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Dalam ruang 2 liter dimasukkan 16 gram gas  dan terurai sebanyak 40% menurut reaksi : Nilai tetapan kesetimbangan  reaksi tersebut adalah .... (Ar S = 32, Ar O = 16)

Pertanyaan

Dalam ruang 2 liter dimasukkan 16 gram gas begin mathsize 14px style S O subscript 3 end style dan terurai sebanyak 40% menurut reaksi :


begin mathsize 14px style 2 S O subscript 3 subscript open parentheses g close parentheses end subscript space rightwards harpoon over leftwards harpoon space 2 S O subscript 2 subscript open parentheses g close parentheses end subscript space plus space O subscript 2 subscript open parentheses g close parentheses end subscript end style


Nilai tetapan kesetimbangan begin mathsize 14px style K subscript c end style reaksi tersebut adalah .... (Ar S = 32, Ar O = 16)space 

Pembahasan Soal:

Nilai tetapan kesetimbangan begin mathsize 14px style K subscript c end style reaksi menyatakan perbandingan hasil kali konsentrasi reaktan dan produk dalam keadaan setimbang dipangkatkan koefisien reaksinya masing-masing.

Dalam menentukan begin mathsize 14px style K subscript c end style dapat dilakukan dengan cara sebagai berikut.

  • Menentukan mol begin mathsize 14px style S O subscript 3 end style mula-mula dan yang bereaksi


begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell Mr space S O subscript 3 end cell equals cell 1 point Ar space S and 3 point Ar space O end cell row blank equals cell 1.32 plus 3.16 end cell row blank equals cell 32 plus 48 end cell row blank equals 80 row blank blank blank row cell n space S O subscript 3 space mula bond mula end cell equals cell fraction numerator massa space S O subscript 3 over denominator Mr space S O subscript 3 end fraction end cell row blank equals cell fraction numerator 16 space g over denominator 80 space begin display style bevelled g over mol end style end fraction end cell row blank equals cell 0 comma 2 space mol end cell row blank blank blank row cell n space S O subscript 3 space yang space bereaksi end cell equals cell alpha space. space n space S O subscript 3 space mula bond mula end cell row blank equals cell 40 percent sign space. space 0 comma 2 space mol end cell row blank equals cell 0 comma 08 space mol end cell end table end style

 

  • Membuat M-R-S untuk menentukan mol semua senyawa saat setimbang


 

  • Menentukan konsentrasi masing-masing senyawa saat setimbang


begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open square brackets... close square brackets end cell equals cell M equals n over V space end cell row blank blank blank row cell open square brackets S O subscript 3 close square brackets end cell equals cell fraction numerator n space S O subscript 3 space over denominator V end fraction end cell row blank equals cell fraction numerator 0 comma 12 space mol over denominator 2 space L end fraction end cell row blank equals cell 0 comma 06 end cell row blank equals cell 6.10 to the power of negative sign 2 end exponent end cell row blank blank blank row cell open square brackets S O subscript 2 close square brackets end cell equals cell fraction numerator n space S O subscript 2 over denominator V end fraction end cell row blank equals cell fraction numerator 0 comma 08 over denominator 2 end fraction end cell row blank equals cell 0 comma 04 end cell row blank equals cell 4.10 to the power of negative sign 2 end exponent end cell row blank blank blank row cell open square brackets O subscript 2 close square brackets end cell equals cell fraction numerator n space O subscript 2 over denominator V end fraction end cell row blank equals cell fraction numerator 0 comma 04 over denominator 2 end fraction end cell row blank equals cell 0 comma 02 end cell row blank equals cell 2.10 to the power of negative sign 2 end exponent end cell end table end style

 

  • Menghitung begin mathsize 14px style K subscript c end style dengan rumus yaitu :


begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell K subscript c end cell equals cell fraction numerator open square brackets S O subscript 2 close square brackets squared space. space open square brackets O subscript 2 close square brackets over denominator open square brackets S O subscript 3 close square brackets squared end fraction end cell row blank equals cell fraction numerator left parenthesis 4.10 to the power of negative sign 2 end exponent right parenthesis squared space. space left parenthesis 2.10 to the power of negative sign 2 end exponent right parenthesis over denominator left parenthesis 6.10 to the power of negative sign 2 end exponent right parenthesis squared end fraction end cell row blank equals cell fraction numerator 16.10 to the power of negative sign 4 end exponent space. space 2.10 to the power of negative sign 2 end exponent over denominator 36.10 to the power of negative sign 4 end exponent end fraction end cell row blank equals cell fraction numerator 8.10 to the power of negative sign 2 end exponent over denominator 9 end fraction end cell row blank equals cell 0 comma 89.10 to the power of negative sign 2 end exponent end cell row blank equals cell 8 comma 9.10 to the power of negative sign 3 end exponent end cell end table end style


Jadi, nilai tetapan kesetimbangan begin mathsize 14px style K subscript c end style reaksi tersebut adalah begin mathsize 14px style 8 comma 9.10 to the power of negative sign 3 end exponent end style.space​​​​​​​ 

Pembahasan terverifikasi oleh Roboguru

Terakhir diupdate 03 April 2021

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