Diketahui data : ∆Hf° CaH2 (s)= -189 Kj/mol ∆Hf° H2O (l) = -285 Kj/mol ∆Hf° Ca(OH)2(s) = -197 Kj/mol Perubahan entalpi dari reaksi : CaH2 (s) + 2H2O (l)
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Diketahui data : ∆Hf° CaH2 (s)= -189 Kj/mol ∆Hf° H2O (l) = -285 Kj/mol ∆Hf° Ca(OH)2(s) = -197 Kj/mol Perubahan entalpi dari reaksi : CaH2 (s) + 2H2O (l) → Ca(OH)2 (s) + 2H2 (g) adalah
1 Jawaban
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1. Jawaban cieciemansyur7
∆Hf° CaH2 (s) = -189 Kj/mol
∆Hf° H2O (l) = -285 Kj/mol
∆Hf° Ca(OH)2(s) = -197 Kj/mol
CaH2 (s) + 2H2O (l) → Ca(OH)2 (s) + 2H2 (g) ∆H = ..... ?
jawab ;
∆H reaksi = [ total ∆H produk ] - [ total ∆H reaktan ]
= [(∆Hf°Ca(OH)2)+ 2(∆Hf° H2) ] - [(∆Hf° CaH2 ) + 2(∆Hf° H2O) ]
= [(-197) + 2 (0) ]-[(-189)+2(-285)]
= [-197]-[-759]
∆H reaksi = + 562 Kj/mol ,,,,