Roboguru

Diketahui data berikut.   Energi ikatan rata-rata:   Tentukan entalpi pembakaran metana membentuk gas karbon dioksida dan uap air berdasarkan: a. entalpi pembentukan dan b. energi ikatan. Apakah Anda menemukan hasil yang sama? Jika tidak, jelaskan mengapa terdapat perbedaan.

Pertanyaan

Diketahui data berikut.

increment H subscript f degree space C H subscript 4 open parentheses italic g close parentheses equals minus sign 75 space kJ space mol to the power of negative sign 1 end exponent increment H subscript f degree space C O subscript 2 open parentheses italic g close parentheses equals minus sign 393 comma 5 space kJ space mol to the power of negative sign 1 end exponent increment H subscript f degree space H subscript 2 O open parentheses italic g close parentheses equals minus sign 242 space kJ space mol to the power of negative sign 1 end exponent 

Energi ikatan rata-rata:

C bond H space colon space 413 space kJ space mol to the power of negative sign 1 end exponent O double bond O space colon space 495 space kJ space mol to the power of negative sign 1 end exponent C double bond O space colon space 799 space kJ space mol to the power of negative sign 1 end exponent O bond H space colon space 463 space kJ space mol to the power of negative sign 1 end exponent 

Tentukan entalpi pembakaran metana membentuk gas karbon dioksida dan uap air berdasarkan:

a. entalpi pembentukan dan

b. energi ikatan.

Apakah Anda menemukan hasil yang sama? Jika tidak, jelaskan mengapa terdapat perbedaan. space 

Pembahasan Soal:

Reaksi pembakaran metana:

C H subscript 4 and 2 O subscript 2 yields C O subscript 2 and 2 H subscript 2 O 

a. Dengan data entalpi pembentukan

table attributes columnalign right center left columnspacing 0px end attributes row cell increment H subscript c end cell equals cell left parenthesis 1 cross times increment H subscript f degree space C O subscript 2 right parenthesis plus left parenthesis 2 cross times increment H subscript f degree space H subscript 2 O right parenthesis minus sign left parenthesis 1 cross times increment H subscript f degree space C H subscript 4 right parenthesis end cell row blank equals cell left parenthesis 1 cross times minus sign 393 comma 5 space kJ space mol to the power of negative sign 1 end exponent right parenthesis plus left parenthesis 2 cross times minus sign 242 space kJ space mol to the power of negative sign 1 end exponent right parenthesis minus sign left parenthesis 1 cross times minus sign 75 space kJ space mol to the power of negative sign 1 end exponent right parenthesis end cell row blank equals cell negative sign 802 comma 5 space kJ space mol to the power of negative sign 1 end exponent end cell end table 

b. Dengan data energi ikatan

table attributes columnalign right center left columnspacing 0px end attributes row cell increment H subscript c end cell equals cell left parenthesis 4 cross times D subscript C bond H end subscript right parenthesis plus left parenthesis 2 cross times D subscript O double bond O end subscript right parenthesis minus sign left parenthesis 2 cross times D subscript C double bond O end subscript right parenthesis minus sign left parenthesis 2 cross times D subscript O bond H end subscript right parenthesis end cell row blank equals cell left parenthesis 4 cross times 413 space kJ space mol to the power of negative sign 1 end exponent right parenthesis plus left parenthesis 2 cross times 495 space kJ space mol to the power of negative sign 1 end exponent right parenthesis minus sign left parenthesis 2 cross times 799 space kJ space mol to the power of negative sign 1 end exponent right parenthesis minus sign left parenthesis 2 cross times 2 cross times 463 space kJ space mol to the power of negative sign 1 end exponent right parenthesis end cell row blank equals cell negative sign 808 space kJ space mol to the power of negative sign 1 end exponent end cell end table  

Kedua hasil memiliki selisih 5,5. Selisih ini kemungkinan karena perbedaan ketelitian ketika melakukan pengukuran.


Jadi, entalpi pembakaran metana sebesar -802,5 kJ/mol atau -808 kJ/mol.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

S. Hidayati

Mahasiswa/Alumni Universitas Indonesia

Terakhir diupdate 03 Mei 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

Diketahui     dan . Tentukan perubahan entalpi pada proses peruraian  dengan reaksi:

Pembahasan Soal:

Perubahan entalpi proses peruraian Ca C O subscript 3 dapat dicari menggunakan data perubahan entalp pembentukan standar seperti yang diketahui disoal, yaitu sebagai berikut:

increment H equals begin inline style sum with blank below end style increment H subscript f degree sesudah minus sign begin inline style sum with blank below end style begin inline style increment end style begin inline style H subscript f end style begin inline style degree end style begin inline style space end style begin inline style sebelum end style begin inline style space end style begin inline style space end style space begin inline style space end style begin inline style space end style begin inline style space space end style begin inline style space end style begin inline style equals end style begin inline style left parenthesis end style begin inline style increment end style begin inline style H subscript f end style begin inline style degree end style begin inline style space end style begin inline style Ca O end style begin inline style plus end style begin inline style increment end style begin inline style H subscript f end style begin inline style degree end style begin inline style space end style begin inline style C O subscript 2 end style begin inline style right parenthesis end style begin inline style negative sign end style begin inline style left parenthesis end style begin inline style increment end style begin inline style H subscript f end style begin inline style degree end style begin inline style space end style begin inline style Ca C O subscript 3 end style begin inline style right parenthesis end style begin inline style space end style begin inline style space end style space begin inline style space end style begin inline style space end style begin inline style space space end style begin inline style space end style begin inline style equals end style begin inline style left parenthesis end style begin inline style negative sign end style begin inline style 635 end style begin inline style comma end style begin inline style 5 end style begin inline style plus end style begin inline style left parenthesis end style begin inline style negative sign end style begin inline style 394 end style begin inline style right parenthesis end style begin inline style right parenthesis end style begin inline style negative sign end style begin inline style left parenthesis end style begin inline style negative sign end style begin inline style 1 end style begin inline style. end style begin inline style 207 end style begin inline style right parenthesis end style begin inline style space end style begin inline style space end style space begin inline style space space end style begin inline style space end style begin inline style space end style begin inline style space end style begin inline style equals end style begin inline style plus end style begin inline style 177 end style begin inline style comma end style begin inline style 5 end style begin inline style space end style begin inline style kJ end style begin inline style forward slash end style begin inline style mol end style 

Jadi, perubahan entalpi peruraian Ca C O subscript 3 adalah +177,5 kJ/mol.

2

Roboguru

Diketahui: Hitunglah besarnya kalor pembentukan !

Pembahasan Soal:

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 

0

Roboguru

Pembakaran  terjadi menurut persamaan reaksi berikut.   Bila entalpi pembentukan standar  dan  berturut-turut adalah -115,3 kJ/mol, -393,5 kJ/mol, dan -296,8 kJ/mol, maka entalpi pembakaran  7,6 gra...

Pembahasan Soal:

Entalpi pembentukan standarbegin mathsize 14px style open parentheses increment H subscript f superscript degree close parentheses end style  adalah 1 mol senyawa terbentuk dari unsur-unsurnya pada kondisi standar. Untuk menentukan perubahan entalpi pada reaksi, jika diketahui data-data entalpi pembentukan begin mathsize 14px style open parentheses increment H subscript f superscript degree close parentheses end style digunakan rumus :

begin mathsize 14px style increment H subscript r equals increment H subscript f superscript degree subscript produk minus sign increment H subscript f superscript degree subscript reaktan end style 

untuk begin mathsize 14px style open parentheses increment H subscript f superscript degree close parentheses end style unsur = 0

Pada reaksi :

begin mathsize 14px style C S subscript 2 subscript open parentheses italic l close parentheses end subscript plus 3 O subscript 2 subscript open parentheses italic g close parentheses end subscript rightwards arrow C O subscript 2 subscript open parentheses italic g close parentheses end subscript plus 2 S O subscript 2 subscript open parentheses italic g close parentheses end subscript end style

increment H subscript r equals left parenthesis 1 cross times increment H subscript f superscript degree C O subscript 2 plus 2 cross times increment H subscript f superscript degree S O subscript 2 right parenthesis minus sign left parenthesis 1 cross times increment H subscript f superscript degree C S subscript 2 right parenthesis space space space space space space space equals left parenthesis 1 cross times open parentheses negative sign 393 comma 5 close parentheses plus 2 cross times open parentheses negative sign 296 comma 8 close parentheses right parenthesis minus sign left parenthesis 1 cross times open parentheses negative sign 115 comma 3 close parentheses right parenthesis space space space space space space space equals space minus sign 987 comma 1 plus 115 comma 3 space space space space space space space equals minus sign 871 comma 8 space kJ forward slash mol 

Entalpi pembakaran begin mathsize 14px style left parenthesis increment H degree subscript c right parenthesis end style 7,6 gram begin mathsize 14px style C S subscript 2 end style (Mr = 76) =increment H subscript r cross times mol equals minus sign 871 comma 8 cross times fraction numerator 7 comma 6 over denominator 76 end fraction equals minus sign 87 comma 18 space kJ  

Jadi, jawaban yang tepat adalah A. 

0

Roboguru

Diketahui: Tentukan nilai kalor bakar dari alkohol !

Pembahasan Soal:

Diketahui:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell italic increment H italic degree subscript italic f italic space C O subscript 2 open parentheses italic g close parentheses end subscript end cell equals cell negative sign 394 space kj space mol to the power of negative sign 1 end exponent end cell row cell italic increment H italic degree subscript italic f italic space H subscript 2 O subscript open parentheses italic g close parentheses end subscript end cell equals cell negative sign 285 space kj space mol to the power of negative sign 1 end exponent end cell row cell italic increment H italic degree subscript italic f italic space C subscript 2 H subscript 5 open parentheses italic l close parentheses end subscript end cell equals cell negative sign 227 space kj space mol to the power of negative sign 1 end exponent end cell end table end style 

Ditanya: nilai kalor bakar dari alkohol begin mathsize 14px style open parentheses C subscript 2 H subscript 5 O H close parentheses end style?

Jawab:

begin mathsize 14px style C subscript 2 H subscript 5 O H and 3 O subscript 2 yields 2 C O subscript 2 and 3 H subscript 2 O end style 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell increment H end cell equals cell sum italic increment H italic degree subscript italic f italic space produk minus sign sum italic increment H italic degree subscript italic f italic space reaktan end cell row cell increment H end cell equals cell open parentheses 2 italic increment H italic degree subscript italic f italic space C O subscript 2 plus 3 italic increment H italic degree subscript italic f italic space H subscript 2 O close parentheses minus sign open parentheses italic increment H italic degree subscript italic f italic space C subscript 2 H subscript 5 O H plus italic increment H italic degree subscript italic f italic space O subscript 2 close parentheses end cell row cell increment H end cell equals cell open parentheses 2 cross times open parentheses negative sign 394 close parentheses plus 3 cross times open parentheses negative sign 285 close parentheses close parentheses minus sign open parentheses negative sign 227 plus 0 close parentheses end cell row cell increment H end cell equals cell negative sign 1416 space kj space mol to the power of negative sign 1 end exponent end cell end table end style 

Tanda negatif (-) pada perubahan entalpi menunjukkan bahwa reaksi pembakaran alkohol termasuk reaksi yang melepaskan kalor.

Untuk menentukan nilai kalor bakar:

MrC2H5OH=(2×ArC)+(6×ArH)+(1×ArO)MrC2H5OH=(2×12)+(6×1)+(1×16)MrC2H5OH=46g/mol

qbakar=MrHqbakar=46g/mol1416kJ/molqbakar=30,78kJ/g

Jadi, nilai kalor bakar alkohol undefined adalah H=30,78kJ/g.

0

Roboguru

Diketahui:, , dan . Tentukan perubahan entalpi pada proses peruraian  dengan reaksi:

Pembahasan Soal:

Diketahui:

begin mathsize 14px style increment H degree subscript f space Ca C O subscript 3 open parentheses italic s close parentheses equals minus sign 1.207 space kJ space mol to the power of negative sign 1 end exponent increment H degree subscript f space Ca O open parentheses italic s close parentheses space space space space equals minus sign 635 comma 5 space kJ space mol to the power of negative sign 1 end exponent increment H degree subscript f space C O subscript 2 open parentheses italic g close parentheses space space space space equals minus sign 394 space kJ space mol to the power of negative sign 1 end exponent end style

Ditanya: begin mathsize 14px style increment H end style reaksi peruraian begin mathsize 14px style Ca C O subscript 3 end style?

Penentuan undefined berdasarkan data perubahan entalpi pembentukan standar (begin mathsize 14px style increment H degree subscript f end style):
begin mathsize 14px style increment H equals left curly bracket left parenthesis increment H degree subscript f space Ca O right parenthesis plus left parenthesis increment H degree subscript f space C O subscript 2 right parenthesis right curly bracket minus sign left curly bracket increment H degree subscript f space Ca C O subscript 3 right curly bracket increment H equals left curly bracket left parenthesis minus sign 635 comma 5 space kJ space mol to the power of negative sign 1 end exponent right parenthesis plus left parenthesis minus sign 394 space kJ space mol to the power of negative sign 1 end exponent right parenthesis right curly bracket minus sign left curly bracket minus sign 1.207 space kJ space mol to the power of negative sign 1 end exponent right curly bracket increment H equals minus sign 1.029 comma 5 space kJ space mol to the power of negative sign 1 end exponent plus 1.207 space kJ space mol to the power of negative sign 1 end exponent increment H equals plus 177 comma 5 space kJ space mol to the power of negative sign 1 end exponent end style 

Jadi, nilai perubahan entalpi untuk reaksi peruraian begin mathsize 14px style Ca C O subscript bold 3 end style adalah begin mathsize 14px style bold plus bold 177 bold comma bold 5 bold space bold kJ bold space bold mol to the power of bold minus sign bold 1 end exponent end style

 

4

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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