Question

1.Data energi ikatan
Diketahui energi ikatan rata-rata sebagai berikut :
C = C = 6,4 kJ/mol; C – C = 348 kJ/mol; C – H = 43 kJ/mol; Cl – Cl = 242 kJ/mol;
C – Cl = 328 kJ/mol
Tentuka ΔHreaksi dari : C2H4 + Cl2 → C2H4Cl
2.Diketahui :
∆Hof CaH2(s) = - 189 kJ/mol
∆Hof H2O(l) = - 285 kJ/mol
∆Hof Ca(OH)2(s) = - 197 kJ/mol
Perubahan entalpi dari reaksi : CaH2(s) + 2H2O(l) → Ca(OH)2(s) + 2H2(g) adalah………….

Answer 1

Jawaban:

1.

C2H4 + Cl2 -> C2H4Cl2

H H H H

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C=C + Cl-Cl -> Cl-C-C-Cl

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H H H H

∆Hreaksi = Jumlah ikatan terputus - Jumlah ikatan terbentuk

∆Hreaksi = [ 1 x C=C + 4 x C-H + 1 x Cl-Cl ] - [ 1 x C-C + 2 x C-Cl + 4 x C-H ]

∆Hreaksi = [ 614 + 4x413 + 242 ] kJ/mol - [ 348 + 2x328 + 4x413 ] kJ/mol

∆Hreaksi = [ 2508 kJ/mol ] - [ 2656 kJ/mol ]

∆Hreaksi = -148 kJ/mol

2.

CaH2(s) + 2H2O(l) → Ca(OH)2(s) + 2H2(g)

∆H = Jumlah ∆H°f produk - jumlah ∆H°f reaktan

∆H = [ ∆H°f Ca(OH)2 + 2. ∆H°f H2 ] - [ ∆H°f CaH2 + 2. ∆H°f H2O ]

∆H = [ -197 + 2.0 ] kJ/mol - [ -189 + 2.(-285) ] kJ/mol

∆H = - 197 kJ/mol - (-759) kJ/mol

∆H = - 197 kJ/mol + 759 kJ/mol

∆H = +562 kJ/mol