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Pada subu tertentu dalam wadah 1 L terjadi reaksi ...

Pada subu tertentu dalam wadah 1 L terjadi reaksi kesetimbangan berikut.


begin mathsize 14px style 2 C O F subscript 2 open parentheses italic g close parentheses equilibrium C O subscript 2 open parentheses italic g close parentheses and C F subscript 4 open parentheses italic g close parentheses end style 


Konsentrasi gas saat setimbang masing-masing sebesar 3 M. Jika ke dalam wadah tersebut ditambahkan 1 mol gas begin mathsize 14px style C O F subscript 2 end style, berapakah konsentrasi gas begin mathsize 14px style C O subscript 2 end style dan begin mathsize 14px style C F subscript 4 end style saat kesetimbangan baru tercapai?undefined 

  1. undefined 

  2. undefined 

Jawaban:

Konsentrasi gas saat setimbang masing-masing sebesar 3 M, ditambahkan 1 mol gas begin mathsize 14px style C O F subscript 2 end style.

begin mathsize 14px style K subscript c space equals space fraction numerator open square brackets C O subscript 2 close square brackets open square brackets C F subscript 4 close square brackets over denominator open square brackets C O F subscript 2 close square brackets squared end fraction K subscript c space equals fraction numerator left parenthesis 3 right parenthesis left parenthesis 3 right parenthesis over denominator 3 squared end fraction K subscript c space equals 1 end style 

ditambahkan 1 mol gas begin mathsize 14px style C O F subscript 2 end style, maka:

 

begin mathsize 14px style K subscript c space equals space fraction numerator open square brackets C O subscript 2 close square brackets open square brackets C F subscript 4 close square brackets over denominator open square brackets C O F subscript 2 close square brackets squared end fraction 1 space equals fraction numerator left parenthesis 3 plus 0 comma 5 x right parenthesis left parenthesis 3 plus 0 comma 5 x right parenthesis over denominator left parenthesis 4 minus sign x right parenthesis squared end fraction square root of 1 equals square root of fraction numerator left parenthesis 3 plus 0 comma 5 x right parenthesis left parenthesis 3 plus 0 comma 5 x right parenthesis over denominator left parenthesis 4 minus sign x right parenthesis squared end fraction end root 1 equals fraction numerator 3 plus 0 comma 5 x over denominator 4 minus sign x end fraction 4 minus sign x equals 3 plus 0 comma 5 x 1 comma 5 x equals 1 x equals 2 over 3  open square brackets C O subscript 2 close square brackets space equals space 3 plus 1 half x 2 over 3 space mol open square brackets C O subscript 2 close square brackets space equals 3 plus 0 comma 33 space mol open square brackets C O subscript 2 close square brackets space equals 3 comma 33 space mol open square brackets C F subscript 4 close square brackets space equals space 3 plus 1 half x 2 over 3 space mol open square brackets C F subscript 4 close square brackets space equals 3 plus 0 comma 33 space mol open square brackets C F subscript 4 close square brackets space equals 3 comma 33 space mol  end style \
 

Maka konsentrasi a 3,33 mol dan b adalah 3,33 mol

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