺪد اﻟﻤﻮﻻت( – أﺿﻔﻨﺎ اﻟﻤﺎء ﻣﻦ أﺟﻞ‬
‫اﻟﺴّﻘﻲ ﻓﻘﻂ ‪.‬‬
‫‪1 (n 2 − n1 ) 1 (5,2 − 3,4) × 10 −4‬‬
‫=‬
‫‪= 1,8 × 10 − 4 mol.L−1 .mn −1‬‬
‫‪V‬‬
‫‪Δt‬‬
‫‪0,1‬‬
‫‪10‬‬

‫= ‪vm‬‬

‫ب( اﻟﺴﺮﻋﺔ اﻟﺤﺠﻤﻴﺔ اﻟﻠﺤﻈﻴﺔ ﻟﺘﺸﻜﻞ ﺛﻨﺎﺋﻲ اﻟﻴﻮد ﻓﻲ اﻟﻠﺤﻈﺔ ‪: t = 15 mn‬‬

‫‪d n( I 2 ) AB 5,3 × 10 −4‬‬
‫=‬
‫=‬
‫‪= 1,76 × 10 −5‬‬
‫‪dt‬‬
‫‪CB‬‬
‫‪6×5‬‬
‫‪1 d n( I 2 ) 1‬‬
‫=‬
‫‪× 1,76 × 10 −5 = 1,76 × 10 − 4 mol.L−1 .mn −1‬‬
‫‪V dt‬‬
‫‪0,1‬‬

‫= ‪v15‬‬

‫أ( ﻳﺤﺪث اﻟﺘﻔﺎﻋﻞ ﺑﻴﻦ اﻟﺜﻨﺎﺋﻴﺘﻴﻦ ‪ I2 / I – :‬و ‪S2O82 - / SO42-‬‬

‫‪-5‬‬

‫اﻟﻤﻌﺎدﻟﺘﺎن اﻟﻨﺼﻔﻴﺘــﺎن ‪:‬‬

‫–‪2I – = I2 + 2 e‬‬
‫ ‪S2O82 - + 2 e– = 2 SO42‬‬‫ﻣﻌﺎدﻟﺔ اﻷآﺴﺪة – إرﺟــﺎع ‪2 I –(aq) + S2O82 –(aq) = I2(aq) + 2 SO42 –(aq) :‬‬
‫ب(‬
‫)‪(aq‬‬

‫–‪2‬‬

‫‪2 SO4‬‬
‫‪0‬‬

‫‪2x‬‬

‫‪+‬‬

‫)‪I2(aq‬‬

‫=‬

‫)‪(aq‬‬

‫‪0‬‬

‫‪x‬‬

‫‪4‬‬

‫–‬

‫‪+ 2I‬‬

‫)‪(aq‬‬

‫–‪2‬‬

‫‪S2O8‬‬

‫)–‪n0 (I‬‬

‫)– ‪n0 (S2O82‬‬

‫‪t=0‬‬

‫‪n0 (I–) - 2 x‬‬

‫‪n0 (S2O82-) - x‬‬

‫‪t‬‬

‫ﻟﺪﻳﻨﺎ ﻓﻲ اﻟﻠﺤﻈﺔ ‪ t‬آﻤﻴﺔ ﻣﺎدة ‪ S2O82-‬ﻓﻲ اﻟﻤﺰﻳﺞ هﻲ ‪:‬‬

‫‪n (S2O82–) = n0 (S2O82-) - x‬‬

‫ﻧﺸﺘﻖ ﻃﺮﻓﻲ هﺬﻩ اﻟﻤﻌﺎدﻟﺔ ﺑﺎﻟﻨﺴﺒﺔ ﻟﻠﺰﻣﻦ ‪:‬‬

‫) ‪d n( S 2 O82−‬‬
‫‪d n0 ( S 2 O82− ) dx‬‬
‫=‬
‫‪−‬‬
‫‪dt‬‬
‫‪dt‬‬
‫‪dt‬‬
‫) ‪d n( S 2 O82−‬‬
‫‪dx‬‬
‫‪=−‬‬
‫وﺑﺎﻟﺘﺎﻟﻲ ‪:‬‬
‫‪dt‬‬
‫‪dt‬‬

‫) ‪d n0 ( S 2 O82−‬‬
‫ﻋﺒﺎرة ﻋﻦ ﺛﺎﺑﺖ ‪ ،‬إذن ﻣﺸﺘﻘﻪ ﺑﺎﻟﻨﺴﺒﺔ ﻷي ﻣﺘﻐﻴﺮ ﻣﻌﺪوم ‪.‬‬
‫‪ ،‬وﻟﺪﻳﻨﺎ‬
‫‪dt‬‬

‫‪ ،‬وﻣﻨﻪ ﺳﺮﻋﺔ اﺧﺘﻔﺎء ‪ S2O82-‬هﻲ ﺳﺮﻋﺔ اﻟﺘﻔﺎﻋﻞ ﺑﺎﻟﻘﻴﻤﺔ اﻟﻤﻄﻠﻘﺔ ‪.‬‬

‫ﺟـ( ﻧﻼﺣﻆ ﻓﻲ اﻟﺴﺆال )ب( أن )‪ ، x = n (I2‬إذن اﻟﺒﻴﺎن )‪ n (I2) = f (t‬هﻮ ﻧﻔﺲ اﻟﺒﻴﺎن )‪. x = f (t‬‬
‫ﺳﺮﻋﺔ اﻟﺘﻔﺎﻋﻞ ﻋﻨﺪ اﻟﻠﺤﻈﺔ ‪ t = 15 mn‬هﻲ ﻣﻴﻞ اﻟﻤﻤﺎس ﻟﻠﺒﻴﺎن ﻓﻲ اﻟﻨﻘﻄﺔ اﻟﺘﻲ ﻓﺎﺻﻠﺘﻬﺎ ‪. 15 mn‬‬

‫) ‪dx d n (I 2‬‬
‫=‬
‫‪= 1, 76 × 10−5 mol .L −1.mn −1‬‬
‫‪dt‬‬
‫‪dt‬‬

‫= ‪v‬‬

‫اﻟﺘﻤﺮﻳﻦ ‪16‬‬

‫‪ – 1‬اﻟﺜﻨﺎﺋﻴﺘﺎن هﻤــﺎ ‪Zn2+ / Zn :‬‬

‫و ‪H+ / H2‬‬

‫اﻟﻤﻌﺎدﻟﺘﺎن اﻟﻨﺼﻔﻴﺘﺎن ‪Zn (s) = Zn2+ (aq) + 2 e– :‬‬
‫)‪2 H+ (aq) + 2 e– = H2 (g‬‬
‫ﻣﻌﺎدﻟﺔ اﻷآﺴﺪة – ارﺟــﺎع ‪Zn (s) + 2 H+ = Zn2+ (aq) + H2 (g) :‬‬
‫‪ – 2‬ﺟﺪول اﻟﺘﻘﺪّم ‪:‬‬

‫)‪H2 (g‬‬

‫‪Zn2+ (aq) +‬‬

‫)‪2 H+ (q‬‬

‫→‬

‫ﻣﻌﺎدﻟﺔ اﻟﺘﻔﺎﻋﻞ‬

‫‪Zn (s) +‬‬
‫اﻟﺘﻘﺪم‬

‫ﺣﺎﻟﺔ اﻟﺠﻤﻠﺔ‬

‫‪0‬‬

‫‪0‬‬

‫)‪n (H+‬‬

‫)‪n (Zn‬‬

‫‪0‬‬

‫اﻻﺑﺘﺪاﺋﻴﺔ‬

‫‪x‬‬

‫‪x‬‬

‫‪n (H+) – 2 x‬‬

‫‪n (Zn) – x‬‬

‫‪x‬‬

‫اﻻﻧﺘﻘﺎﻟﻴﺔ‬

‫‪x max‬‬

‫‪x max‬‬

‫‪n (H+) – 2 x max‬‬

‫‪n (Zn) – x max‬‬

‫‪x max‬‬

‫اﻟﻨﻬﺎﺋﻴﺔ‬

‫آﻤﻴﺔ اﻟﻤـــﺎدة‬

‫)‪(mol‬‬

‫ﺗﻌﻴﻴﻦ اﻟﻤﺘﻔﺎﻋﻞ اﻟﻤﺤ ّﺪ ‪:‬‬

‫‪2,3‬‬
‫‪m‬‬
‫=‬
‫‪= 3,5 × 10 − 2 mol‬‬
‫‪M 65,4‬‬

‫= )‪n( Zn‬‬

‫‪n( H + ) = CA V = 0,2 × 0,1 = 2,0 × 10 −2 mol‬‬
‫اﻟﻘﻴﻤﺔ اﻷﺻﻐﺮ ﻟـ ‪ x‬ﻓﻲ ﺣﻞ اﻟﻤﻌﺎدﻟﺘﻴﻦ اﻟﺘﺎﻟﻴﺘﻴﻦ ﺗﻮاﻓﻖ اﻟﻤﺘﻔﺎﻋﻞ اﻟﻤﺤ ّﺪ ‪:‬‬

‫‪3,5 × 10 −2 − x = 0 ⇒ x = 3,5 × 10 −2 mol‬‬
‫‪2,0 × 10 −2 − 2 x = 0 ⇒ x = 1,0 × 10 −2 mol‬‬
‫إذن اﻟﻤﺘﻔﺎﻋﻞ اﻟﻤﺤ ّﺪ هﻮ ﺣﻤﺾ آﻠﻮر اﻟﻬﻴﺪروﺟﻴﻦ )ﻻ ﺗﻨﺲ أن )‪. ( n(H+) = n(Cl–) = n(HCl‬‬
‫ﻣﻦ اﻟﺠﺪول ﻟﺪﻳﻨﺎ ‪ ، n (Zn2+) = x‬وﺑﺎﻟﺘﺎﻟﻲ‬

‫‪، [Zn2+] V = x‬‬

‫اﻟﻌﻼﻗﺔ اﻟﻤﻄﻠﻮﺑﺔ هﻲ ‪:‬‬

‫‪ – 3‬زﻣﻦ ﻧﺼﻒ اﻟﺘﻔﺎﻋﻞ هﻮ اﻟﻤﺪّة اﻟﻼزﻣﺔ ﻟﺒﻠﻮغ اﻟﺘﻔﺎﻋﻞ ﻧﺼﻒ ﺗﻘﺪّﻣﻪ اﻟﻨﻬﺎﺋﻲ ‪.‬‬
‫إذا آﺎن هﺬا اﻟﺘﻔﺎﻋﻞ ﺗــﺎﻣﺎ ﻳﻜﻮن هﺬا اﻟﺰﻣﻦ ﻻزﻣﺎ ﻻﺳﺘﻬﻼك ﻧﺼﻒ آﻤﻴﺔ ﻣﺎدة اﻟﻤﺘﻔﺎﻋﻞ اﻟﻤﺤ ّﺪ ‪.‬‬

‫‪5‬‬

‫]‪x = 0,1 [Zn2+‬‬

‫ﻟﺪﻳﻨﺎ‬

‫‪ ، [Zn2+] max = 0,1 mol/L‬وﻣﻨﻪ ‪. xmax = 1,0×10-2 mol‬‬

‫‪ ، xmax = 0,1 [Zn2+] max‬وﻣﻦ اﻟﺒﻴﺎن ﻟﺪﻳﻨﺎ‬

‫‪x‬‬
‫‪0,5 x max 5 × 10 −3‬‬
‫=‬
‫‪ . max = 5,00 × 10 −3 mol‬هﺬﻩ اﻟﻘﻴﻤﺔ ﺗﻮاﻓﻖ ﻋﻠﻰ اﻟﺒﻴﺎن ‪= 50 × 10 −3 = 50 mmol / L‬‬
‫‪V‬‬
‫‪0,1‬‬
‫‪2‬‬

‫ﻓﺎﺻﻠﺔ هﺬﻩ اﻟﻘﻴﻤﺔ ﻟﻠﺘﺮآﻴﺰ اﻟﻤﻮﻟﻲ ﺗﻮاﻓﻖ ﺣﻮاﻟﻲ ‪ ، t = 4,9 mn‬أي ‪) t1/ 2 = 4,5 mn‬اﻟﺸﻜﻞ – ‪(1‬‬
‫ﻣﻼﺣﻈﺔ ‪:‬‬
‫‪2+‬‬

‫آﺎن ﻣﻦ اﻟﻤﻤﻜﻦ ﺗﻘﺴﻴﻢ اﻟﺘﺮآﻴﺰ اﻟﻤﻮﻟﻲ ﻟـ ‪ Zn‬ﻋﻠﻰ ‪ 2‬واﺳﺘﻨﺘﺎج زﻣﻦ ﻧﺼﻒ اﻟﺘﻔﺎﻋﻞ ﻣﺒﺎﺷﺮة ‪ ،‬ﻟﻜﻨﻲ ﻓﺼﻠﺖ ذﻟﻚ ﻟﻬﺪف ﻣﻨﻬﺠﻲ ‪.‬‬
‫)‪[Zn2+] (mmol.L-1‬‬

‫‪ – 4‬ﺗﺮآﻴﺐ اﻟﻮﺳﻂ اﻟﺘﻔﺎﻋﻠﻲ ﻋﻨﺪ ‪: t 1/2 = 4,9 mn‬‬

‫ﻟﺪﻳﻨﺎ ﻋﻨﺪ هﺬﻩ اﻟﻠﺤﻈﺔ ‪[Zn2+] = 50 × 10 –3 mol/ L :‬‬

‫‪100‬‬

‫وﻣﻨﻪ ‪n (Zn2+) = 50 × 10 –3 × 0,1 = 5,00 × 10 –3 mol‬‬
‫وﻟﺪﻳﻨﺎ ﻣﻦ اﻟﺠﺪول ‪ ، n(Zn2+) = x :‬وﻣﻨﻪ ‪:‬‬
‫‪ ، x = 5 × 10 −3 mol‬وﺑﺎﻟﺘﺎﻟﻲ ‪:‬‬

‫‪50‬‬

‫‪n(Zn) = 3,5 × 10–2 – 5 × 10 –3 = 3,00 × 10 -3 mol‬‬
‫‪n(H+) = 2 × 10–2 – 10 × 10 –3 =1,0 × 10 -2 mol‬‬
‫ﺗﺮآﻴﺐ اﻟﻮﺳﻂ اﻟﺘﻔﺎﻋﻠﻲ ﻋﻨﺪ ‪: t = tf‬‬

‫)‪t (mn‬‬

‫‪4,9‬‬

‫‪20‬‬

‫ﻟﺪﻳﻨﺎ ﻣﻦ اﻟﺒﻴﺎن ‪ . [Zn2+] = 0,1 mol/ L‬ﻧﺤﺴﺐ ﻋﺪد ﻣﻮﻻت هﺬﻩ اﻟﺸﺎردة ﻓﻲ ﺣﺠﻢ اﻟﻤﺰﻳﺞ ‪. 100 mL‬‬

‫اﻟﺸﻜﻞ ‪1 -‬‬

‫‪ ، n(Zn2+) = [Zn2+] V = 0,1 × 0,1 = 10 –2 mol‬وﻣﻨﻪ ‪ ، x = 10 −2 mol / L‬إذن ‪:‬‬

‫‪n(Zn) = 3,5 × 10 –2 – 10 –2 = 2,5 × 10 –2 mol‬‬
‫‪n(H+) = 2 × 10 –2 – 2 × 10 –2 = 0‬‬
‫‪ - 5‬ﺣﺘﻰ ﻳﻜﻮن اﻟﺮﺳﻢ واﺿﺤﺎ ﻓﺼﻠﻨﺎ آﻞ ﺟﺰء ﻟﻮﺣﺪﻩ ‪.‬‬
‫ﻓﻲ اﻟﻠﺤﻈﺔ ‪: t = 0‬‬
‫‪dx AB 9 × 10 −3‬‬
‫=‬
‫=‬
‫‪= 1,2 × 10 −3‬‬
‫‪dt CB‬‬
‫‪7,5‬‬
‫‪1 dx 1,2 × 10 −3‬‬
‫=‬
‫‪= 1,2 × 10 − 2 mol.L−1 .mn −1‬‬
‫‪V dt‬‬
‫‪0,1‬‬

‫) ‪x (mmol‬‬

‫) ‪x (mmol‬‬
‫•‪A‬‬
‫•‪E‬‬

‫‪10‬‬

‫=‪v‬‬

‫‪10‬‬

‫ﻓﻲ اﻟﻠﺤﻈﺔ ‪: t1/2‬‬

‫‪dx EF 8 × 10 −3‬‬
‫=‬
‫=‬
‫‪= 8,9 × 10 − 4‬‬
‫‪dt GF‬‬
‫‪9‬‬
‫‪1 dx 8,9 × 10 −4‬‬
‫=‬
‫‪= 8,9 × 10 −3 mol.L−1 .mn −1‬‬
‫‪V dt‬‬
‫‪0,1‬‬

‫‪5‬‬

‫=‪v‬‬

‫‪•B‬‬

‫)‪t (mn‬‬

‫‪10‬‬

‫اﻟﺸﻜﻞ ‪2 -‬‬
‫‪6‬‬

‫•‪C‬‬

‫‪•F‬‬

‫)‪t (mn‬‬

‫•‬

‫‪G‬‬
‫‪4,5‬‬

‫‪20‬‬

‫اﻟﺸﻜﻞ ‪3 -‬‬


Aperçu du document ExoLivre%20U1B.pdf - page 1/6

Aperçu du document ExoLivre%20U1B.pdf - page 2/6

Aperçu du document ExoLivre%20U1B.pdf - page 3/6

Aperçu du document ExoLivre%20U1B.pdf - page 4/6

Aperçu du document ExoLivre%20U1B.pdf - page 5/6

Aperçu du document ExoLivre%20U1B.pdf - page 6/6




Télécharger le fichier (PDF)


ExoLivre%20U1B.pdf (PDF, 365 Ko)

Télécharger
Formats alternatifs: ZIP Texte



Documents similaires


exolivre u1b
exolivre 20u1b
couleur ions
exolivre u1a
exolivre 20u1a
solind1

Sur le même sujet..




🚀  Page générée en 0.01s