Roboguru

Pada suhu tertentu dalam ruang tertutup yang bertekanan 10 atm terdapat dalam keadaan setimbang 0,3 mol gas ; 0,1 mol gas  dan 0,1 mol gas  dengan reaksi:   Harga  pada suhu tersebut adalah ...

Pertanyaan

Pada suhu tertentu dalam ruang tertutup yang bertekanan 10 atm terdapat dalam keadaan setimbang 0,3 mol gas S O subscript 2; 0,1 mol gas S O subscript 3 dan 0,1 mol gas O subscript 2 dengan reaksi:

2 S O subscript 3 open parentheses italic g close parentheses equilibrium 2 S O subscript 2 open parentheses italic g close parentheses and O subscript 2 open parentheses italic g close parentheses 

Harga K subscript p pada suhu tersebut adalah ...space space space

  1. 36 atmspace space space

  2. 18 atmspace space space

  3. 9 atmspace space space

  4. 4,5 atmspace space space

  5. 0,05 atmspace space space

Pembahasan Soal:

Tetapan kesetimbangan open parentheses K subscript c close parentheses adalah perbandingan konsentrasi produk terhadap pereaksi, masing-masing dipangkatkan dengan koefisien reaksinya, dan merupakan bilangan konstan. 

Selain tetapan kesetimbangan yang berdasarkan konsentrasi, tetapan kesetimbangan untuk sistem kesetimbangan gas juga dapat dinyatakan berdasarkan tekanan parsial gas. Tetapan  kesetimbangan yang berdasarkan tekanan parsial disebut tetapan kesetimbangan tekanan parsial dan dinyatakan dengan K subscript P, yaitu perbandingan tekanan parsial dari produk terhadap pereaksi, masing-masing dipangkatkan dengan koefisien reaksinya.

Untuk reaksi:

2 S O subscript 3 open parentheses italic g close parentheses equilibrium 2 S O subscript 2 open parentheses italic g close parentheses and O subscript 2 open parentheses italic g close parentheses 

tetapan kesetimbangan tekanannya open parentheses K subscript P close parentheses yaitu:

K subscript P equals fraction numerator open parentheses P subscript S O subscript 2 end subscript close parentheses squared open parentheses P subscript O subscript 2 end subscript close parentheses over denominator open parentheses P subscript S O subscript 3 end subscript close parentheses squared end fraction 

Berikut langkah-langkah menghitung nilai K subscript P:

  1. Menghitung tekanan parsial masing-masing zat 

    n space total double bond n space S O subscript 3 and n space S O subscript 2 and n space O subscript 2 n space total equals left parenthesis 0 comma 1 plus 0 comma 3 plus 0 comma 1 right parenthesis space mol n space total equals 0 comma 5 space mol   P subscript S O subscript 3 end subscript equals fraction numerator n space S O subscript 3 over denominator n space total end fraction cross times P space total P subscript S O subscript 3 end subscript equals fraction numerator 0 comma 1 space mol over denominator 0 comma 5 space mol end fraction cross times 10 space atm P subscript S O subscript 3 end subscript equals 2 space atm   P subscript S O subscript 2 end subscript equals fraction numerator n space S O subscript 2 over denominator n space total end fraction cross times P space total P subscript S O subscript 2 end subscript equals fraction numerator 0 comma 3 space mol over denominator 0 comma 5 space mol end fraction cross times 10 space atm P subscript S O subscript 2 end subscript equals 6 space atm   P subscript O subscript 2 end subscript equals fraction numerator n space O subscript 2 over denominator n space total end fraction cross times P space total P subscript O subscript 2 end subscript equals fraction numerator 0 comma 1 space mol over denominator 0 comma 5 space mol end fraction cross times 10 space atm P subscript O subscript 2 end subscript equals 2 space atm 
     
  2. Menghitung nilai tetapan kesetimbangan tekanan open parentheses K subscript P close parentheses  

    K subscript P equals fraction numerator open parentheses P subscript S O subscript 2 end subscript close parentheses squared open parentheses P subscript O subscript 2 end subscript close parentheses over denominator open parentheses P subscript S O subscript 3 end subscript close parentheses squared end fraction K subscript P equals fraction numerator open parentheses 6 space atm close parentheses squared up diagonal strike open parentheses 2 space atm close parentheses end strike over denominator open parentheses 2 space atm close parentheses to the power of up diagonal strike 2 end exponent end fraction K subscript P equals fraction numerator 36 space atm squared over denominator 2 space atm end fraction K subscript P equals 18 space atm 
     


Jadi, jawaban yang tepat adalah B.space space space space

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

S. Lubis

Mahasiswa/Alumni Universitas Sumatera Utara

Terakhir diupdate 13 September 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

Di dalam ruangan tertutup yang volumenya 10 liter dipanaskan 5 gram . Pada 700 K,  terurai membentuk reaksi kesetimbangan: Volume  yang terdapat pada saat seimbang adalah 22,4 mL. Hitunglah derajat...

Pembahasan Soal:

table attributes columnalign right center left columnspacing 0px end attributes row cell Volume space C O subscript 2 space open parentheses Setimbang close parentheses end cell equals cell 22 comma 4 space mL end cell row blank equals cell 0 comma 0224 space L end cell row blank blank blank row V bold equals cell italic n bold space italic x bold space bold 22 bold comma bold 4 bold space italic L bold forward slash bold mol end cell row blank blank blank row n equals cell fraction numerator V over denominator 22 comma 4 space L forward slash mol end fraction end cell row blank equals cell fraction numerator 0 comma 0224 space L over denominator 22 comma 4 space L forward slash mol end fraction end cell row blank equals cell 0 comma 001 space mol end cell end table

 

table attributes columnalign right center left columnspacing 0px end attributes row cell mol space Ca C O subscript 3 end cell equals cell fraction numerator massa over denominator Mr space end fraction end cell row blank equals cell fraction numerator 5 space g over denominator 100 space g forward slash mol end fraction end cell row blank equals cell 0 comma 05 space mol end cell row blank blank blank end table

table attributes columnalign right center left columnspacing 0px end attributes row alpha equals cell fraction numerator mol space zat space terurai over denominator mol space zat space awal end fraction end cell row blank equals cell fraction numerator 0 comma 001 space mol over denominator 0 comma 05 space mol end fraction end cell row blank equals cell 0 comma 02 end cell end table

 

table attributes columnalign right center left columnspacing 0px end attributes row cell K subscript italic c end cell italic equals cell open square brackets C O subscript 2 close square brackets end cell row blank equals cell open parentheses fraction numerator 0 comma 001 space italic m italic o italic l over denominator 10 space italic L end fraction close parentheses end cell row blank equals cell 10 to the power of negative sign 4 end exponent end cell row blank blank blank row cell K italic p end cell equals cell Kc space open parentheses RT close parentheses to the power of increment n end exponent space end cell row blank equals cell 10 to the power of negative sign 4 end exponent left parenthesis 0 comma 082 space L space atm space mol to the power of negative sign 1 end exponent space K to the power of negative sign 1 end exponent space x space 700 space K right parenthesis to the power of 1 space end cell row blank equals cell 5 comma 74 x 10 to the power of negative sign 3 end exponent end cell end table

 

Jadi, derajat disosiasi open parentheses alpha close parentheses pada reaksi di atas adalah 0,02 dan nilai tetapan kesetimbangan open parentheses K subscript italic p close parentheses adalah 5 comma 74 x 10 to the power of negative sign 3 end exponent.

Roboguru

Dalam ruang 10 liter direaksikan 0,5 mol gas  dan 1,5 mol gas  hingga terjadi reaksi setimbang: Pada saat setimbang terdapat 0,25 mol gas  dan tekanannya 3 atm. Hitung nilai Kc dan Kp.

Pembahasan Soal:

Nilai Kc dan Kp dapat ditentukan melalui langkah sebagai berikut:

  • Menentukan Kc
    • menentukan mol saat setimbang
       
    • menentukan nilai Kc
      table attributes columnalign right center left columnspacing 0px end attributes row Kc equals cell fraction numerator open square brackets N H subscript 3 close square brackets squared over denominator open square brackets N subscript 2 close square brackets open square brackets H subscript 2 close square brackets cubed end fraction end cell row blank equals cell fraction numerator open parentheses begin display style fraction numerator 0 comma 5 over denominator 10 end fraction end style close parentheses squared over denominator open parentheses begin display style fraction numerator 0 comma 25 over denominator 10 end fraction end style close parentheses open parentheses begin display style fraction numerator 0 comma 75 over denominator 10 end fraction end style close parentheses cubed end fraction end cell row blank equals cell fraction numerator 2 comma 5 space x space 10 to the power of negative sign 3 end exponent over denominator 2 comma 5 space x space 10 to the power of negative sign 2 end exponent.4 comma 2 space x space 10 to the power of negative sign 4 end exponent end fraction end cell row blank equals cell fraction numerator 2 comma 5 space x space 10 to the power of negative sign 3 end exponent over denominator 1 comma 05 space x space 10 to the power of negative sign 5 end exponent end fraction end cell row blank equals 238 end table
  • Menentukan Kp
    • ​​​​​​​n space total space equals 0 comma 25 plus 0 comma 75 plus 0 comma 5 space equals space 1 comma 5 space mol
    • P space N subscript 2 equals fraction numerator n space N subscript 2 over denominator n subscript total end fraction space x space P subscript total equals fraction numerator 0 comma 25 over denominator 1 comma 5 end fraction space x space 3 space atm equals 2 space atm
    • P space H subscript 2 equals fraction numerator n space H subscript 2 over denominator n subscript total end fraction space x space P subscript total equals fraction numerator 0 comma 75 over denominator 1 comma 5 end fraction space x space 3 space atm equals 1 comma 5 space atm
    • P space N H subscript 3 equals fraction numerator n space N H subscript 3 over denominator n subscript total end fraction space x space P subscript total equals fraction numerator 0 comma 5 over denominator 1 comma 5 end fraction space x space 3 space atm equals 1 space atm
    • Kp equals fraction numerator open parentheses P space N H subscript 3 close parentheses squared over denominator P space N subscript 2 point open parentheses P space H subscript 2 close parentheses cubed end fraction equals fraction numerator open parentheses 1 close parentheses squared over denominator open parentheses 2 close parentheses open parentheses 1 comma 5 close parentheses cubed end fraction equals 1 comma 48 space x space 10 to the power of negative sign 1 end exponent

Jadi, nilai Kc sebesar 238 dan nilai Kp sebesar bold 1 bold comma bold 48 bold space italic x bold space bold 10 to the power of bold minus sign bold 1 end exponent.space​​​​​​​

Roboguru

Sebanyak 5,60 g karbon padat ditempatkan dalam wadah hampa udara bervolume 2,5 L. Kemudian ke dalam wadah tersebut dialirkan gas karbon dioksida dengan tekanan 1,50 atm pada 298 K. a. Hitunglah jumla...

Pembahasan Soal:

a point space jumlah space mol space C equals fraction numerator 5 comma 60 space g over denominator 12 comma 01 space g space mol to the power of negative sign 1 end exponent end fraction equals 0 comma 466 space mol 

Jumlah space mol space C O subscript 2 n equals fraction numerator P over denominator V RT end fraction equals fraction numerator left parenthesis 1 comma 5 space atm right parenthesis left parenthesis 2 comma 5 space L right parenthesis over denominator left parenthesis 0 comma 082 space L space atm space mol to the power of negative sign 1 end exponent space K to the power of negative sign 1 end exponent right parenthesis left parenthesis 298 space K right parenthesis end fraction n equals 0 comma 268 space mol 

b point space i point space P subscript 2 double bond P subscript 1 left parenthesis T subscript 2 over T subscript 1 right parenthesis space space space space space space space P subscript 2 equals 1 comma 5 space atm space open parentheses fraction numerator 1.100 space K over denominator 298 space K end fraction close parentheses space space space space space space space P subscript 2 equals 5 comma 54 space atm space space space ii point space P subscript t equals 1 comma 75 space left parenthesis 5 comma 54 space atm right parenthesis space space space space space space space P subscript t equals 9 comma 70 space atm space space space space space space space n subscript t equals fraction numerator P over denominator V RT end fraction space space space space space space space n subscript t equals fraction numerator left parenthesis 9 comma 70 space atm right parenthesis left parenthesis 2 comma 5 space L right parenthesis over denominator left parenthesis 0 comma 082 space L space atm space mol to the power of negative sign 1 end exponent space K to the power of negative sign 1 end exponent right parenthesis left parenthesis 1100 space K right parenthesis end fraction space space space space space space space n subscript t equals 0 comma 2680 space mol 
 

 


Pada saat kesetimbangan:


0 comma 268 equals left parenthesis 0 comma 153 minus sign italic x right parenthesis plus 2 italic x italic x equals 0 comma 115 space mol n space C O subscript 2 equals 0 comma 153 minus sign 0 comma 115 equals 0 comma 038 space mol n space C O equals 2 left parenthesis 0 comma 115 right parenthesis equals 0 comma 230 space mol  P subscript C O subscript 2 end subscript double bond X subscript C O subscript 2 end subscript point P subscript total P subscript C O subscript 2 end subscript equals open parentheses fraction numerator 0 comma 038 space mol over denominator 0 comma 268 space mol end fraction close parentheses left parenthesis 9 comma 70 space atm right parenthesis equals 1 comma 37 space atm  P subscript CO double bond P subscript t bond P subscript C O subscript 2 end subscript P subscript CO equals 9 comma 70 space atm minus sign 1 comma 37 space atm P subscript CO equals 8 comma 33 space atm 

iii point space K subscript p equals open parentheses P subscript CO close parentheses squared over P subscript C O subscript 2 end subscript  

iv point space K subscript p equals open parentheses P subscript CO close parentheses squared over P subscript C O subscript 2 end subscript equals fraction numerator left parenthesis 8 comma 33 right parenthesis squared over denominator left parenthesis 1 comma 37 right parenthesis end fraction equals 50 comma 65 


Jadi, jawaban yang paling tepat seperti cara di atas.space 

Roboguru

Dala ruang yang tekanannya 3 atm, dipanaskan 0,5 mol gas  dan 1,5 mol gas . Pada suhu 400 K, terjadi kesetimbangan:     Ternyata saat setimbang terdapat gas  sebanyak 0,25 mol. Hitunglah Kp dan Kc ...

Pembahasan Soal:

Pertama kita tentukan dulu nilai mol zat masing-masing pada saat kesetimbangan yaitu:
 


 

Selanjutnya kita tentukan nilai tekanan parsial zat masing-masing yaitu:
 

P space N subscript 2 double bond P subscript total cross times fraction numerator mol space N subscript 2 over denominator mol space total end fraction equals 3 cross times fraction numerator 0 comma 25 over denominator 1 comma 5 end fraction equals 0 comma 5 space atm P space H subscript 2 double bond P subscript total cross times fraction numerator mol space H subscript 2 over denominator mol space total end fraction equals 3 cross times fraction numerator 0 comma 75 over denominator 1 comma 5 end fraction equals 1 comma 5 space atm P space N H subscript 3 double bond P subscript total cross times fraction numerator mol space N H subscript 3 over denominator mol space total end fraction equals 3 cross times fraction numerator 0 comma 5 over denominator 1 comma 5 end fraction equals 1 space atm
 

Maka nilai Kp reaksi tersebut adalah:
 

table attributes columnalign right center left columnspacing 0px end attributes row Kp equals cell fraction numerator open parentheses P space N H subscript 3 close parentheses squared over denominator open parentheses P space N subscript 2 close parentheses middle dot open parentheses P space H subscript 2 close parentheses cubed end fraction end cell row blank equals cell fraction numerator open parentheses 1 close parentheses squared over denominator open parentheses 0 comma 5 close parentheses middle dot open parentheses 1 comma 5 close parentheses cubed end fraction end cell row blank equals cell 0 comma 59 end cell end table
 

Selanjutnya kita tentukan nilai Kc nya berdasarkan hubungan Kc dan Kp yaitu:
 

table attributes columnalign right center left columnspacing 0px end attributes row Kp equals cell Kc middle dot open parentheses R middle dot T close parentheses to the power of increment n end exponent end cell row Kc equals cell Kp over open parentheses R middle dot T close parentheses to the power of increment n end exponent end cell row blank equals cell fraction numerator 0 comma 59 over denominator open parentheses 0 comma 082 middle dot 400 close parentheses to the power of negative sign 2 end exponent end fraction end cell row blank equals cell 0 comma 59 middle dot left parenthesis 0 comma 082 middle dot 400 right parenthesis squared equals 0 comma 59 middle dot left parenthesis 1075 comma 84 right parenthesis end cell row blank equals cell 634 comma 7 end cell end table
 

Dengan demikian maka nilai Kp dan Kc secara berurutan adalah 0,59 dan 634,7.space

Roboguru

Besi karbonil,  adalah gas yang dalam wadah tertutup dapat terurai sesuai dengan persamaan reaksi kesetimbangan berikut ini:   Ke dalam suatu wadah dimasukan sejumlah  sehingga konsentrasinya 0,47 M. ...

Pembahasan Soal:

a. Mengidentifikasi tekanan awal pada reaksi tersebut.


P subscript awal double bond MRT P subscript awal equals left parenthesis 0 comma 47 M right parenthesis left parenthesis 0 comma 082 L point atm point mol to the power of negative sign 1 end exponent K to the power of negative sign 1 end exponent right parenthesis left parenthesis 298 K right parenthesis P subscript awal equals 11 comma 5 space atm 


b. Mengidentifikasi left square bracket Fe subscript 2 open parentheses C O close parentheses subscript 9 right square bracket pada keadaan kesetimbangan.


 

K subscript c equals fraction numerator left square bracket Fe subscript 2 open parentheses C O close parentheses subscript 9 right square bracket open square brackets C O close square brackets over denominator left square bracket Fe open parentheses C O close parentheses subscript 5 right square bracket to the power of 2 end exponent end fraction 9 equals fraction numerator left parenthesis 0 comma 5 italic x right parenthesis left parenthesis 0 comma 5 italic x right parenthesis over denominator left parenthesis 0 comma 47 minus sign italic x right parenthesis squared end fraction 3 equals fraction numerator left parenthesis 0 comma 5 italic x right parenthesis over denominator left parenthesis 0 comma 47 minus sign italic x right parenthesis end fraction italic x equals 0 comma 403 space M 
 

Dengan demikian, pada keadaan setimbang:


left square bracket Fe subscript 2 open parentheses C O close parentheses subscript 9 left parenthesis italic g right parenthesis right square bracket equals left square bracket C O open parentheses italic g close parentheses right square bracket equals 0 comma 5 italic x 0 comma 5 italic x equals 0 comma 5 cross times 0 comma 403 space M equals 0 comma 2 space M 


c. Pada keadaan setimbang:


M space Fe open parentheses C O close parentheses subscript 5 equals M subscript awal bond M subscript reaksi M space Fe open parentheses C O close parentheses subscript 5 equals 0 comma 470 space M minus sign 0 comma 403 space M M space Fe open parentheses C O close parentheses subscript 5 equals 0 comma 067 space M  K subscript p double bond K subscript c open parentheses RT close parentheses to the power of increment n end exponent space K subscript p double bond K subscript c open parentheses RT close parentheses to the power of left parenthesis 1 plus 1 right parenthesis minus sign 2 end exponent K subscript p double bond K subscript c open parentheses RT close parentheses to the power of 0 space K subscript p double bond K subscript c K subscript p equals 9 


d. Pada keadaan setimbang: 


P subscript total double bond M subscript total RT P subscript total equals left parenthesis left square bracket 0 comma 2 plus 0 comma 2 plus 0 comma 067 right square bracket M right parenthesis left parenthesis 0 comma 082 begin inline style fraction numerator L point atm over denominator mol point K end fraction end style right parenthesis left parenthesis 298 K right parenthesis P subscript total equals 11 comma 5 space atm 


Jadi, jawaban yang paling tepat seperti cara di atas.space 

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

Tapi Roboguru masih mau belajar. Menurut kamu pembahasan kali ini sudah membantu, belum?

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