Roboguru

Diketahui  dari dan  adalah - 46,2 , + 90,4 dan - ...

Diketahui begin mathsize 14px style increment H subscript f end style dari begin mathsize 14px style N H subscript 3 open parentheses italic g close parentheses comma space N O open parentheses italic g close parentheses space end styledan begin mathsize 14px style H subscript 2 O open parentheses italic l close parentheses end style adalah - 46,2 , + 90,4 dan - 286 kJ/mol. Maka begin mathsize 14px style increment H subscript reaksi end style oksidasi amonia: begin mathsize 14px style 4 N H subscript 3 open parentheses italic g close parentheses and 5 O subscript 2 open parentheses italic g close parentheses yields 4 N O open parentheses italic g close parentheses and 6 H subscript 2 O open parentheses italic l close parentheses end style adalah ....

  1. - 149,4 kJundefined 

  2. + 330,2 kJundefined 

  3. - 597,6 kJundefined 

  4. - 1.169,6 kJundefined 

  5. -1.892,8 kJundefined 

Jawaban:

Reaksi yang diminta adalah:


begin mathsize 14px style 4 N H subscript 3 open parentheses italic g close parentheses and 5 O subscript 2 open parentheses italic g close parentheses yields 4 N O open parentheses italic g close parentheses and 6 H subscript 2 O open parentheses italic l close parentheses end style


Sehingga:
 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell increment H subscript reaksi end cell equals cell sum with blank below increment H subscript f degree subscript produk minus sign sum with blank below increment H subscript f degree subscript reaktan end cell row cell increment H subscript reaksi end cell equals cell open parentheses 4 increment H degree subscript f space N O plus 6 increment H degree subscript f space H subscript 2 O close parentheses minus sign open parentheses 4 increment H degree subscript f space N H subscript 3 plus 5 increment H degree subscript f space O subscript 2 close parentheses end cell row blank equals cell left parenthesis left parenthesis 4 cross times 90 comma 4 right parenthesis plus left parenthesis 6 cross times left parenthesis minus sign 286 right parenthesis right parenthesis right parenthesis space kJ minus sign left parenthesis left parenthesis 4 cross times left parenthesis minus sign 46 comma 2 right parenthesis right parenthesis plus left parenthesis 5 cross times 0 right parenthesis right parenthesis space kJ end cell row blank equals cell left parenthesis left parenthesis 361 comma 6 minus sign 1716 right parenthesis plus 184 comma 8 right parenthesis space kJ end cell row blank equals cell negative sign 1169 comma 6 space kJ end cell end table end style


Berdasarkan perhitungan tersebut, jawaban yang tepat adalah D

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