Roboguru

Diketahui: 2P(s)+O2​(g)+3Cl2​(g)→2POCl3​(g)△H=−1.150kJH2​(g)+Cl2​(g)→2HCl(g)△H=−184kJ2P(s)+5Cl2​(g)→2PCl5​(g)△H=−640kJ2H2​(g)+O2​(g)→2H2​O(g)△H=−482kJ  Hitunglah △H untuk reaksi: PCl5​(g)+H2​O(g)→POCl3​(g)+2HCl(g)

Pertanyaan

Diketahui:

2 P open parentheses italic s close parentheses and O subscript 2 open parentheses italic g close parentheses and 3 Cl subscript 2 open parentheses italic g close parentheses yields 2 P O Cl subscript 3 open parentheses italic g close parentheses space increment H equals minus sign 1.150 space kJ H subscript 2 open parentheses italic g close parentheses and Cl subscript 2 open parentheses italic g close parentheses yields 2 H Cl open parentheses italic g close parentheses space space space space space space space space space space space space space space space space space space space increment H equals minus sign 184 space kJ 2 P open parentheses italic s close parentheses and 5 Cl subscript 2 open parentheses italic g close parentheses yields 2 P Cl subscript 5 open parentheses italic g close parentheses space space space space space space space space space space space space space space space space increment H equals minus sign 640 space kJ 2 H subscript 2 open parentheses italic g close parentheses and O subscript 2 open parentheses italic g close parentheses yields 2 H subscript 2 O open parentheses italic g close parentheses space space space space space space space space space space space space space space space space space space increment H equals minus sign 482 space kJ 

Hitunglah increment H untuk reaksi:

P Cl subscript 5 open parentheses italic g close parentheses and H subscript 2 O open parentheses italic g close parentheses yields P O Cl subscript 3 open parentheses italic g close parentheses and 2 H Cl open parentheses italic g close parentheses 

Pembahasan Soal:

Untuk menentukan increment H reaksi tersebut, menggunakan data perubahan entalpi pembentukan standar (increment H subscript f degree).  Data-data yang diketahui di soal bukan increment H subscript f degree, tapi perubahan entalpi pada pembentukan 2 mol untuk masing - masing zat. increment H subscript f degree untuk masing - masing zat yaitu:

P open parentheses italic s close parentheses plus begin inline style 1 half end style O subscript 2 open parentheses italic g close parentheses plus begin inline style 3 over 2 end style Cl subscript 2 open parentheses italic g close parentheses yields P O Cl subscript 3 open parentheses italic g close parentheses space increment H equals minus sign 575 space kJ space mol to the power of negative sign 1 end exponent begin inline style 1 half end style H subscript 2 open parentheses italic g close parentheses plus begin inline style 1 half end style Cl subscript 2 open parentheses italic g close parentheses yields H Cl open parentheses italic g close parentheses space space space space space space space space space space space space space space space increment H equals minus sign 92 space kJ space mol to the power of negative sign 1 end exponent P open parentheses italic s close parentheses plus begin inline style 5 over 2 end style Cl subscript 2 open parentheses italic g close parentheses yields P Cl subscript 5 open parentheses italic g close parentheses space space space space space space space space space space space space space space space space space space increment H equals minus sign 320 space kJ space mol to the power of negative sign 1 end exponent H subscript 2 open parentheses italic g close parentheses plus begin inline style 1 half end style O subscript 2 open parentheses italic g close parentheses yields H subscript 2 O open parentheses italic g close parentheses space space space space space space space space space space space space space space space space space space increment H equals minus sign 241 space kJ space mol to the power of negative sign 1 end exponent 

Kemudian,  increment H untuk reaksinya dapat dicari seperti berikut:

increment H equals begin inline style sum with blank below end style increment H subscript f degree space 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 begin inline style space end style begin inline style space end style begin inline style space end style begin inline style space end style space space space space space space 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 P O Cl subscript 3 end style begin inline style plus end style begin inline style 2 end style begin inline style cross times 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 H Cl 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 P Cl subscript 5 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 H subscript 2 end style begin inline style O end style right parenthesis begin inline style space end style begin inline style space end style begin inline style 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 left parenthesis end style begin inline style negative sign end style begin inline style 575 end style begin inline style plus end style begin inline style 2 end style begin inline style left parenthesis end style begin inline style negative sign end style begin inline style 92 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 320 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 241 end style begin inline style right parenthesis end style begin inline style right parenthesis end style space begin inline style space end style begin inline style space end style begin inline style 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 negative sign end style begin inline style 667 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 561 end style begin inline style right parenthesis end style space begin inline style space end style begin inline style space end style begin inline style 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 negative sign end style begin inline style 106 end style begin inline style space end style kJ 

Jadi, increment H dari reaksi tersebut adalah -106 kJ.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

S. Lubis

Mahasiswa/Alumni Universitas Sumatera Utara

Terakhir diupdate 07 Oktober 2021

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Pertanyaan yang serupa

Hitung kalor pembentukan anilin (C6​H7​N) jika diketahui data berikut. △H pembakaran anilin =−3.504,1kJmol−1  pembentukan CO2​ =−393,7kJmol−1  pembentukan H2​O =−285,85kJmol−1

Pembahasan Soal:

Kalor pembentukan anilin dapat diperoleh dari data entalpi pembentukan senyawa yang ada pada reaksi.

Reaksi yang terjadi pada pembentukan anilin (C subscript 6 H subscript 7 N) adalah:


C subscript 6 H subscript 7 N open parentheses italic g close parentheses and 35 O subscript 2 open parentheses italic g close parentheses yields 24 C O subscript 2 open parentheses italic g close parentheses and 4 N O subscript 2 open parentheses italic g close parentheses and 14 H subscript 2 O open parentheses italic l close parentheses


Berdasarkan data entalpi pembentukan standar increment H N O subscript 2equals 33 comma 2 space kJ space mol to the power of negative sign 1 end exponent


table attributes columnalign right center left columnspacing 0px end attributes row cell increment H end cell equals cell total space increment H subscript produk italic bond total space increment H subscript italic reaktan end cell row cell negative sign 3.504 comma 1 space kJ space mol to the power of negative sign 1 end exponent end cell equals cell left parenthesis left parenthesis 14 cross times minus sign 285 comma 85 right parenthesis plus left parenthesis 4 cross times 33 comma 2 right parenthesis plus left parenthesis 24 cross times minus sign 393 comma 7 right parenthesis right parenthesis minus sign left parenthesis 4 cross times increment H subscript C subscript 6 H subscript 7 N end subscript right parenthesis end cell row cell 4 cross times increment H subscript C subscript 6 H subscript 7 N end subscript end cell equals cell left parenthesis minus sign 4.001 comma 9 plus 132 comma 8 minus sign 9.448 comma 8 plus 3.504 comma 1 right parenthesis space kJ space mol to the power of negative sign 1 end exponent end cell row cell increment H subscript C subscript 6 H subscript 7 N end subscript end cell equals cell negative sign 2.453 comma 45 space kJ space mol to the power of negative sign 1 end exponent end cell end table


Jadi, kalor pembentukan anilin C subscript 6 H subscript 7 N adalah negative sign 2.453 comma 45 space kJ space mol to the power of negative sign 1 end exponent.

0

Roboguru

Dari reaksi: 2C2​H2​(g)+5O2​(g)→4CO2​(g)+2H2​O(l) Bila △Hf∘​C2​H2​(g)=+226,9kJ          △Hf∘​CO2​(g)=−393,3kJ△Hf∘​H2​O(l)=−285,8kJ maka besarnya △H reaksi = ....

Pembahasan Soal:

Perubahan entalpi dapat dihitung dengan data entalpi pembentukan standar, yaitu dengan mengurangkan jumlah entalpi pembentukan standar produk terhadap jumlah entalpi pembentukan standar reaktan.

2 C subscript 2 H subscript 2 open parentheses italic g close parentheses and 5 O subscript 2 open parentheses italic g close parentheses yields 4 C O subscript 2 open parentheses italic g close parentheses and 2 H subscript 2 O open parentheses italic l close parentheses

increment H double bond jumlah space increment H subscript f degree produk bond jumlah space increment H subscript f degree reaktan increment H equals open square brackets 4 open parentheses increment H subscript f degree C O subscript 2 close parentheses plus 2 open parentheses increment H subscript f degree H subscript 2 O close parentheses close square brackets minus sign open square brackets 2 open parentheses increment H subscript f degree C subscript 2 H subscript 2 close parentheses plus 5 open parentheses increment H subscript f degree O subscript 2 close parentheses close square brackets increment H equals open square brackets 4 open parentheses negative sign 393 comma 3 close parentheses plus 2 open parentheses negative sign 285 comma 8 close parentheses close square brackets minus sign open square brackets 2 open parentheses 226 comma 9 close parentheses plus 5 open parentheses 0 close parentheses close square brackets increment H equals minus sign 2.144 comma 8 minus sign 453 comma 8 increment H equals minus sign 2.598 comma 6 space kJ

Jadi, besarnya increment H reaksi adalah -2.598,6 kJ.

Jadi, jawaban yang tepat adalah C.space

1

Roboguru

Diketahui: △Hf∘​C2​H4​(g)=52molkJ​△Hf∘​C2​H6​(g)=−85molkJ​​ Hitunglah besarnya △H pada reaksi: C2​H4​(g)+H2​(g)→C2​H6​(g)

Pembahasan Soal:

Untuk menyelesaikan soal diatas dapat digunakan rumus perubahan entalpi berdasarkan entalpi pembentukan standar (increment H subscript italic f degree) yaitu:


table attributes columnalign right center left columnspacing 0px end attributes row cell increment H end cell equals cell sum increment H subscript italic f italic degree subscript italic p italic r italic o italic d italic u italic k end subscript italic space italic minus sign italic space sum increment H subscript italic f italic degree subscript italic r italic e italic a italic k italic tan end subscript end cell end table


Nilai perubahan entalpi standar (increment H subscript italic f degree) unsur bebas, dalam hal ini H subscript 2 adalah 0, maka:


table attributes columnalign right center left columnspacing 0px end attributes row cell increment H end cell equals cell sum increment H subscript italic f italic degree subscript italic p italic r italic o italic d italic u italic k end subscript italic space italic minus sign italic space sum increment H subscript italic f italic degree subscript italic r italic e italic a italic k italic tan end subscript end cell row blank equals cell increment H blank subscript italic f italic degree italic space C subscript 2 H subscript 6 space end subscript minus sign space left parenthesis increment H blank subscript italic f italic degree italic space C blank subscript 2 H blank subscript 4 space end subscript space plus space increment H blank subscript italic f italic degree italic space H subscript 2 right parenthesis end cell row blank equals cell negative sign 85 space minus sign space left parenthesis 52 plus 0 right parenthesis end cell row blank equals cell negative sign 85 minus sign 52 end cell row blank equals cell negative sign 137 space bevelled kJ over mol end cell end table


Jadi, besarnya increment H pada reaksi tersebut adalah negative sign 137 space bevelled kJ over mol.space 

0

Roboguru

Diketahui △Hf​ dari NH3​(g),NO(g),danH2​O(l) adalah –46,2 , +90,4 dan –286 kj/mol. Maka △H reaksi oksidasi amonia : 4NH3​(g)+5O2​(g)→4NO(g)+6H2​O(l) adalah...

Pembahasan Soal:

Menentukan increment H space reaksi dengan persamaan:

begin mathsize 14px style increment H space reaksi space equals space sum with blank below increment H space produk space minus sign space begin inline style sum with blank below end style increment H space pereaksi space end style 

begin mathsize 14px style increment H space produk space equals space 4 left parenthesis increment H space N O right parenthesis space plus space 6 left parenthesis increment H space H subscript 2 O right parenthesis space end style

Error converting from MathML to accessible text. 4 . 90,4 + 6 . -286

Error converting from MathML to accessible text.361,6 + -1716

Error converting from MathML to accessible text.-1354,4  kJ/mol

 

begin mathsize 14px style increment H space pereaksi space equals space 4 space left parenthesis increment H space N H subscript 3 right parenthesis end style

Error converting from MathML to accessible text. 4 . -46,2

Error converting from MathML to accessible text.-184,8 kJ/mol 

 

begin mathsize 14px style increment H space reaksi space equals space sum with blank below increment H space produk space minus sign space begin inline style sum with blank below end style increment H space pereaksi space end style 

begin mathsize 14px style increment H space reaksi space equals space end style -1354,4 - (-184,8)

begin mathsize 14px style increment H space reaksi space equals space end style-1169,6 kJ/mol

Jadi, jawaban yang tepat adalah D.

0

Roboguru

Diketahui :  △H∘fH2​O(g)=−242kJ/mol△H∘fCO2​(g)=−394kJ/mol△H∘fCH4​(g)=−84kJ/mol  Maka perubahan entalpi pembakaran (△Hc​)CH4​ adalah . . .

Pembahasan Soal:

Reaksi pembakaran begin mathsize 14px style C H subscript 4 end style adalah :undefined 

begin mathsize 14px style C H subscript 4 italic open parentheses italic g italic close parentheses space plus space 2 O subscript 2 italic open parentheses italic g italic close parentheses space rightwards arrow space C O subscript 2 italic open parentheses italic g italic close parentheses space plus space 2 H subscript 2 O italic open parentheses italic g italic close parentheses end style 

Maka penentuan entalpi pembakarannya yaitu :undefined 

begin mathsize 14px style increment Hc space equals space sum with blank below and blank on top increment H degree f space produk space minus sign space sum with blank below and blank on top increment H degree f space reaktan end style 

begin mathsize 14px style increment Hc space equals space left square bracket increment H degree f space C O subscript 2 space plus space 2 left parenthesis increment H degree f space H subscript 2 O right parenthesis right square bracket space minus sign space left square bracket increment H degree f space C H subscript 4 right square bracket end style 

begin mathsize 14px style increment Hc space equals space left square bracket minus sign 394 space plus space 2 left parenthesis minus sign 242 right parenthesis right square bracket space minus sign space left square bracket minus sign 84 right square bracket increment Hc space equals space minus sign 878 space plus space 84 increment Hc space equals space minus sign 794 space kJ forward slash mol end style  

Jadi, jawaban yang benar adalah -794 kj/mol

0

Roboguru

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