Sebanyak 2 mol N2​O4​ dipanaskan dalam suatu ruangan 1 Liter sehingga sebagian berubah menjadi NO2​ menurut reaksi kesetimbangan berikut.   N2​O4​(g) ⇌ 2NO2​(g)     Pada suhu T K tercapai suatu kesetimbangan di mana terdapat 1 mol . Tentukan susunan kesetimbangan!

Pertanyaan

Sebanyak 2 mol N subscript 2 O subscript 4 dipanaskan dalam suatu ruangan 1 Liter sehingga sebagian berubah menjadi N O subscript 2 menurut reaksi kesetimbangan berikut.
 

N subscript 2 O subscript 4 open parentheses italic g close parentheses space rightwards harpoon over leftwards harpoon space 2 N O subscript 2 open parentheses italic g close parentheses space 
 

Pada suhu T K tercapai suatu kesetimbangan di mana terdapat 1 mol N O subscript 2. Tentukan susunan kesetimbangan!

N. Puspita

Master Teacher

Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah table attributes columnalign right center left columnspacing 0px end attributes row cell N subscript 2 O subscript 4 open parentheses italic g close parentheses end cell rightwards harpoon over leftwards harpoon cell 2 N O subscript 2 open parentheses italic g close parentheses end cell row cell Mula bond mula space end cell equals cell space space space space space space space space space 2 space mol space space space space space space space space space space space space minus sign end cell row cell Reaksi space space space space space space space space space end cell equals cell space space space space space space space space space 0 comma 5 space mol space space space space space space space 1 space mol end cell row cell Setimbang space space space end cell equals cell space space space space space space space space space 1 comma 5 space mol space space space space space space space 1 space mol space space end cell end tablespace 

Pembahasan

Pembahasan

Diketahui :

Mula-mula N subscript 2 O subscript 4 space(2 mol) dipanaskan dalam ruangan 1 liter
N subscript 2 O subscript 4 open parentheses italic g close parentheses space rightwards harpoon over leftwards harpoon space 2 N O subscript 2 open parentheses italic g close parentheses space

Saat setimbang N O subscript 2 spaceterdapat 1 mol

Ditanya : susunan kesetimbangan ?

Jawab :

Karena pada saat setimbang terdapat 1 mol N O subscript 2 space, maka :

table attributes columnalign right center left columnspacing 0px end attributes row cell Saat space bereaksi space N subscript 2 O subscript 4 space end cell equals cell space fraction numerator Koefisien space N subscript 2 O subscript 4 over denominator Koefisien space N O subscript 2 end fraction cross times mol space N O subscript 2 space saat space bereaksi end cell row blank equals cell 1 half cross times 1 space mol end cell row blank equals cell 0 comma 5 space mol end cell row blank blank blank row cell Saat space setimbang space N subscript 2 O subscript 4 end cell equals cell left parenthesis mol space mula bond mula space N subscript 2 O subscript 4 right parenthesis minus sign open parentheses mol space bereaksi space N subscript 2 O subscript 4 close parentheses end cell row blank equals cell 2 space mol minus sign 0 comma 5 space mol end cell row blank equals cell 1 comma 5 space mol space end cell row blank blank blank row cell Maka space susunan space kesetimbangannya space end cell equals blank row cell N subscript 2 O subscript 4 open parentheses italic g close parentheses end cell rightwards harpoon over leftwards harpoon cell 2 N O subscript 2 open parentheses italic g close parentheses end cell row cell Mula bond mula space end cell equals cell space space space space space space space space space 2 space mol space space space space space space space space space space space space minus sign end cell row cell Reaksi space space space space space space space space space end cell equals cell space space space space space space space space space 0 comma 5 space mol space space space space space space space 1 space mol end cell row cell Setimbang space space space end cell equals cell space space space space space space space space space 1 comma 5 space mol space space space space space space space 1 space mol space end cell row blank blank blank end table

Jadi, jawaban yang tepat adalah table attributes columnalign right center left columnspacing 0px end attributes row cell N subscript 2 O subscript 4 open parentheses italic g close parentheses end cell rightwards harpoon over leftwards harpoon cell 2 N O subscript 2 open parentheses italic g close parentheses end cell row cell Mula bond mula space end cell equals cell space space space space space space space space space 2 space mol space space space space space space space space space space space space minus sign end cell row cell Reaksi space space space space space space space space space end cell equals cell space space space space space space space space space 0 comma 5 space mol space space space space space space space 1 space mol end cell row cell Setimbang space space space end cell equals cell space space space space space space space space space 1 comma 5 space mol space space space space space space space 1 space mol space space end cell end tablespace 

107

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Rieke Nur Salsabilatul Khusna

Makasih ❤️

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Setarakan persamaan kimia berikut.​ a. Ca(OH)2​ (s) + HBr (aq) → CaBr2​ (aq) + H2​O (l)b. HgS (s) + HNO3​ (aq) + HCl (aq) → HgCl2​ (aq) + NO (g) + H2​O (l) + S (s)

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