Iklan

Iklan

Pertanyaan

Dalam volume 1 liter dipanaskan gas NH 3 hingga terjadi reaksi: 2 N H 3 ( g ) ​ ⇄ N 2 ( g ) ​ + 3 H 2 ( g ) ​ Data yang diperoleh sebagai berikut: Harga Kc kesetimbangan tersebut adalah...

Dalam volume 1 liter dipanaskan gas NH3 hingga terjadi reaksi:

Data yang diperoleh sebagai berikut:

Harga Kc kesetimbangan tersebut adalah...

  1. K c space equals space fraction numerator open parentheses 0 comma 2 close parentheses over denominator open parentheses 0 comma 6 close parentheses squared open parentheses 0 comma 6 close parentheses cubed end fraction

  2. K c space equals space fraction numerator open parentheses 0 comma 2 close parentheses open parentheses 0 comma 6 close parentheses cubed over denominator open parentheses 0 comma 6 close parentheses to the power of blank end fraction

  3. K c space equals space fraction numerator open parentheses 0 comma 2 close parentheses open parentheses 0 comma 6 close parentheses cubed over denominator open parentheses 0 comma 6 close parentheses squared end fraction

  4. K c space equals space fraction numerator open parentheses 0 comma 6 close parentheses open parentheses 0 comma 6 close parentheses to the power of blank over denominator open parentheses 0 comma 2 close parentheses to the power of blank end fraction

  5. K c space equals space fraction numerator open parentheses 0 comma 6 close parentheses cubed over denominator open parentheses 0 comma 6 close parentheses squared left parenthesis 0 comma 2 right parenthesis end fraction

Iklan

S. Lubis

Master Teacher

Mahasiswa/Alumni Universitas Sumatera Utara

Jawaban terverifikasi

Iklan

Pembahasan

Dalam volume 1 L

table row cell 2 N H subscript 3 open parentheses g close parentheses end subscript end cell row cell S space equals space 0 comma 6 m o l end cell end table table row rightwards arrow over leftwards arrow row blank end table table row cell N subscript 2 open parentheses g close parentheses end subscript end cell row cell 0 comma 2 space m o l end cell end table table row plus row blank end table table row cell 3 space H subscript 2 open parentheses g close parentheses end subscript end cell row cell 0 comma 6 space m o l end cell end table

Dalam volume 1 L

K c space equals space open square brackets p r o d u k close square brackets to the power of k o e f i s i e n end exponent over open square brackets r e a k tan close square brackets to the power of k o e f i s i e n end exponent space equals space fraction numerator open square brackets N subscript 2 close square brackets to the power of 1 x open square brackets H subscript 2 close square brackets cubed over denominator open square brackets N H subscript 3 close square brackets squared end fraction space equals space fraction numerator open square brackets 0 comma 2 close square brackets to the power of 1 cross times open square brackets 0 comma 6 close square brackets cubed over denominator open square brackets 0 comma 6 close square brackets squared end fraction

Perdalam pemahamanmu bersama Master Teacher
di sesi Live Teaching, GRATIS!

22

Iklan

Iklan

Pertanyaan serupa

Jika molaritas zat dalam reaksi kesetimbangan: A ( g ) + 2 B ( g ) ⇄ 2 C ( g ) adalah [ A ] = 2 , 4 × 1 0 − 2 M ; [ B ] = 4 , 6 × 1 0 − 3 M ; [ C ] = 6 , 2 × 1 0 − 3 M , maka hitung nilai tetapa...

8

4.0

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

Hubungi Kami

Ruangguru WhatsApp

+62 815-7441-0000

Email info@ruangguru.com

[email protected]

Contact 02140008000

02140008000

Ikuti Kami

©2024 Ruangguru. All Rights Reserved PT. Ruang Raya Indonesia