Roboguru
SD
SMP
SMA
UTBK/SNBT

Iklan

Iklan

Pertanyaan

Diketahui beberapa reaksi pembentukan sebagai berikut. Berdasarkan data, reaksi berikut yang mempunyai entalpi 2 e − ( 2 1 ​ b + c ) adalah ...

Diketahui beberapa reaksi pembentukan sebagai berikut.

N subscript 2 open parentheses italic g close parentheses plus begin inline style 3 over 2 end style O subscript 2 open parentheses italic g close parentheses yields N subscript 2 O subscript 3 open parentheses italic g close parentheses space increment H subscript f degree space equals space a space kJ space mol to the power of negative sign 1 end exponent space 2 H subscript 2 open parentheses italic g close parentheses and O subscript 2 open parentheses italic g close parentheses yields 2 H subscript 2 O open parentheses italic g close parentheses space increment H subscript f degree space equals space b space kJ space mol to the power of negative sign 1 end exponent space space space N subscript 2 open parentheses italic g close parentheses plus begin inline style 5 over 2 end style O subscript 2 open parentheses italic g close parentheses yields N subscript 2 O subscript 5 open parentheses italic g close parentheses space increment H subscript f degree space equals space c space kJ space mol to the power of negative sign 1 end exponent space begin inline style 1 half end style H subscript 2 open parentheses italic g close parentheses plus begin inline style 1 half end style N subscript 2 open parentheses italic g close parentheses and O subscript 2 open parentheses italic g close parentheses yields H N O subscript 2 open parentheses italic l close parentheses space increment H subscript f degree space equals space d space kJ space mol to the power of negative sign 1 end exponent space begin inline style 1 half end style begin inline style begin inline style H end style subscript 2 end style begin inline style left parenthesis end style begin inline style italic g end style begin inline style right parenthesis end style begin inline style plus end style begin inline style 1 half end style begin inline style begin inline style N end style subscript 2 end style begin inline style left parenthesis end style begin inline style italic g end style begin inline style right parenthesis end style begin inline style plus end style begin inline style 3 over 2 end style begin inline style begin inline style O end style subscript 2 end style begin inline style left parenthesis end style begin inline style italic g end style begin inline style right parenthesis end style begin inline style rightwards arrow end style begin inline style begin inline style H N O end style subscript 3 end style begin inline style left parenthesis end style begin inline style italic l end style begin inline style right parenthesis end style begin inline style space end style begin inline style increment end style begin inline style begin inline style H end style subscript f end style begin inline style degree end style begin inline style space end style begin inline style equals end style begin inline style space end style begin inline style e end style begin inline style space end style begin inline style kJ end style begin inline style space end style begin inline style begin inline style mol end style to the power of negative sign 1 end exponent end style begin inline style space end style begin inline style space end style begin inline style space end style 

Berdasarkan data, reaksi berikut yang mempunyai entalpi  adalah ...

  1. 2 N subscript 2 O subscript 5 open parentheses italic g close parentheses yields 2 N subscript 2 open parentheses italic g close parentheses and 5 O subscript 2 open parentheses italic g close parentheses 

  2. N subscript 2 O subscript 5 open parentheses italic g close parentheses and H subscript 2 O open parentheses italic g close parentheses yields 2 H N O subscript 3 open parentheses italic l close parentheses  

  3. 2 N subscript 2 O subscript 3 open parentheses italic g close parentheses yields 2 N subscript 2 open parentheses italic g close parentheses and 3 O subscript 2 open parentheses italic g close parentheses  

  4. N subscript 2 O subscript 3 open parentheses italic g close parentheses and H subscript 2 O open parentheses italic g close parentheses yields 2 H N O subscript 2 open parentheses italic l close parentheses  

  5. H subscript 2 O open parentheses italic g close parentheses yields H subscript 2 open parentheses italic g close parentheses and begin inline style 1 half end style O subscript 2 open parentheses g close parentheses 

Ke roboguru plus

Iklan

A. Nurul

Master Teacher

Mahasiswa/Alumni UIN Syarif Hidayatullah Jakarta

Jawaban terverifikasi

Jawaban

jawaban benar adalah B.

jawaban benar adalah B.

Iklan

Pembahasan

menentukan reaksi 2e : menentukan reaksi (- b) : menentukan reaksi -c: menentukan reaksi yang terbentuk dengan hukum Hess : Jadi, jawaban benar adalah B.

menentukan reaksi 2e :

bottom enclose 1 half begin inline style H subscript 2 end style begin inline style left parenthesis end style begin inline style italic g end style begin inline style right parenthesis end style begin inline style plus end style begin inline style 1 half end style begin inline style N subscript 2 end style begin inline style left parenthesis end style begin inline style italic g end style begin inline style right parenthesis end style begin inline style plus end style begin inline style 3 over 2 end style begin inline style O subscript 2 end style begin inline style left parenthesis end style begin inline style italic g end style begin inline style right parenthesis end style begin inline style rightwards arrow end style begin inline style H N O subscript 3 end style begin inline style left parenthesis end style begin inline style italic l end style begin inline style right parenthesis end style begin inline style space end style begin inline style increment end style begin inline style H subscript f end style begin inline style degree end style begin inline style space end style begin inline style equals end style begin inline style space end style begin inline style e end style begin inline style space end style begin inline style kJ end style begin inline style space end style begin inline style mol to the power of negative sign 1 end exponent end style end enclose begin inline style left parenthesis cross times 2 right parenthesis end style begin inline style H subscript 2 open parentheses italic g close parentheses and N subscript 2 open parentheses italic g close parentheses and O subscript 2 open parentheses italic g close parentheses yields 2 H N O subscript 3 open parentheses italic l close parentheses space increment H subscript f degree space equals space 2 e space kJ space mol to the power of negative sign 1 end exponent end style 

menentukan reaksi (-begin inline style 1 half end styleb) :

bottom enclose 2 H subscript 2 open parentheses italic g close parentheses and O subscript 2 open parentheses italic g close parentheses yields 2 H subscript 2 O open parentheses italic g close parentheses space increment H subscript f degree space equals space b space kJ space mol to the power of negative sign 1 end exponent end enclose left parenthesis cross times minus sign begin inline style 1 half end style right parenthesis H subscript 2 O open parentheses italic g close parentheses yields H subscript 2 open parentheses italic g close parentheses and begin inline style 1 half end style O subscript 2 open parentheses italic g close parentheses space increment H subscript f degree space equals space minus sign begin inline style 1 half end style b space kJ space mol to the power of negative sign 1 end exponent

menentukan reaksi -c :

bottom enclose N subscript 2 open parentheses italic g close parentheses plus begin inline style 5 over 2 end style O subscript 2 open parentheses italic g close parentheses yields N subscript 2 O subscript 5 open parentheses italic g close parentheses space increment H subscript f degree space equals space c space kJ space mol to the power of negative sign 1 end exponent end enclose space left parenthesis cross times minus sign 1 right parenthesis N subscript 2 O subscript 5 open parentheses italic g close parentheses yields N subscript 2 open parentheses italic g close parentheses and begin inline style 5 over 2 end style O subscript 2 open parentheses italic g close parentheses space increment H subscript f degree space equals minus sign c space kJ space mol to the power of negative sign 1 end exponent

menentukan reaksi yang terbentuk dengan hukum Hess :

up diagonal strike H subscript 2 end strike left parenthesis italic g right parenthesis plus up diagonal strike N subscript 2 end strike left parenthesis italic g right parenthesis plus up diagonal strike O subscript 2 end strike left parenthesis italic g right parenthesis yields 2 H N O subscript 3 open parentheses italic l close parentheses space increment H subscript italic f degree equals 2 e space kJ space mol to the power of negative sign 1 end exponent H subscript 2 O open parentheses italic g close parentheses yields up diagonal strike H subscript 2 end strike left parenthesis italic g right parenthesis plus up diagonal strike 1 half O subscript 2 end strike left parenthesis italic g right parenthesis space increment H subscript italic f degree equals minus sign begin inline style 1 half end style b space kJ space mol to the power of negative sign 1 end exponent bottom enclose N subscript 2 O subscript 5 open parentheses italic g close parentheses yields up diagonal strike N subscript 2 end strike left parenthesis italic g right parenthesis plus up diagonal strike 5 over 2 O subscript 2 end strike left parenthesis italic g right parenthesis space increment H subscript italic f degree equals minus sign c space kJ space mol to the power of italic minus sign italic 1 end exponent end enclose N subscript 2 O subscript 5 open parentheses italic g close parentheses and H subscript 2 O open parentheses italic g close parentheses yields 2 H N O subscript 3 open parentheses italic l close parentheses 

Jadi, jawaban benar adalah B.

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

1

Farraskanz Algibra Rachmat

harus ny pada reaksi 2e itu 3O² ka, disitu hanya O² saja

Iklan

Iklan

Gold Icon

Klaim Gold gratis sekarang!

Dengan Gold kamu bisa tanya soal ke Forum sepuasnya, lho.

Pertanyaan serupa

Jika diketahui perubahan entalpi pembentukan standar aluminium oksida dan tembaga(II)oksida berturut-turut -1.676 kJ/mol dan -155 kJ/mol, maka perubahan entalpi padareaksi: 2 Al ( s ) + 3 CuO ( s...

12

4.3

Jawaban terverifikasi

Diketahui : △ H f ​ CO 2 ​ △ H f ​ H 2 ​ O CH 4 ​ + 2 O 2 ​ ​ = = → ​ − 393 , 5 kJ − 242 kJ 2 H 2 ​ O + CO 2 ​ △ H = − 677 kJ ​ Maka besarnya △ H f ​ CH 4 ​ adalah ....

5

5.0

Jawaban terverifikasi

Iklan

Iklan

Diketahui data dan persamaan reaksi berikut. Nilai adalah...

5

4.0

Jawaban terverifikasi

Diketahui: △ H f ∘ ​ H 2 ​ O ( g ) = − 242 k J / m o l △ H f ​ C O 2 ​ ( g ) = − 394 k J / m o l △ H f ∘ ​ C 2 ​ H 2 ​ ( g ) = + 52 k J / m o l Jika 5,2 gram C 2 ​ H 2 ​ dibakar secara sempur...

35

4.5

Jawaban terverifikasi

Diketahui entalpi pembakaran karbon (grafit), hidrogen, dan etanol ( C 2 ​ H 6 ​ O ) berturut-turut, yaitu − 393 , 5 kJ mol − 1 ; − 286 kJ mol − 1 ; dan − 1.367 kJ mol − 1 . Tentukan entalpi pembentuk...

13

4.8

Jawaban terverifikasi
Ruangguru

RUANGGURU HQ

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

Coba GRATIS Aplikasi Roboguru

Download di Google Play

Coba GRATIS Aplikasi Ruangguru

Download di Google PlayDownload di AppstoreDownload di App Gallery

Produk Ruangguru

Produk Lainnya

Bantuan & Panduan

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