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Pada suhu dan volume tertentu terbentuk kesetimban...

Pada suhu dan volume tertentu terbentuk kesetimbangan betikut.

begin mathsize 14px style 2 H Br open parentheses italic g close parentheses equilibrium H subscript 2 open parentheses italic g close parentheses and Br subscript 2 open parentheses italic g close parentheses end style

Jika derajat disosiasi HBr adalab 0,2 dan tekanan total 0,5 atm, tetapan kesetimbangan (Kp) reaksi tersebut adalah . . . .

  1. begin mathsize 14px style 1 comma 25 cross times 10 to the power of negative sign 2 end exponent end style 

  2. begin mathsize 14px style 1 comma 5625 cross times 10 to the power of negative sign 2 end exponent end style 

  3. begin mathsize 14px style 2 comma 50 cross times 10 to the power of negative sign 2 end exponent end style 

  4. begin mathsize 14px style 3 comma 125 cross times 10 to the power of negative sign 2 end exponent end style 

  5. begin mathsize 14px style 6 comma 25 cross times 10 to the power of negative sign 2 end exponent end style 

Jawaban:

Dimisalkan mol mula-mula HBr adalah 1 mol, maka

 

 begin mathsize 14px style alpha equals fraction numerator space mol space zat space terurai over denominator mol space zat space mula bond mula end fraction mol space zat space terurai space equals alpha space x space mol space zat space mula bond mula mol space zat space terurai space equals 0 comma 2 space x space 1 space mol mol space zat space terurai space equals 0 comma 2 space mol end style

begin mathsize 14px style n space total space equals space nHBr plus space n space H subscript 2 space plus space n space Br subscript 2 n space total space equals space 0 comma 1 plus space 0 comma 1 plus space 0 comma 8 space mol n space total space equals space 1 space mol P space total space equals space 0 comma 5 space atm P subscript HBr space equals fraction numerator mol space H Br over denominator mol space total end fraction x space P subscript tot space space space space space space space space space equals fraction numerator 0 comma 8 over denominator 1 end fraction x space 0 comma 5 space atm space space space space space space space space space equals 0 comma 4 space atm space P subscript H subscript 2 end subscript equals fraction numerator mol space H subscript 2 over denominator mol space total end fraction x space P subscript tot space space space space space space space space space equals fraction numerator 0 comma 1 over denominator 1 end fraction x space 0 comma 5 space atm space space space space space space space space space equals 0 comma 05 space atm P subscript Br subscript 2 end subscript equals fraction numerator mol space Br subscript 2 over denominator mol space total end fraction x space P subscript tot space space space space space space space space space equals fraction numerator 0 comma 1 over denominator 1 end fraction x space 0 comma 5 space atm space space space space space space space space space equals 0 comma 05 space atm space space end style 

 

begin mathsize 14px style K subscript italic p italic space equals fraction numerator P subscript H subscript 2 end subscript space x space P subscript Br subscript 2 end subscript over denominator P subscript Hbr squared end fraction K subscript italic p italic space equals fraction numerator 0 comma 05 space x space 0 comma 05 over denominator 0 comma 4 squared end fraction K subscript italic p italic space equals space 0 comma 015625 end style 
 

Jadi, jawaban yang benar adalah B.

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