Roboguru

Diketahui kalor pembakaran  adalah  serta kalor pembentukan  dan  berturut-turut  dan . Berdasarkan data tersebut, pernyataan berikut yang benar adalah ....

Pertanyaan

Diketahui kalor pembakaran begin mathsize 14px style C H subscript 3 O H end style adalah begin mathsize 14px style negative sign 726 space kJ space mol to the power of negative sign 1 end exponent end style serta kalor pembentukan begin mathsize 14px style C O subscript 2 open parentheses italic g close parentheses end style dan begin mathsize 14px style H subscript 2 O open parentheses italic g close parentheses end style berturut-turut begin mathsize 14px style negative sign 393 comma 5 space kJ space mol to the power of negative sign 1 end exponent end style dan begin mathsize 14px style negative sign 285 comma 8 space kJ space mol to the power of negative sign 1 end exponent end style. Berdasarkan data tersebut, pernyataan berikut yang benar adalah ....

  1. Kalor pembentukan standar metanol sebesar begin mathsize 14px style negative sign 239 comma 1 space kJ space mol to the power of negative sign 1 end exponent end style.

  2. Penguraian standar gas begin mathsize 14px style C O subscript 2 end style membutuhkan kalor sebesar 39,35 kJ.

  3. Pembentukan 36 gram begin mathsize 14px style H subscript 2 O end style melepaskan kalor sebesar 285,8 kJ.

  4. Persamaan termokimia pembakaran metanol adalah: begin mathsize 14px style C H subscript 3 O H open parentheses italic l close parentheses and O subscript 2 open parentheses italic g close parentheses yields C O subscript 2 open parentheses italic g close parentheses and H subscript 2 O open parentheses italic l close parentheses space increment H equals plus 726 space kJ space mol to the power of negative sign 1 end exponent end style 

  5. Pembakaran 80 gram undefined melepaskan kalor sebesar 1.452 kJ

Pembahasan Soal:

Reaksi pembakaran begin mathsize 14px style C H subscript 3 O H end style 

begin mathsize 14px style C H subscript 3 O H plus begin inline style bevelled 3 over 2 end style O subscript 2 yields C O subscript 2 and 2 H subscript 2 O space space increment H equals minus sign 726 space kJ space mol to the power of negative sign 1 end exponent end style 

Pilihan jawaban D salah, karena reaksi belum setara dan begin mathsize 14px style increment H end style reaksi seharusnya bernilai negatif.

Pilihan jawaban E salah, seharusnya 1,815 kJ, karena 80 gram undefined adalah 2,5 mol.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row mol equals cell massa over M subscript r end cell row blank equals cell 80 over 32 end cell row blank equals cell 2 comma 5 space mol end cell end table end style 

Maka,

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell increment H end cell equals cell 2 comma 5 cross times left parenthesis minus sign 726 space kJ space mol to the power of negative sign 1 end exponent right parenthesis end cell row blank equals cell negative sign 1.815 space kJ end cell end table end style 

Reaksi pembentukan begin mathsize 14px style C O subscript 2 end style 

begin mathsize 14px style C and O subscript 2 yields C O subscript 2 space space increment H equals minus sign 393 comma 5 space kJ space mol to the power of negative sign 1 end exponent end style 

Pilihan jawaban B salah, seharusnya begin mathsize 14px style plus 393 comma 5 space kJ space mol to the power of negative sign 1 end exponent end style, karena reaksi penguraiannya adalah kebalikan dari reaksi pembentukan.

begin mathsize 14px style C O subscript 2 yields C and O subscript 2 space space increment H equals plus 393 comma 5 space kJ space mol to the power of negative sign 1 end exponent end style 

Reaksi pembentukan begin mathsize 14px style H subscript 2 O end style 

begin mathsize 14px style H subscript 2 plus begin inline style bevelled 1 half end style O subscript 2 yields H subscript 2 O space space increment H equals minus sign 285 comma 8 space kJ space mol to the power of negative sign 1 end exponent end style 

Pilihan jawaban C salah, seharusnya begin mathsize 14px style 571 comma 6 space kJ space mol to the power of negative sign 1 end exponent end style, karena 36 gram begin mathsize 14px style H subscript 2 O end style adalah 2 mol.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row mol equals cell massa over M subscript r end cell row blank equals cell 36 over 18 end cell row blank equals cell 2 space mol end cell end table end style 

Maka,

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell increment H end cell equals cell 2 cross times left parenthesis minus sign 285 comma 8 space kJ space mol to the power of negative sign 1 end exponent right parenthesis end cell row blank equals cell negative sign 571 comma 6 space kJ space mol to the power of negative sign 1 end exponent end cell end table end style 

Pilihan A:

Pembentukan standar metanol:

begin mathsize 14px style C plus begin inline style bevelled 1 half end style O subscript 2 and 2 H subscript 2 yields C H subscript 3 O H space space space increment H equals... ? end style 

 begin mathsize 14px style C H subscript 3 O H plus begin inline style bevelled 3 over 2 end style O subscript 2 yields C O subscript 2 and 2 H subscript 2 O space space increment H equals minus sign 726 space kJ space mol to the power of negative sign 1 end exponent open parentheses dibalik close parentheses C and O subscript 2 yields C O subscript 2 space space increment H equals minus sign 393 comma 5 space kJ space mol to the power of negative sign 1 end exponent open parentheses tetap close parentheses H subscript 2 plus begin inline style bevelled 1 half end style O subscript 2 yields H subscript 2 O space space increment H equals minus sign 285 comma 8 space kJ space mol to the power of negative sign 1 end exponent left parenthesis cross times 2 right parenthesis end style 

Sehingga:

begin mathsize 14px style bottom enclose C O subscript 2 and 2 H subscript 2 O yields C H subscript 3 O H and begin inline style bevelled 3 over 2 end style O subscript 2 space space increment H equals plus 726 space kJ space mol to the power of negative sign 1 end exponent C and O subscript 2 yields C O subscript 2 space space increment H equals minus sign 393 comma 5 space kJ space mol to the power of negative sign 1 end exponent 2 H subscript 2 and O subscript 2 yields 2 H subscript 2 O space space increment H equals minus sign 571 comma 6 space kJ space mol to the power of negative sign 1 end exponent end enclose C plus begin inline style bevelled 1 half end style O subscript 2 and 2 H subscript 2 yields C H subscript 3 O H space space space increment H equals minus sign 239 comma 1 space kJ space mol to the power of negative sign 1 end exponent end style 

Jadi, jawaban yang benar adalah A.   

Pembahasan terverifikasi oleh Roboguru

Terakhir diupdate 12 Maret 2021

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