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Friday, August 19, 2016

MULOK

kita di tanyakan tentang siapa yang tidak mengerjakan tugas yg di berikan pada minggu lalu
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Contoh Soal :

Perhatikan contoh perhitungan berikut.
Contoh Soal (6) :
CH4(g) + 2O2(g) → CO2(g) + 2 H2O(l) ΔH = - 802 kJ.
Berdasarkan entalpi pembentukan standar, hitunglah ΔHf CH4(g).
Jawaban :
ΔHR = [1 ΔHf CO2 + 2 ΔHf H2O] – [ ΔHf CH4 + 3 ΔHf O2)
- 802 kJ = [1(- 393,51) + 2 (-285,83)] – [ ΔHf CH4 + 3 . 0] kJ
- 802 kJ = [- 393,51 + (-571,66)] kJ – [ ΔHf CH4] kJ
ΔHf CH4 = - 163,17 kJ
Jadi, entalpi pembentukannya adalah - 163,17 kJ.
 
Tentukan entalpi pembakaran dari H2S(g), bila entalpi pembentukan H2S, H2O, dan SO2, berturut-turut = 20,6 kJ/mol; - 241,81 kJ/mol; dan – 296,81 kJ/mol.
Pembahasan :
Reaksi pembakaran H2S adalah :
H2S(g) + ½ O2(g) → H2O(g) + SO2(g)
ΔHR = [ΔHf H2O(g) + ΔHf SO2(g)] – [ΔHf H2S + ΔHf O2]
= [- 241,81 + (- 296,81)] kJ – [(-20,6) + 0] kJ
= 518,02 kJ
Jadi, entalpi pembakarannya adalah 518,02 kJ.
 
Hitunglah besamya energi ikatan rata-rata (energi disosiasi) dari N-H dalam molekul NH3 bila ΔHd = 46,11 kJ; lkatan energi H-H = 436 kJ; dan NºN = 945,9 kJ. 
Jawaban : 
ΔHd NH3 = 46,11 kJ 
Reaksi desosiasi NH3 adalah: 
NH3(g) → ½ N2(g) + 3/2 H2(g)    ΔHd = 46,11 kJ
NH3(g) → ½ N2(g) + 3/2 H2(g)    ΔHd = 46,11 kJ
ΔHd NH3 = Energi ikatan yang putus dari NH3 - Energi yang terbentuk dari ½ NºN + 3/2 H-H 
46,11 kJ = DNH3 - ( ½ x 945,3 + 3/2 (436) kJ 
46,11 kJ = (DNH3 - 1126,6) kJ 
DNH3 = (1116,6 + 46,11) kJ = 1172,71 kJ 
Energi ikat rata-rata N-H = 1/3 x 1172,71 kJ = 390,9 kJ 
Jadi, energi ikatan rata-rata dari N-H adalah 390,9 kJ.
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