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Diketahui nilai △ H c 0 ​ CH 4 ​ ( g ) sebesar -889 kJ/mol, △ H f 0 ​ CH 4 ​ ( g ) bernilai -74.85 kJ/mol, △ H f 0 ​ C 3 ​ H 8 ​ ( g ) -103.85 kJ/mol, -285.83 kJ/mol maka nilai △ H c 0 ​ C 3 ​ H 8 ​ ( g ) sebesar…

Diketahui nilai  sebesar -889 kJ/mol,  bernilai -74.85 kJ/mol,  -103.85 kJ/mol, begin mathsize 14px style increment straight H subscript straight c superscript 0 space straight H subscript 2 open parentheses straight g close parentheses end style -285.83 kJ/mol maka nilai  sebesar…

  1. -2216.04 kJ/mol

  2. -2501.87 kJ/mol

  3. -1930.21 kJ/mol

  4. -2423.74 kJ/mol

  5. -2112.19 kJ/mol

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S. Hidayati

Master Teacher

Mahasiswa/Alumni Universitas Indonesia

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Pembahasan

Sebelum mengerjakan soal, tuliskan dulu reaksi dari perubahan entalpi yang diketahui Reaksi yang diinginkan adalah pembakaran standar propana yang reaksinya diperlukan di sisi reaktan, perlu membalik reaksi 3 diperlukan 3 mol di sisi produk, perlu kalikan 3 pada reaksi 1 tidak diperlukan, perlu kalikan 3 pada reaksi 2 untuk mengeliminasi kelebihan 2 di sisi produk, perlu membalik dan kalikan 2 pada reaksi 4 Setiap spesi sudah sesuai posisi dan jumlahnya, selanjutnya reaksi dijumlah sehingga ΔH ikut dijumlah ΔH = +103.85 kJ/mol + (-2667) kJ/mol + (-224.55) kJ/mol + 571.66 kJ/mol ΔH = -2216.04 kJ/mol

Sebelum mengerjakan soal, tuliskan dulu reaksi dari perubahan entalpi yang diketahui

begin mathsize 14px style increment straight H subscript straight c superscript 0 space CH subscript 4 left parenthesis straight g right parenthesis space space space space space CH subscript 4 left parenthesis straight g right parenthesis plus 2 space straight O subscript 2 left parenthesis straight g right parenthesis rightwards arrow CO subscript 2 left parenthesis straight g right parenthesis plus 2 space straight H subscript 2 straight O left parenthesis straight l right parenthesis space space space space space space space space ΔH equals negative 889 kJ divided by mol increment straight H subscript straight f superscript 0 space CH subscript 4 left parenthesis straight g right parenthesis space space space space space straight C left parenthesis straight s right parenthesis plus 2 space straight H subscript 2 left parenthesis straight g right parenthesis rightwards arrow CH subscript 4 left parenthesis straight g right parenthesis space space space space space space space space space space space space space space space space space space space space space space space space space space space space space ΔH equals negative 74.85 kJ divided by mol increment straight H subscript straight f superscript 0 space straight C subscript 3 straight H subscript 8 left parenthesis straight g right parenthesis space space space 3 space straight C left parenthesis straight s right parenthesis plus 4 space straight H subscript 2 left parenthesis straight g right parenthesis rightwards arrow straight C subscript 3 straight H subscript 8 left parenthesis straight g right parenthesis space space space space space space space space space space space space space space space space space space space space space space space space space ΔH equals negative 103.85 kJ divided by mol increment straight H subscript straight c superscript 0 space straight H subscript 2 open parentheses straight g close parentheses space space space space space space space straight H subscript 2 left parenthesis straight g right parenthesis plus 1 half straight O subscript 2 left parenthesis straight g right parenthesis rightwards arrow straight H subscript 2 straight O left parenthesis straight l right parenthesis space space space space space space space space space space space space space space space space space space space space space space space space space space space space ΔH equals negative 285.83 kJ divided by mol end style 

Reaksi yang diinginkan adalah pembakaran standar propana yang reaksinya

begin mathsize 14px style straight C subscript 3 straight H subscript 8 left parenthesis straight g right parenthesis space plus space 5 space straight O subscript 2 left parenthesis straight g right parenthesis rightwards arrow space 3 space CO subscript 2 left parenthesis straight g right parenthesis space plus space 4 space straight H subscript 2 straight O left parenthesis straight l right parenthesis end style 

begin mathsize 14px style straight C subscript 3 straight H subscript 8 end style diperlukan di sisi reaktan, perlu membalik reaksi 3

begin mathsize 14px style straight C subscript 3 straight H subscript 8 left parenthesis straight g right parenthesis rightwards arrow space 3 space straight C left parenthesis straight s right parenthesis space plus space 4 space straight H subscript 2 left parenthesis straight g right parenthesis space space space space space space space space space space space ΔH space equals space plus 103.85 space kJ divided by mol end style 

undefined diperlukan 3 mol di sisi produk, perlu kalikan 3 pada reaksi 1

begin mathsize 14px style 3 space CH subscript 4 left parenthesis straight g right parenthesis space plus space 6 space straight O subscript 2 left parenthesis straight g right parenthesis rightwards arrow space 3 space CO subscript 2 left parenthesis straight g right parenthesis space plus space 6 space straight H subscript 2 straight O left parenthesis straight l right parenthesis space space space space space space space space space ΔH space equals space minus 2667 space kJ divided by mol end style 

undefined  tidak diperlukan, perlu kalikan 3 pada reaksi 2 untuk mengeliminasi undefined 

begin mathsize 14px style 3 space straight C left parenthesis straight s right parenthesis space plus space 6 space straight H subscript 2 left parenthesis straight g right parenthesis rightwards arrow space 3 space CH subscript 4 left parenthesis straight g right parenthesis space space space space space space space space space space space space ΔH space equals space minus 224.55 space kJ divided by mol end style 

undefined kelebihan 2 di sisi produk, perlu membalik dan kalikan 2 pada reaksi 4

begin mathsize 14px style 2 space straight H subscript 2 straight O left parenthesis straight l right parenthesis rightwards arrow space 2 space straight H subscript 2 left parenthesis straight g right parenthesis space plus space straight O subscript 2 left parenthesis straight g right parenthesis space space space space space space space space space space ΔH space equals space plus 571.66 space kJ divided by mol end style 

Setiap spesi sudah sesuai posisi dan jumlahnya, selanjutnya reaksi dijumlah sehingga ΔH ikut dijumlah

undefined 

 

ΔH = +103.85 kJ/mol + (-2667) kJ/mol + (-224.55) kJ/mol + 571.66 kJ/mol

ΔH = -2216.04 kJ/mol

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