Roboguru

Pertanyaan

Dalam suatu wadah yang volumenya 2 liter dimasukkan 0,1 mol gas begin mathsize 14px style N H subscript 3 end style menurut reaksi :

begin mathsize 14px style 2 N H subscript 3 open parentheses italic g close parentheses equilibrium N subscript 2 open parentheses italic g close parentheses and 3 H subscript 2 open parentheses italic g close parentheses end style. Jika gas begin mathsize 14px style N subscript 2 end style yang terbentuk adalah 0,02 mol dan gas begin mathsize 14px style H subscript 2 end style adalah 0,03 mol. Tentukan nilai

  1. ...undefined

  2. ...undefined

N. Puspita

Master Teacher

Jawaban terverifikasi

Jawaban

nilai begin mathsize 14px style K subscript italic c end style untuk persamaan reaksi kesetimbangan tersebut adalah begin mathsize 14px style bold 2 bold comma bold 5 bold cross times bold 10 to the power of bold minus sign bold 5 end exponent end style.

Pembahasan

begin mathsize 14px style M subscript N H subscript 3 end subscript equals n subscript N H subscript 3 end subscript over V M subscript N H subscript 3 end subscript equals fraction numerator 0 comma 08 space mol over denominator 2 space L end fraction M subscript N H subscript 3 end subscript equals 4 cross times 10 to the power of negative sign 2 end exponent space M end style 

begin mathsize 14px style M subscript N subscript 2 end subscript equals n subscript N subscript 2 end subscript over V M subscript N subscript 2 end subscript equals fraction numerator 0 comma 02 space mol over denominator 2 space L end fraction M subscript N subscript 2 end subscript equals 10 to the power of negative sign 2 end exponent space M end style

begin mathsize 14px style M subscript H subscript 2 end subscript equals n subscript H subscript 2 end subscript over V M subscript H subscript 2 end subscript equals fraction numerator 0 comma 03 space mol over denominator 2 space L end fraction M subscript H subscript 2 end subscript equals 1 comma 5 cross times 10 to the power of negative sign 2 end exponent space M end style

begin mathsize 14px style K subscript c equals fraction numerator open square brackets N subscript 2 close square brackets 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 K subscript c equals fraction numerator open parentheses 10 to the power of negative sign 2 end exponent close parentheses open parentheses 1 comma 5 cross times 10 to the power of negative sign 2 end exponent close parentheses cubed over denominator open parentheses 4 cross times 10 to the power of negative sign 2 end exponent close parentheses squared end fraction K subscript c equals 2 comma 1 cross times 10 to the power of negative sign 5 end exponent end style

Jadi, nilai begin mathsize 14px style K subscript italic c end style untuk persamaan reaksi kesetimbangan tersebut adalah begin mathsize 14px style bold 2 bold comma bold 5 bold cross times bold 10 to the power of bold minus sign bold 5 end exponent end style.

85

0.0 (0 rating)

Pertanyaan serupa

Dalam ruang yang volumenya 1 L, terdapat reaksi kesetimbangan: 4NH3​(g)+3O2​(g)⇌2N2​(g)+6H2​O(g)  Bila pada keadaan awal terdapat 1 mol NH3​ dan 1 mol O2​ dan pada keadaan akhir jumlah mol  berkuran...

145

0.0

Jawaban terverifikasi

RUANGGURU HQ

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

Coba GRATIS Aplikasi Ruangguru

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