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Sebanyak 1 mol A dan 1 mol B direaksikan hingga te...

Sebanyak 1 mol A dan 1 mol B direaksikan hingga tercapai kesetimbangan. Pada kesetimbangan diperoleh zat A sebanyak 0,33 mol, maka harga tetapan kesetimbangan dari reaksi begin mathsize 14px style A and B equilibrium C and D end style adalah...undefined 

  1. 0,25undefined 

  2. 0,33undefined 

  3. 0,67undefined 

  4. 2,00undefined 

  5. 4,00undefined 

Jawaban:

Gunakan persamaan m, r, s untuk menentukan konsentrasi pada kondisi setimbang:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell space space space space space space space space A space space space space space space space plus space space space space space space space B space space space space space space space end cell rightwards harpoon over leftwards harpoon cell space space space space space space space C space space space space space space plus space space space space space space D end cell row cell mula bond mula end cell equals cell space space space space space 1 comma 0 space mol space space space space space space space 1 comma 0 space mol end cell row cell reaksi space end cell equals cell space minus sign 0 comma 67 space mol space space minus sign 0 comma 67 space mol space space plus 0 comma 67 space mol space space space plus 0 comma 67 space mol end cell row cell setimbang space end cell equals cell space space space space 0 comma 33 space mol space space space space space 0 comma 33 space mol space space space space space 0 comma 67 space mol space space space space space space space 0 comma 67 space mol end cell end table end style   

Dimisalkan volumenya adalah satu, maka 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open square brackets C close square brackets space end cell equals cell open square brackets D close square brackets equals space fraction numerator 0 comma 67 space mol over denominator 1 space L end fraction space equals 0 comma 67 space M end cell row cell open square brackets A close square brackets space end cell equals cell space open square brackets B close square brackets equals fraction numerator 0 comma 33 space mol over denominator 1 space L end fraction space equals 0 comma 33 space M end cell row blank blank blank row cell K subscript c space end cell equals cell space fraction numerator open square brackets C close square brackets open square brackets D close square brackets over denominator open square brackets A close square brackets open square brackets B close square brackets end fraction end cell row cell K subscript c space end cell equals cell space fraction numerator 0 comma 67 cross times 0 comma 67 over denominator 0 comma 33 cross times 0 comma 33 end fraction end cell row cell K subscript c space end cell equals cell 4 comma 12 end cell end table end style 

Jadi, jawaban yang paling mendekati adalah E.

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