Roboguru

Diketahui persamaan termokimia sebagai berikut: Berdasarkan data di atas maka perubahan entalpi untuk reaksi:

Diketahui persamaan termokimia sebagai berikut:


begin mathsize 14px style C open parentheses italic s close parentheses and O subscript 2 open parentheses italic g close parentheses yields C O subscript 2 open parentheses italic g close parentheses space increment H equals minus sign 393 comma 5 space kJ 2 Ca O open parentheses italic s close parentheses yields 2 Ca open parentheses italic s close parentheses and O subscript 2 open parentheses italic g close parentheses space increment H equals plus 1269 space kJ Ca open parentheses italic s close parentheses and C open parentheses italic s close parentheses plus begin inline style bevelled 3 over 2 end style O subscript 2 open parentheses italic g close parentheses yields Ca C O subscript 3 open parentheses italic s close parentheses space increment H equals minus sign 1207 comma 5 space kJ end style


Berdasarkan data di atas maka perubahan entalpi untuk reaksi:


begin mathsize 14px style Ca C O subscript 3 open parentheses italic s close parentheses yields Ca O open parentheses italic s close parentheses and C O subscript 2 open parentheses italic g close parentheses end style

  1. begin mathsize 14px style plus 1.028 comma 0 space kJ space mol to the power of negative sign 1 end exponent end style

  2. begin mathsize 14px style negative sign 179 comma 5 space kJ space mol to the power of negative sign 1 end exponent end style

  3. begin mathsize 14px style negative sign 1.028 comma 0 space kJ space mol to the power of negative sign 1 end exponent end style

  4. begin mathsize 14px style plus 1.207 comma 5 space kJ space mol to the power of negative sign 1 end exponent end style

  5. begin mathsize 14px style plus 179 comma 5 space kJ space mol to the power of negative sign 1 end exponent end style

Jawaban:

Reaksi yang ditanyakan: begin mathsize 14px style Ca C O subscript 3 open parentheses italic s close parentheses yields Ca O open parentheses italic s close parentheses and C O subscript 2 open parentheses italic g close parentheses end style

Nilai begin mathsize 14px style increment H end style untuk reaksi tersebut dapat ditentukan berdasarkan reaksi yang diketahui:


begin mathsize 14px style C open parentheses italic s close parentheses and O subscript 2 open parentheses italic g close parentheses yields C O subscript 2 open parentheses italic g close parentheses space increment H equals minus sign 393 comma 5 space kJ Ca open parentheses italic s close parentheses plus begin inline style bevelled 1 half end style O subscript 2 open parentheses italic g close parentheses yields Ca O open parentheses italic s close parentheses space increment H equals begin inline style fraction numerator negative sign 1.269 over denominator 2 end fraction end style space kJ equals minus sign 634 comma 5 space kJ Ca C O subscript 3 open parentheses italic s close parentheses yields Ca open parentheses italic s close parentheses and C open parentheses italic s close parentheses and begin inline style bevelled 3 over 2 end style O subscript 2 open parentheses italic g close parentheses space increment H equals plus 1.207 comma 5 space kJ end style


Maka nilai begin mathsize 14px style increment H end style reaksi tersebut adalah:


begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell increment H end cell equals cell left parenthesis left parenthesis minus sign 393 comma 5 right parenthesis plus left parenthesis minus sign 634 comma 5 right parenthesis plus left parenthesis 1 comma 207 comma 5 right parenthesis right parenthesis space kJ end cell row blank equals cell plus 179 comma 5 space kJ end cell end table end style


Jadi, jawaban yang benar adalah E.

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