Jika diketahui: △Hf∘​ CH4​(g)=−75 kJ mol−1△Hf∘​ CO2​(g)=−393 kJ mol−1△Hf∘​ H2​O(g)=−242 kJ mol−1  Maka △H reaksi pembakaran gas CH4​ menurut reaksi: CH4​(g)+2O2​(g)→CO2​(g)+2H2​O(g)  △Hf∘​=−226,7 kJ/mol   adalah ....

Pertanyaan

Jika diketahui:


begin mathsize 14px style 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 space kJ space mol to the power of negative sign 1 end exponent increment H subscript italic 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 end style 


Maka begin mathsize 14px style increment H end style reaksi pembakaran gas begin mathsize 14px style C H subscript 4 end style menurut reaksi:


begin mathsize 14px style C H subscript 4 open parentheses italic g close parentheses and 2 O subscript 2 open parentheses italic g close parentheses yields C O subscript 2 open parentheses italic g close parentheses and 2 H subscript 2 O open parentheses italic g close parentheses space space increment H subscript f degree equals minus sign 226 comma 7 space kJ forward slash mol end style  


adalah ....

  1. -1.040 kJ/mol

  2. -802 kJ/mol

  3. -445 kJ/mol

  4. +890 kJ/mol

  5. +1.040 kJ/mol

Y. Rochmawatie

Master Teacher

Jawaban terverifikasi

Jawaban

jawaban yang benar adalah B.space 

Pembahasan

Pembahasan

Pada soal diketahui data-data sebagai berikut:

  • begin mathsize 14px style 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 space kJ space mol to the power of negative sign 1 end exponent increment H subscript italic 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 end style
  • Reaksi pembakaran yang terjadi:

begin mathsize 14px style C H subscript 4 open parentheses italic g close parentheses and 2 O subscript 2 open parentheses italic g close parentheses yields C O subscript 2 open parentheses italic g close parentheses and 2 H subscript 2 O open parentheses italic g close parentheses space space increment H subscript f degree equals minus sign 226 comma 7 space kJ forward slash mol end style

Data yang akan dihitung adalah begin mathsize 14px style increment H end style reaksi pembakaran begin mathsize 14px style C H subscript 4 end style (begin mathsize 14px style increment H subscript c degree end style). Nilai begin mathsize 14px style increment H subscript c degree end style dapat dihitung berdasarkan persamaan berikut:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell increment H subscript c degree end cell equals cell open parentheses begin inline style sum with space below end style increment H subscript f degree space produk close parentheses minus sign open parentheses begin inline style sum with space below increment H subscript f degree space reaktan end style close parentheses end cell row cell increment H subscript c degree end cell equals cell open parentheses begin inline style increment H subscript f degree space C O subscript 2 plus 2 cross times increment H subscript f degree space H subscript 2 O end style close parentheses minus sign open parentheses begin inline style increment H subscript f degree space C H subscript 4 plus increment H subscript f degree space O subscript 2 end style close parentheses end cell row cell increment H subscript c degree end cell equals cell open parentheses begin inline style negative sign 393 plus 2 cross times open parentheses negative sign 242 close parentheses end style close parentheses minus sign open parentheses begin inline style negative sign 75 plus 0 end style close parentheses space kJ space mol to the power of negative sign 1 end exponent end cell row cell increment H subscript c degree end cell equals cell begin inline style negative sign end style begin inline style 393 end style begin inline style negative sign end style begin inline style 484 end style begin inline style plus end style begin inline style 75 end style begin inline style plus end style begin inline style 0 end style space kJ space mol to the power of negative sign 1 end exponent end cell row cell increment H subscript c degree end cell equals cell begin inline style negative sign end style 802 space kJ space mol to the power of negative sign 1 end exponent end cell row blank blank blank end table end style 

Pada perhitungan di atas, begin mathsize 14px style increment H subscript f degree space O subscript 2 end style adalah 0, karena begin mathsize 14px style O subscript 2 end style termasuk unsur bebas. Berdasarkan perhitungan yang telah dilakukan, nilai begin mathsize 14px style increment H end style reaksi pembakaran begin mathsize 14px style C H subscript 4 end style (begin mathsize 14px style increment H subscript c degree end style) adalah begin mathsize 14px style negative sign 802 space kJ space mol to the power of negative sign 1 end exponent end style.

Jadi, jawaban yang benar adalah B.space 

258

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