Roboguru

Reaksi yang terjadi antara gas karbon monoksida dengan uap air dituliskan sebagai berikut. CO(g)+H2​O(g)→CO2​(g)+H2​(g) Diketahui nilai : CO(g)=−110kJmol−1 H2​O(g)=−242kJmol−1 CO2​(g)=−394kJmol−1 H2​(g)=0kJmol−1 Apabila gas hidrogen yang dihasilkan sebanyak 11,2 L (STP), nilai △H untuk reaksi tersebut adalah ....

Pertanyaan

Reaksi yang terjadi antara gas karbon monoksida dengan uap air dituliskan sebagai berikut.

CO(g)+H2O(g)CO2(g)+H2(g)

Diketahui nilai increment H subscript f superscript degree:

CO(g)=110kJmol1

H2O(g)=242kJmol1

CO2(g)=394kJmol1

H2(g)=0kJmol1

Apabila gas hidrogen yang dihasilkan sebanyak 11,2 L (STP), nilai H untuk reaksi tersebut adalah ....

  1. -42 kJ

  2. -21 kJ

  3. + 21 kJ

  4. +42 kJ

  5. +56 kJ

Pembahasan Soal:

CO(g)+H2O(g)CO2(g)+H2(g)

increment H equals sum with blank below increment H subscript f superscript degree subscript produk minus sign sum with blank below increment H subscript f superscript degree subscript reaksi

table attributes columnalign right center left columnspacing 0px end attributes row cell increment H end cell equals cell left parenthesis increment H subscript f superscript degree C O subscript 2 plus increment H subscript f superscript degree space H subscript 2 right parenthesis minus sign left parenthesis increment H subscript f superscript degree C O plus increment H subscript f superscript degree H subscript 2 O right parenthesis end cell row blank equals cell left parenthesis minus sign 394 plus 0 right parenthesis minus sign left parenthesis minus sign 110 plus left parenthesis minus sign 242 right parenthesis right parenthesis end cell row blank equals cell negative sign 394 minus sign left parenthesis minus sign 352 right parenthesis end cell row blank equals cell negative sign 394 plus 352 end cell row blank equals cell negative sign 42 space kJ space mol to the power of negative sign 1 end exponent end cell end table

 

Penentuan mol gas hidrogen

VmolH2====nx22,4Lmol122,4Lmol1V22,4Lmol111,2L0,5mol

Penentuan H untuk 0,5 mol gas hidrogen

H==1mol0,5molx(42kJ)21kJ

Sehingga nilai H untuk reaksi tersebut adalah -21 kJ.

Jadi, jawaban yang tepat adalah B.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

G. Suprobo

Mahasiswa/Alumni Institut Pertanian Bogor

Terakhir diupdate 30 Agustus 2021

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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.

Roboguru

Diketahui reaksi: jika perubahan entalpi ; ; ; dan . reaksi sebesar.....kJ

Pembahasan Soal:

Perubahan entalpi dari suatu reaksi dapat ditentukan dari perubahan entalpi pembentukan produk (kanan) dikurang dengan perubahan entalpi reaktan (kiri)

table attributes columnalign right center left columnspacing 0px end attributes row cell increment H end cell equals cell Perubahan space Entalpi space kanan bond Perubahan space Entalpi space kiri end cell row cell increment H end cell equals cell open square brackets left parenthesis 4 middle dot increment H thin space N O right parenthesis plus left parenthesis 6 middle dot increment H thin space H subscript 2 O right parenthesis close square brackets minus sign open square brackets left parenthesis 4 middle dot increment H thin space N H subscript 3 right parenthesis plus left parenthesis 5 middle dot increment H space O subscript 2 right parenthesis close square brackets end cell row cell increment H end cell equals cell left parenthesis 4 c and 6 d right parenthesis minus sign left parenthesis 4 a and 5 b right parenthesis space kJ end cell row cell increment H end cell equals cell left parenthesis 4 c and 6 d minus sign 4 a minus sign 5 b right parenthesis space kJ end cell end table


Jadi, bold increment H reaksi sebesar bold left parenthesis bold 4 italic c bold plus bold 6 italic d bold minus sign bold 4 italic a bold minus sign bold 5 italic b bold right parenthesis bold space bold kJ.

Roboguru

Diketahui:   Hitunglah  untuk reaksi:

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.

Roboguru

Diketahui entalpi pembentukan uap air ; entalpi pembentukan gas karbon dioksida  Jika 4,2 gram gas propena dibakar secara sempurna sesuai persamaan reaksi:   akan dihasilkan kalor sebesar .

Pembahasan Soal:

Reaksi:
begin mathsize 14px style 2 C subscript 3 H subscript 6 open parentheses italic g close parentheses and 9 O subscript 2 open parentheses italic g close parentheses yields 6 C O subscript 2 open parentheses italic g close parentheses and 6 H subscript 2 O open parentheses italic g italic close parentheses end style

Diasumsikan entalpi pembentukan C subscript 3 H subscript 6 equals minus sign 106 space kJ point mol to the power of negative sign 1 end exponent

increment H equals begin inline style sum with blank below end style increment H subscript f space produk 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 space end style begin inline style reaktan 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 6 end style begin inline style. end style begin inline style increment end style begin inline style H subscript f end style begin inline style space end style begin inline style C O subscript 2 end style begin inline style plus end style begin inline style 6 end style begin inline style. end style begin inline style increment end style begin inline style H subscript f end style begin inline style space end style begin inline style H subscript 2 end style begin inline style O 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 2 end style begin inline style. end style begin inline style increment end style begin inline style H subscript f end style begin inline style space end style begin inline style C subscript 3 end style begin inline style H subscript 6 end style begin inline style plus end style begin inline style 9 end style begin inline style. end style begin inline style increment end style begin inline style H subscript f end style begin inline style space end style begin inline style O subscript 2 end style begin inline style right parenthesis 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 6 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 plus end style begin inline style 6 end style begin inline style left parenthesis end style begin inline style negative sign end style begin inline style 242 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 2 end style begin inline style left parenthesis end style begin inline style negative sign end style begin inline style 106 end style begin inline style right parenthesis end style begin inline style plus end style begin inline style 9 end style begin inline style left parenthesis end style begin inline style 0 end style begin inline style right parenthesis end style begin inline style right parenthesis 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 space begin inline style equals end style begin inline style negative sign end style begin inline style 3604 end style begin inline style space end style begin inline style kJ end style 

Kalor setiap 4,2 gram gas propena:

n space C subscript 3 H subscript 6 equals massa over Mr space space space space space space space space space space space equals fraction numerator 4 comma 2 over denominator 42 end fraction equals 0 comma 1 space mol q equals fraction numerator 0 comma 1 space mol over denominator 2 space mol end fraction x increment H space C subscript 3 H subscript 6 space equals fraction numerator 0 comma 1 space mol over denominator 2 space mol end fraction x left parenthesis minus sign 3604 right parenthesis space kJ space equals minus sign 180 comma 2 space kJ   

Jadi, kalor yang dihasilkan sebesar 180,2 kJ.

Roboguru

Jika diketahui ⧍H pembentukan gas karbon dioksida , uap air  dan gas propana berturut-turut yaitu –393,5 kJ,  –242 kJ, dan 104 kJ. Tentukan banyak panas yang dibebaskan pada pembakaran 100 gr propana ...

Pembahasan Soal:

Data dari entalpi pembentukan standar dapat juga digunakan untuk menghitung ΔH reaksi (ΔHr). Zat-zat pereaksi mengurai membentuk unsur-unsurnya, kemudian unsur-unsur hasil uraian tersebut membentuk zat baru. Rumus yang digunakan adalah :

increment H subscript r equals sum increment H subscript f superscript degree hasil space reaksi minus sign sum increment H subscript f superscript degree pereaksi 

Untuk menentukan banyak panas yang dibebaskan pada pembakaran 100 gr gas propana open parentheses C subscript 3 H subscript 8 close parentheses dari data increment H subscript f superscript degree diketahui, dengan menggunakan rumus di atas.

 C subscript 3 H subscript 8 subscript open parentheses italic l close parentheses end subscript plus 5 O subscript 2 subscript open parentheses italic g close parentheses end subscript rightwards arrow 3 C O subscript 2 subscript open parentheses italic g close parentheses end subscript plus 4 H subscript 2 O subscript open parentheses italic g close parentheses end subscript 

increment H subscript r equals open parentheses 3 cross times increment H subscript f superscript degree C O subscript 2 plus 4 cross times increment H subscript f superscript degree H subscript 2 O close parentheses minus sign left parenthesis 1 cross times increment H subscript f superscript degree C subscript 3 H subscript 8 plus 5 cross times increment H subscript f superscript degree O subscript 2 right parenthesis space space space space space space space equals open parentheses 3 cross times open parentheses negative sign 393 comma 5 close parentheses plus 4 cross times left parenthesis minus sign 242 right parenthesis close parentheses minus sign left parenthesis 1 cross times 104 plus 5 cross times 0 right parenthesis space space space space space space space equals open parentheses negative sign 1180 comma 5 minus sign 968 close parentheses minus sign 104 space space space space space space space equals minus sign 2252 comma 5 space kkal forward slash space mol 

mol space C subscript 3 H subscript 8 equals fraction numerator massa space C subscript 3 H subscript 8 over denominator Mr end fraction Mr space C subscript 3 H subscript 8 equals 3 cross times Ar space C plus 8 cross times Ar space H space space space space space space space space space space space space space space equals 3 cross times 12 plus 8 cross times 1 space space space space space space space space space space space space space space equals 44 mol space C subscript 3 H subscript 8 equals 100 over 44 space mol 

increment H equals fraction numerator negative sign Q over denominator mol end fraction minus sign 2252 comma 5 equals fraction numerator negative sign Q over denominator begin display style bevelled 100 over 44 end style end fraction Q space equals 5119 comma 32 space Kj 

Maka, banyak panas yang dibebaskan pada pembakaran 100 gr propana adalah 5119,32 Kj. 

Roboguru

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