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Diketahui reaksi kesetimbangan: Satu mol  terurai dalam ruang 5 liter. Bila , tentukan  bila .

Pertanyaan

Diketahui reaksi kesetimbangan:

2 I Cl open parentheses italic g close parentheses yields and is yielded by I subscript 2 open parentheses italic g close parentheses and Cl subscript 2 open parentheses italic g close parentheses

Satu mol I Cl terurai dalam ruang 5 liter. Bila K subscript italic c equals 0 comma 25, tentukan K subscript italic p bila P subscript italic t italic o italic t italic a italic l end subscript equals 2 space atm.space

Pembahasan Soal:

  • Menentukan mol dari persamaan reaksi

  • mencari nilai x dari nilai K subscript italic c equals 0 comma 25

table attributes columnalign right center left columnspacing 0px end attributes row cell K subscript italic c italic space end cell equals cell fraction numerator open square brackets I subscript 2 close square brackets open square brackets Cl subscript 2 close square brackets over denominator open square brackets I C I close square brackets squared end fraction end cell row cell 0 comma 25 end cell equals cell fraction numerator open square brackets x over 5 close square brackets open square brackets x over 5 close square brackets over denominator open square brackets fraction numerator 1 minus sign 2 x over denominator 5 end fraction close square brackets squared end fraction end cell row cell 0 comma 25 space left parenthesis 1 minus sign 4 x plus 4 x squared right parenthesis end cell equals cell x squared end cell row cell 0 comma 25 space minus sign space x space plus space x squared end cell equals cell x squared end cell row cell x space end cell equals cell 0 comma 25 end cell end table

  • Menentukan nilai Mol I subscript 2 dan Cl subscript 2
    Pada saat reaksi setara mol I subscript 2 = x dan Cl subscript 2 = x, sehingga nilai Mol I subscript 2 dan Cl subscript 2 sama yaitu sebesar 0,25
     
  • Menentukan nilai K subscript italic p jika mol  I subscript 2 dan Cl subscript 2 sebesar 0,25

P space I C I space equals fraction numerator mol space I C I over denominator mol space total end fraction space x space P subscript italic t italic o italic t italic a italic l end subscript italic space equals fraction numerator 0 comma 5 over denominator 1 end fraction x 2 space equals 1 P space I subscript 2 space equals fraction numerator mol space I subscript 2 over denominator mol space total end fraction space x space P subscript italic t italic o italic t italic a italic l end subscript italic space equals fraction numerator 0 comma 25 over denominator 1 end fraction x 2 space equals 0 comma 5 P space Cl subscript 2 space equals fraction numerator mol space Cl subscript 2 over denominator mol space total end fraction space x space P subscript italic t italic o italic t italic a italic l end subscript italic space equals fraction numerator 0 comma 25 over denominator 1 end fraction x 2 space equals 0 comma 5

table attributes columnalign right center left columnspacing 0px end attributes row cell K subscript italic p end cell equals cell fraction numerator open square brackets P subscript I subscript 2 end subscript close square brackets open square brackets P subscript Cl subscript 2 end subscript close square brackets over denominator open square brackets P subscript ICI close square brackets squared end fraction end cell row cell K subscript italic p end cell equals cell fraction numerator open square brackets 0 comma 5 close square brackets open square brackets 0 comma 5 close square brackets over denominator open square brackets 1 close square brackets squared end fraction end cell row cell K subscript italic p end cell equals cell 0 comma 25 end cell end table
 

Jadi, nilai K subscript italic p yang terbentuk sebesar 0,25.space

Pembahasan terverifikasi oleh Roboguru

Terakhir diupdate 02 Juni 2021

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