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increment H subscript pembakaran italic space C subscript 2 H subscript 5 O H equals minus sign 1.368 space bevelled kJ over mol increment H subscript italic f degree space C O subscript 2 open parentheses italic g close parentheses equals minus sign 394 space bevelled kJ over mol increment H subscript italic f degree space H subscript 2 O open parentheses italic l close parentheses equals minus sign 286 space bevelled kJ over mol


Hitunglah besarnya kalor pembentukan C subscript 2 H subscript 5 O H!space 

S. Susanti

Master Teacher

Mahasiswa/Alumni Universitas Jayabaya

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 besarnya kalor pembentukan C subscript 2 H subscript 5 O H adalah table attributes columnalign right center left columnspacing 0px end attributes row blank blank minus sign end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank 278 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank space end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell bevelled kJ over mol end cell end table.space 

Pembahasan

Untuk menyelesaikan soal diatas dapat digunakan konsep penentuan perubahan entalpi berdasarkan data perubahan entalpi pembentukan standar (increment H subscript italic f degree). Langkah-langkah penyelesaiannya adalah sebagai berikut.

  • Membuat persamaan termokimia dari  data increment H subscript pembakaran italic space C subscript 2 H subscript 5 O H equals negative sign 1.368 space bevelled kJ over mol. Pembakaran senyawa hidrokarbon akan menghasilkan C O subscript 2 dan H subscript 2 O, maka persamaan reaksinya yaitu:


C subscript 2 H subscript 5 O H open parentheses italic l close parentheses space plus space 3 O subscript 2 open parentheses italic g close parentheses space rightwards arrow space 2 C O subscript 2 open parentheses italic g close parentheses space plus space 3 H subscript 2 O open parentheses italic l close parentheses space space space space space space space space increment H equals minus sign 1.368 space bevelled kJ over mol

 

  • Menentukan kalor pembentukan increment H subscript italic f degree dari C subscript 2 H subscript 5 O H


table attributes columnalign right center left columnspacing 0px end attributes row cell increment H subscript italic r italic e italic a italic k italic s italic i end subscript end cell equals cell sum increment H subscript italic f degree subscript italic p italic r italic o italic d italic u italic k end subscript space minus sign space sum increment H subscript italic f degree subscript italic r italic e italic a italic k italic t italic a italic n end subscript end cell row cell increment H subscript italic r italic e italic a italic k italic s italic i end subscript end cell equals cell left parenthesis 2 cross times increment H subscript italic f degree subscript C O subscript 2 end subscript space plus space 3 cross times increment H subscript italic f italic degree subscript H subscript 2 O end subscript right parenthesis space minus sign space left parenthesis increment H subscript italic f degree subscript C subscript 2 H subscript 5 O H end subscript space plus space 3 cross times increment H subscript italic f degree subscript O subscript 2 end subscript right parenthesis end cell row cell negative sign 1.368 end cell equals cell left parenthesis 2 cross times left parenthesis minus sign 394 right parenthesis space plus space 3 cross times left parenthesis minus sign 286 right parenthesis right parenthesis space minus sign space left parenthesis increment H subscript italic f degree subscript C subscript 2 H subscript 5 O H end subscript space plus space 3 cross times 0 right parenthesis end cell row cell negative sign 1.368 end cell equals cell left parenthesis minus sign 788 plus left parenthesis minus sign 858 right parenthesis right parenthesis space minus sign space increment H subscript italic f degree subscript C subscript 2 H subscript 5 O H end subscript end cell row cell negative sign 1.368 end cell equals cell negative sign 1.646 space minus sign space increment H subscript italic f degree subscript C subscript 2 H subscript 5 O H end subscript end cell row cell increment H subscript italic f degree subscript C subscript 2 H subscript 5 O H end subscript end cell equals cell negative sign 1.646 plus 1.368 end cell row cell increment H subscript italic f degree subscript C subscript 2 H subscript 5 O H end subscript end cell equals cell negative sign 278 space bevelled kJ over mol end cell end table


Dengan demikian, besarnya kalor pembentukan C subscript 2 H subscript 5 O H adalah table attributes columnalign right center left columnspacing 0px end attributes row blank blank minus sign end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank 278 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank space end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell bevelled kJ over mol end cell end table.space 

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