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DASAR REAKSI ANORGANIK

DASAR REAKSI ANORGANIK. Sayekti Wahyuningsih JURUSAN KIMIA Universitas Sebelas Maret w.Sayekti @ yahoo.com. Prinsip dasar reaksi kimia. Konsep energi ikatan dan struktur molekul H (g) + H (g) H 2 (g)  H = -436 kJ/mol ( entalpi pembentukan H2)

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DASAR REAKSI ANORGANIK

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  1. DASAR REAKSI ANORGANIK SayektiWahyuningsih JURUSAN KIMIA UniversitasSebelasMaret w.Sayekti @yahoo.com

  2. Prinsipdasarreaksikimia Konsepenergiikatandanstrukturmolekul H(g) + H(g) H2 (g)H = -436 kJ/mol (entalpipembentukan H2) H2 (g) H(g) + H(g)H = 436 kJ/mol ( energiikat H2)

  3. Energiikat Energi yang dibutuhkanuntukmenarik atom bebasdaripengaruhmasing-masing atom terhadap atom lain, dankarenaitumerupakanenergi yang mengikat atom secarabersama-sama. Perubahanentalpipadadissosiasimolekul gas padatemperatur 278 K menjadi atom-atom gas ground state

  4. Bond energy vs mean interatomic distance

  5. Bond length vs bond strength

  6. Bagaimanakorelasiantara bond length dengan bond energy?

  7. Bond energy

  8. Mengapa multiple bond memilikienergiikatlebihbesar?

  9. O-H bond formation

  10. Energiikat kekuatanikatdankestabilan Muatanberlawanan ? Elektronbebas ? F F F N O F N O F F 2(278) + 222 + 190 kJ/mol = 968 kJ/mol 3(278) kJ/mol +222 kJ/mol = 1056kJ/mol Ikatankimiamenjagamolekul-molekul, kristal, dan gas-gas diatomikuntuktetapbersama. Selainituikatankimiajugamenentukanstruktursuatuzat.

  11. Pertemuankedua

  12. Lewis Structure Counting electrons • The total number of electrons represented in a Lewis structure is equal to the sum of the numbers of valence electrons on each individual atom. Non-valence electrons are not represented in Lewis structures. • The octet rule states atoms with eight electrons in their valence shell will be stable, regardless of whether these electrons are bonding or nonbonding. The rule applies well to acidic compounds. The 18-Electron rule is operative on atoms from period 4, which have to achieve 18 electrons to their orbitals and achieve a stable configuration which has the same electron configuration as a Noble gas. Similarly from period 6, the atoms have to achieve 32 electrons to fill their orbitals. Placing electrons • Lewis structures for oxygen, fluorine, the hydrogen sulfate anion, and formamide • Once the total number of available electrons has been determined, electrons must be placed into the structure. They should be placed initially as lone pairs: one pair of dots for each pair of electrons available. Lone pairs should initially be placed on outer atoms (other than hydrogen) until each outer atom has eight electrons in bonding pairs and lone pairs; extra lone pairs may then be placed on the central atom. When in doubt, lone pairs should be placed on more electronegative atoms first. • Once all lone pairs are placed, atoms, especially the central atoms, may not have an octet of electrons. In this case, the atoms must form a double bond; a lone pair of electrons is

  13. Thermodinamikaenergiikat Kecenderungansatugrup energiikatnaik Kecenderungandalamsatuperiode C-C 347 Si-Si 340 Ge-Ge Sn-Sn C-C 347 N-N 160 O-O 146 Energiikatturun, panjangikatanbertambah

  14. Ikatanrangkaphomonuclear • Padaikatanrangkap, ikatanterdiridarisatuikatandansatuatauduaikatan. Energiikatdariikatanrangkapmerupakanhasilkontribusidariikatandanikatan. • Energiikatikatanrangkaptergantungpada order ikatan, ukuranmolekul, danterdapatnyapasanganelektronbebas.

  15. Bond order Energiikat H2+ = 255/2 = 216 kJ/mol (kenapamenjadi 255 kJ/mol? Energitolakanpasanganelektron= 255-216 kJ/mol Ordeikatan (B.O.) = (jumlahelektronpada orbital bonding - jumlahelektronpada orbital non bonding) /2. Bagaimanahubunganantarapanjangikatan, bond order danenergiikat?

  16. (8-3)/2 =2,5

  17. Pada O2 estimasikontribusidariikatandanikatandapatdihitungdaripersamaan: E = Es (ds / dm) Eadalahenergidarikontribusiikatan, Esadalahenergiikatantunggal, dsadalahpanjangikatantunggal, sedangkan dmadalahpanjangikatanrangkap. Eakanberhargamaksimumjikatidakterdapatpasanganelektronbebas (group 14). Sedangkanjikaterdapatpasanganelektronbebas (group 15, dan 16) akibatadanyatolakanpasanganelektron, harga Emenjaditidakmaksimal.

  18. Ikatanrangkapheteronuclear Energiikat heteroatom Denganmenggunakanpendekatan yang dikemukakanoleh Pauling padapembentukansenyawa X-Z dariunsur X dan Z, energiikat X-Z, E(X-Z), harusmemenuhiduaketentuanyaitu: • (1). E(X-Z) > [E(X-X) . E(Z-Z)]1/2, dan • (2). E(X-Z) > [E(X-X) + E(Z-Z)]/2 [E(X-X) . E(Z-Z)]1/2 disebutsebagaireratageometri [E(X-X) + E(Z-Z)]/2 disebutrerataaritmetika

  19. Strukturresonansiionikmemberikankontribusipadaenergiikat heteroatom. Energiresonansiionik, , adalahsebesar:  = E(X-Z) - [E(X-X) . E(Z-Z)]1/2atau  = E(X-Z) - [E(X-X) + E(Z-Z)] / 2 sehinggaenergiikat heteroatom adalah E(X-Z) =  + [E(X-X) . E(Z-Z)]1/2atau E(X-Z) =  + [E(X-X) + E(Z-Z)] / 2 1/2akanberhargasamadenganyaituperbedaanelektronegatifitasantara X dan Z

  20. metodeestimasikarakterionik  = a + 2bq asumsi : (X) = (Z) a(X) + 2b(X).q(X) = a(Z) + 2b(Z).q(Z) q(X) = q(Z) • maka: a(X) + 2b(X).q(Z) = a(Z) + 2b(Z).q(Z) q(Z) = [a(X) –a(Z)] / 2[b(X) + b(Z)]

  21. Contoh: Muatanparsial K : q(K) = [7,59-2,42] / [2(4,22 + 1,92)] = +0,421 DistribusimuatanparsialKBradalan K +0,421 Br-0,421 KarakterionikKBr = 42,1% KarakterkovalenKBr = 57,9%

  22. Faktor- faktor yang mempengaruhienergiikat heteroatom 1. Kombinasifaktorsterikdanfaktorhibridisasi. • 2. Ukuranmolekul • E(H2O) > E(H2S) > E(H2Se) > E(H2Te) • 3. Tolakanpasanganelektron, tolakanpasanganelektrondapat • menyebabkandestabilisasiikatanheteroatom.

  23. 4. Efekresonansiionik Jikadilihatdarienergidisosiasi, menurutpauling rata ratageometridarienergiikatCl-Cldan F-F adalah: (154 x 240 )1/2 = 192 kJ/mol namunfaktadarienergiikatClF > 192 kJ/mol ( adatambahansekitar 30%). Perbedaansebesar (251-192) kJ/mol ( )merupakantambahanenergidariefekresonansiionik. Energiikat yang sesungguhnyamerupakanenergi total darikontribusiikatankovalendanikatanionik.  E(X-Z) =  + [E(X-X) . E(Z-Z)]1/2 251 kJ/mol = 59 kJ/mol + 192 kJ/mol

  24. PrinsipEntalpidankekuatanikatankimia Entalpi, H, adalahjumlahdarienergi internal danenergilainnyadidalamsistem. H = E + PV (jikakerja lain PV saja) sedangkan E = q - PV sehinggapadatekanantetap (reaksikabanyakandilaboratorium) H = ( q - PV ) + PV atau H = qp

  25. Padaprosesreaksidengandisertaiperubahanmol: PV = PVf - PVi PV = P(Vf - Vi ) PV = nfRT - ni RT PV = (nf - ni ) RT = nRT Reaksipadatekanantetap: H E. Perubahanentalpi (H) dapatdihitungdariperbedaanhargaentalpidandariperbedaanenergiikatantarareaktanprodukmaka: H = [Hproduk -Hreaktan] atau H = -[ Eikatproduk -Eikatreaktan]

  26. H = -[ Eikatproduk -Eikatreaktan]

  27. Hess's Law Cycle ΔH1= ΔH(A to B) Pathway 1 the most direct route with no intermediate stages, orΔH1 = ΔH2 + ΔH3 pathway 2 involving one set of intermediates C, or orΔH1 = ΔH4 + ΔH5 + ΔH6 pathway 3 involving two sets of intermediates D and E etc. etc. - there is no limit to the complexity of the Hess's Law Cycle as long as A and B are constant.

  28. Konsepentalpi, entropidanenergibebaspadakespontananreaksi Hubungan G, H, danS Entalpi H : H = E + PV (jikakerja lain PV saja) ReaksispontandidukungdarihargaH negatifdanhargaentropiS positif. Energibebas Gibbs merupakanukurandarikespontananreaksi yang besarannyatergantungpadahargaH, T, danS: G = H - TS JikaH positifapakahreaksiselalutidakspontan? JikaS negatifapakahreaksiselalutidakspontan?

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