Roboguru
SD

Dalam ruang 10 liter direaksikan 0,5 mol gas N2​ dan 1,5 mol gas H2​ hingga terjadi reaksi setimbang: N2​(g) + 3H2​(g) ⇌ 2NH3​(g)  Pada saat setimbang terdapat 0,25 mol gas dan tekanannya 3 atm. Hitunglah nilai Kc​ dan Kp​.

Pertanyaan

Dalam ruang 10 liter direaksikan 0,5 mol gas begin mathsize 14px style N subscript 2 end style dan 1,5 mol gas begin mathsize 14px style H subscript 2 end style hingga terjadi reaksi setimbang:


begin mathsize 14px style N subscript 2 italic open parentheses italic g italic close parentheses space plus space 3 H subscript 2 italic open parentheses italic g italic close parentheses space rightwards harpoon over leftwards harpoon space 2 N H subscript 3 italic open parentheses italic g italic close parentheses end style 


Pada saat setimbang terdapat 0,25 mol gas undefined dan tekanannya 3 atm. Hitunglah nilai begin mathsize 14px style K subscript c end style dan begin mathsize 14px style K subscript p end style.

N. Puspita

Master Teacher

Jawaban terverifikasi

Jawaban

nilai Kc dan Kp berturut-turut adalah table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell bold 237 bold comma bold 037 bold space italic M to the power of bold minus sign bold 2 end exponent end cell end table dan bold 0 bold comma bold 593 bold space bold atm to the power of bold minus sign bold 2 end exponent.space

Pembahasan

Kc merupakan tetapan kesetimbangan kosentrasi sehingga diperlukan konsentrasi setiap zat saat kesetimbangan.

  • Menentukan reaksi mrs
    space space space space space space space space space space space space space space space space space space space space space space space space space N subscript 2 open parentheses italic g close parentheses space plus space 3 H subscript 2 open parentheses italic g close parentheses space rightwards harpoon over leftwards harpoon space 2 N H subscript 3 open parentheses italic g close parentheses Mula bond mula space space colon 0 comma 5 mol space space space space 1 comma 5 mol bottom enclose Reaksi space space space space space space space space space space space colon 0 comma 25 mol space space 0 comma 75 mol space space space 0 comma 5 mol end enclose bottom enclose space space plus end enclose Setimbang space space space space colon 0 comma 25 mol space space 0 comma 75 mol space space space 0 comma 5 mol  
     
  • Menentukan konsentrasi zat saat kesetimbangan
    table attributes columnalign right center left columnspacing 0px end attributes row cell M subscript N subscript 2 end subscript end cell equals cell n over v end cell row blank equals cell fraction numerator 0 comma 25 over denominator 10 end fraction end cell row blank equals cell 0 comma 025 M end cell row cell M subscript H subscript 2 end subscript end cell equals cell n over v end cell row blank equals cell fraction numerator 0 comma 75 over denominator 10 end fraction end cell row blank equals cell 0 comma 075 M end cell row cell M subscript N H subscript 3 end subscript end cell equals cell n over v end cell row blank equals cell fraction numerator 0 comma 5 over denominator 10 end fraction end cell row blank equals cell 0 comma 05 M end cell end table 
     
  • Menentukan nilai Kc
    table attributes columnalign right center left columnspacing 0px end attributes row cell K subscript c end cell 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 left square bracket 0 comma 05 space M right square bracket squared over denominator left square bracket 0 comma 025 space M right square bracket left square bracket 0 comma 075 space M right square bracket cubed end fraction end cell row blank equals cell 237 comma 037 space M to the power of negative sign 2 end exponent end cell end table 
     
  • Menentukan tekanan parsial masing-masing gas
    table attributes columnalign right center left columnspacing 0px end attributes row cell mol space total end cell equals cell mol space N subscript 2 and mol space H subscript 2 and mol space N H subscript 3 end cell row blank equals cell 0 comma 25 mol plus 0 comma 75 mol plus 0 comma 5 mol end cell row blank equals cell 1 comma 5 mol end cell end table
    table attributes columnalign right center left columnspacing 0px end attributes row cell P subscript N subscript 2 end subscript end cell equals cell fraction numerator mol space N subscript 2 over denominator mol space total end fraction cross times P space total end cell row blank equals cell fraction numerator 0 comma 25 over denominator 1 comma 5 end fraction cross times 3 space atm end cell row blank equals cell 0 comma 5 space atm end cell row cell P subscript H subscript 2 end subscript end cell equals cell fraction numerator 0 comma 75 over denominator 1 comma 5 end fraction cross times 3 space atm end cell row blank equals cell 1 comma 5 space atm end cell row cell P subscript N H subscript 3 end subscript end cell equals cell fraction numerator 0 comma 5 over denominator 1 comma 5 end fraction cross times 3 space atm end cell row blank equals cell 1 space atm end cell end table
     
  • Menentukan nilai Kp
    table attributes columnalign right center left columnspacing 0px end attributes row cell K subscript p end cell equals cell fraction numerator P subscript N H subscript 3 end subscript squared over denominator P subscript N subscript 2 end subscript cross times P subscript H subscript 2 end subscript cubed end fraction end cell row blank equals cell fraction numerator left parenthesis 1 space atm right parenthesis squared over denominator left parenthesis 0 comma 5 space atm right parenthesis cross times left parenthesis 1 comma 5 space atm right parenthesis cubed end fraction end cell row blank equals cell 0 comma 593 space atm to the power of negative sign 2 end exponent end cell end table  

Jadi, nilai Kc dan Kp berturut-turut adalah table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell bold 237 bold comma bold 037 bold space italic M to the power of bold minus sign bold 2 end exponent end cell end table dan bold 0 bold comma bold 593 bold space bold atm to the power of bold minus sign bold 2 end exponent.space

6rb+

4.6 (6 rating)

Pertanyaan serupa

Gas N2​O4​ dimasukkan dalam ruang yang volumenya 5 liter sehingga mengalami disosiasi sebagai berikut. N2​O4​(g)⇌2NO2​(g)  Jika  mula-mula 1 mol; saat setimbang terdapat 0,8 mol gas NO2​; tentukan...

352

5.0

Jawaban terverifikasi

RUANGGURU HQ

Jl. Dr. Saharjo No.161, Manggarai Selatan, Tebet, Kota Jakarta Selatan, Daerah Khusus Ibukota Jakarta 12860

Coba GRATIS Aplikasi Roboguru

Coba GRATIS Aplikasi Ruangguru

Download di Google PlayDownload di AppstoreDownload di App Gallery

Produk Ruangguru

Produk Lainnya

Hubungi Kami

Ruangguru WhatsApp

081578200000

Email info@ruangguru.com

info@ruangguru.com

Contact 02140008000

02140008000

Ikuti Kami

©2022 Ruangguru. All Rights Reserved PT. Ruang Raya Indonesia