Diketahui , dan. Tentukan perubahan entalpi pada pembakaran .

Pertanyaan

Diketahui $increment H subscript f degree C subscript 4 H subscript 10 equals minus sign 126 comma 5 kJ forward slash mol$ , $increment H subscript f degree C O subscript 2 equals minus sign 393 comma 5 kJ forward slash mol$ dan $increment H subscript f degree H subscript 2 O equals minus sign 285 comma 6 kJ forward slash mol$ . Tentukan perubahan entalpi pada pembakaran $11 comma 6 space gram space C subscript 4 H subscript 10$ .

Pembahasan Soal:

Perubahan entalpi suatu reaksi dapat juga ditentukan melalui data perubahan entalpi pembentukan standar reaktan dan produknya, yaitu dari reaktan selisih harga $increment H subscript f degree$ reaktan dan $increment H subscript f degree$ produk.

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

Reaksi pembakaran $C subscript 2 H subscript 10$ :

$C subscript 4 H subscript 10 open parentheses italic a italic q close parentheses plus begin inline style 13 over 2 end style O subscript 2 open parentheses italic g close parentheses yields 4 C O subscript 2 open parentheses italic g close parentheses and 5 H subscript 2 O open parentheses italic l close parentheses$

Menghitung nilai perubahan entalpi:

$table attributes columnalign right center left columnspacing 0px end attributes row cell increment H end cell equals cell sum increment H subscript f degree space hasil space reaksi minus sign sum increment H subscript f degree pereaksi end cell row blank equals cell open curly brackets left parenthesis 4 cross times ΔH subscript f degree space C O subscript 2 right parenthesis plus left parenthesis 5 cross times ΔH subscript f degree space H subscript 2 O right parenthesis close curly brackets minus sign open curly brackets ΔH subscript f degree space C subscript 4 H subscript 10 and ΔH subscript f degree space begin inline style 13 over 2 end style O subscript 2 close curly brackets end cell row blank equals cell open curly brackets 4 cross times left parenthesis minus sign 393 comma 5 space kJ forward slash mol right parenthesis right parenthesis plus left parenthesis 5 cross times left parenthesis minus sign 285 comma 6 space kJ forward slash mol right parenthesis close curly brackets minus sign open curly brackets negative sign 126 comma 5 space kJ forward slash mol plus 0 close curly brackets end cell row blank equals cell left parenthesis minus sign 3002 space kJ forward slash mol right parenthesis plus 126 comma 5 space kJ forward slash mol end cell row blank equals cell negative sign 2875 comma 5 space kJ forward slash mol end cell end table$

Menghitung jumlah mol $C subscript 4 H subscript 10$ :

$table attributes columnalign right center left columnspacing 0px end attributes row cell M subscript r space C subscript 4 H subscript 10 end cell equals cell left parenthesis 4 cross times A subscript r space C right parenthesis plus left parenthesis 10 cross times A subscript r space H right parenthesis end cell row blank equals cell left parenthesis 4 cross times 12 right parenthesis plus left parenthesis 10 cross times 1 right parenthesis space space space space space space space space space space space space space space space space space end cell row blank equals cell 48 plus 10 space space space space space space space space space space space space space space space space space end cell row blank equals cell space 58 end cell row blank blank blank row cell n space C subscript 4 H subscript 10 end cell equals cell massa over M subscript r space end cell row blank equals cell fraction numerator 11 comma 6 space g over denominator 58 end fraction end cell row blank equals cell 0.2 space mol end cell end table$

Menghitung perubahan entalpi pada pembakaran:

$table attributes columnalign right center left columnspacing 0px end attributes row cell ΔH subscript c degree end cell equals cell mol space cross times ΔH space space space space space space space space end cell row blank equals cell 0 comma 2 space mol cross times left parenthesis minus sign 2875 comma 5 space kJ forward slash mol right parenthesis space space space space space space space space end cell row blank equals cell negative sign 575 comma 1 space kJ forward slash mol end cell end table$

Jadi, perubahan entalpi pada pembakaran $11 comma 6 space gram space C subscript 4 H subscript 10$ adalah $negative sign 575 comma 1 space kJ forward slash mol$ .

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

B. Rohmawati

Mahasiswa/Alumni Universitas Negeri Semarang

Terakhir diupdate 30 Agustus 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

Tapi Roboguru masih mau belajar. Menurut kamu pembahasan kali ini sudah membantu, belum?

Membantu

Kurang Membantu

Apakah pembahasan ini membantu?

Belum menemukan yang kamu cari?

Post pertanyaanmu ke Tanya Jawab, yuk

Mau Bertanya

Pertanyaan yang serupa

Diketahui , , dan . Tentukan nilai kalor bakar dari alkohol .

Pembahasan Soal:

Persamaan reaksi:

$C subscript 2 H subscript 5 O H and 3 O subscript 2 yields 2 C O subscript 2 and 3 H subscript 2 O$

Rumus:

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

Perhitungan:

$table attributes columnalign right center left columnspacing 0px end attributes row cell increment H end cell equals cell open parentheses left parenthesis 2 x increment H subscript f superscript degree C O subscript 2 right parenthesis plus left parenthesis 3 x increment H subscript f superscript degree H subscript 2 O right parenthesis close parentheses minus sign end cell row blank blank cell open parentheses increment H subscript f superscript degree C subscript 2 H subscript 5 O H plus left parenthesis 3 x increment H subscript f superscript degree O subscript 2 right parenthesis close parentheses end cell row blank equals cell open parentheses left parenthesis 2 x left parenthesis minus sign 394 right parenthesis plus left parenthesis 3 x minus sign 285 right parenthesis close parentheses minus sign open parentheses negative sign 227 close parentheses end cell row blank equals cell open parentheses negative sign 788 minus sign 855 close parentheses plus 227 end cell row blank equals cell negative sign 1416 space kJ space mol to the power of negative sign 1 end exponent end cell end table$

Jadi, nilai kalor bakar dari alkohol $open parentheses C subscript 2 H subscript 5 O H close parentheses$ adalah -1416 $kJ space mol to the power of negative sign 1 end exponent$ .

Roboguru

Jika diketahui berturut turut , dan . Hitunglah .

Pembahasan Soal:

Perubahan entalpi suatu reaksi dapat juga ditentukan melalui data perubahan entalpi pembentukan standar $open parentheses increment H subscript f degree close parentheses$ reaktan dan produknya, yaitu dari reaktan selisih harga $increment H subscript f degree$ reaktan dan $increment H subscript f degree$ produk.

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

Reaksi pembakaran $C S subscript 2$ :

$C S subscript 2 open parentheses italic g close parentheses and 3 O subscript 2 open parentheses italic g close parentheses yields C O subscript 2 open parentheses italic g close parentheses and 2 S O subscript 2 open parentheses italic g close parentheses$

Menghitung perubahan entalpi pembakaran $C S subscript 2$ :

$table attributes columnalign right center left columnspacing 0px end attributes row cell increment H end cell equals cell sum increment H subscript f degree space hasil space reaksi minus sign sum increment H subscript f degree pereaksi end cell row blank equals cell open curly brackets increment H subscript f degree space C O subscript 2 plus 2 left parenthesis increment H subscript f degree space S O subscript 2 right parenthesis close curly brackets minus sign open curly brackets increment H subscript f degree C S subscript 2 plus 3 left parenthesis increment H subscript f degree space O subscript 2 right parenthesis close curly brackets end cell row blank equals cell open curly brackets left parenthesis minus sign 394 right parenthesis plus 2 left parenthesis minus sign 297 right parenthesis close curly brackets minus sign open curly brackets left parenthesis plus 89 comma 5 right parenthesis plus 3 left parenthesis 0 right parenthesis close curly brackets end cell row blank equals cell negative sign 1077 comma 5 space kJ forward slash mol end cell end table$

Jadi, $increment H subscript c degree space C S subscript 2$ adalah $negative sign 1077 comma 5 space kJ forward slash mol$ .

Roboguru

Diketahui Hitunglah kalor yang dilepas pada pembakaran 6,72 liter gas pada suhu , 1 atm. (Ar C = 12, H = 1)

Pembahasan Soal:

Reaksi pembakaran $C subscript 2 H subscript 4$ :

$C subscript 2 H subscript 4 and 3 O subscript 2 yields 2 C O subscript 2 and 2 H subscript 2 O$

Perubahan entalpinya yaitu:

$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 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 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 C O subscript 2 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 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 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 subscript 2 end style begin inline style H subscript 4 end style begin inline style plus end style begin inline style 3 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 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 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 2 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 2 end style begin inline style left parenthesis end style begin inline style negative sign end style begin inline style 285 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 52 end style begin inline style plus end style 0 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 space begin inline style space end style begin inline style space space space space end style begin inline style equals end style begin inline style negative sign end style begin inline style 1358 end style begin inline style negative sign end style begin inline style 52 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 negative sign end style begin inline style 1410 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$

Jumlah mol gas $C subscript 2 H subscript 4$ yaitu:

$n equals fraction numerator V over denominator 22 comma 4 end fraction space space equals fraction numerator 6 comma 72 over denominator 22 comma 4 end fraction space space equals 0 comma 3 space mol$

Maka, jumlah kalor yang dilepaskan yaitu:

$q equals minus sign 1410 space kJ forward slash mol cross times 0 comma 3 space mol space space equals minus sign 423 space kJ$

Tanda negatif,menunjukan kalor dilepas dari sistem ke lingkungan.

Jadi, besarnya kalor yang dilepas adalah 423 kJ.

Roboguru

Pada pembakaran metanol dibebaskan kalor sebesar . Jika diketahui dan . a. tuliskan persamaan termokimianya. b. hitung .

Pembahasan Soal:

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

a. Reaksi pembakaran membutuhkan oksigen, oleh karena itu, persamaan termokimianya adalah

$C subscript 2 H subscript 5 O H and 3 O subscript 2 space yields 2 C O subscript 2 and 3 H subscript 2 O space space ΔH space equals space minus sign 1364 space Kj forward slash mol$

b. Menghitung $increment H subscript f degree space H subscript 2 O$ :

$table attributes columnalign right center left columnspacing 0px end attributes row cell increment H end cell equals cell sum increment H subscript f degree space hasil space reaksi minus sign sum increment H subscript f degree pereaksi end cell row cell increment H end cell equals cell open curly brackets 2 left parenthesis increment H subscript f degree space C O subscript 2 right parenthesis plus 3 left parenthesis increment H subscript f degree space H subscript 2 O right parenthesis close curly brackets minus sign open curly brackets left parenthesis increment H subscript f degree space C subscript 2 H subscript 5 O H right parenthesis plus 3 left parenthesis increment H subscript f degree space O subscript 2 right parenthesis close curly brackets end cell row cell 1364 space kJ forward slash mol end cell equals cell open curly brackets 2 left parenthesis minus sign 393 right parenthesis plus 3 left parenthesis increment H subscript f degree space H subscript 2 O right parenthesis close curly brackets kJ forward slash mol minus sign left parenthesis minus sign 277 kJ forward slash mol right parenthesis end cell row cell 3 left parenthesis increment H subscript f degree space H subscript 2 O right parenthesis end cell equals cell 1364 kJ forward slash mol minus sign 1063 kJ forward slash mol end cell row cell increment H subscript f degree space H subscript 2 O end cell equals cell 301 over 3 kJ forward slash mol end cell row blank equals cell 100 comma 33 space kJ forward slash mol end cell end table$

Dengan demikian, persamaan termokimia yang terjadi adalah $C subscript 2 H subscript 5 O H and 3 O subscript 2 space yields 2 C O subscript 2 and 3 H subscript 2 O space space ΔH space equals space minus sign 1364 space Kj forward slash mol$ dan $increment H subscript f degree space H subscript 2 O$ adalah $100 comma 33 space kj forward slash mol$ .

Roboguru

Diketahui: Jika 5,2 gram dibakar secara sempurna sesuai dengan persamaan: Akan dihasilkan kalor sebesar .... (Ar C=12, Ar H=1)

Pembahasan Soal:

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

$table attributes columnalign right center left columnspacing 0px end attributes row cell increment H end cell equals cell sum with blank below increment H subscript f superscript degree produk minus sign sum with blank below increment H subscript f superscript degree reaktan end cell row blank equals cell left parenthesis 4 x increment H subscript f superscript degree C O subscript 2 and 2 x increment H subscript f superscript degree H subscript 2 O right parenthesis minus sign left parenthesis 2 x increment H subscript f superscript degree C subscript 2 H subscript 2 and 5 x increment H subscript f superscript degree O subscript 2 right parenthesis end cell row blank equals cell left parenthesis 4 x left parenthesis minus sign 394 right parenthesis plus left parenthesis 2 x left parenthesis minus sign 242 right parenthesis right parenthesis minus sign left parenthesis 2 x 52 plus 0 right parenthesis end cell row blank equals cell left parenthesis minus sign 1576 minus sign 484 right parenthesis minus sign 104 end cell row blank equals cell negative sign 2.164 space kJ forward slash mol end cell end table$

Koefisien $C subscript 2 H subscript space 2 end subscript$ adalah 2 maka:

$table attributes columnalign right center left columnspacing 0px end attributes row cell increment H subscript c superscript degree end cell equals cell negative sign Q over n end cell row blank equals cell negative sign open parentheses fraction numerator negative sign 2164 space kJ over denominator 2 space mol end fraction close parentheses end cell row blank equals cell negative sign 1.082 space kJ forward slash mol end cell end table$

Setiap pembakaran 1 mol $C subscript 2 H subscript space 2 end subscript$ dilepaskan kalor sebesar 1.082 $kJ forward slash mol$ .

Perhitungan mol etuna

$Mr C_{2} H_{2} mol = = = = = = = (2 xAr C) + (2 xAr H) (2 x 12) + 2 x 1) 24 + 2 26 g / mol \frac{massa C _{2} H _{2}}{Mr C _{2} H _{2}} \frac{5 , 2 g}{26 g / mol} 0, 2 mol$

$△ H_{c} = = 0, 2 mol x - 1.082 kJ / mol - 216, 4 kJ$

Dengan demikian, perubahan entalpi pembakaran 52 gram $C subscript 2 H subscript space 2 end subscript$ adalah sebesar -216,4 kJ yang berarti kalor yang dihasilkan sebesar 216,4 kJ.

Jadi, tidak ada jawaban yang tepat.

Roboguru