Pertanyaan

Dalam volume 2 liter dipanaskan 1 mol gas SO 3 sehingga terjadi kesetimbangan: 2 SO 3 (g) ⇄ 2SO 2 (g) + O 2 (g) Keadaan Zat Mol SO 3 Mol SO 2 Mol O 2 Mula-mula 1 - - Reaksi 0,50 0,50 0,25 Setimbang 0,50 0,50 0,25 Harga Kc pada kesetimbangan tersebut adalah ...

Dalam volume 2 liter dipanaskan 1 mol gas SO₃sehingga terjadi kesetimbangan:

2 SO₃ (g) $⇄$ 2SO₂(g) + O₂(g)

Keadaan Zat	Mol SO₃	Mol SO₂	Mol O₂
Mula-mula	1	-	-
Reaksi	0,50	0,50	0,25
Setimbang	0,50	0,50	0,25

Harga Kc pada kesetimbangan tersebut adalah ...

$text Kc= end text fraction numerator open square brackets 0 comma 5 divided by 2 close square brackets over denominator open square brackets 0 comma 5 divided by 2 close square brackets open square brackets 0 comma 25 divided by 2 close square brackets end fraction$
$text Kc= end text fraction numerator open square brackets 0 comma 5 divided by 2 close square brackets open square brackets 0 comma 5 divided by 2 close square brackets over denominator open square brackets 0 comma 25 divided by 2 close square brackets end fraction$
$text Kc= end text fraction numerator open square brackets 0 comma 5 close square brackets squared open square brackets 0 comma 25 close square brackets over denominator open square brackets 0 comma 5 close square brackets squared end fraction$
$text Kc= end text fraction numerator open square brackets 0 comma 5 close square brackets over denominator open square brackets 0 comma 5 close square brackets end fraction$
$text Kc= end text fraction numerator open square brackets 0 comma 5 divided by 2 close square brackets squared open square brackets 0 comma 25 divided by 2 close square brackets over denominator open square brackets 0 comma 5 divided by 2 close square brackets squared end fraction$

M. Robo

Master Teacher

Jawaban terverifikasi

Pembahasan

Harga tetapan kesetimbangan (Kc), dihitung saat keadaan setimbang. Sehingga :

$text Kc end text equals fraction numerator left square bracket S O subscript 2 right square bracket squared right parenthesis left square bracket O subscript 2 right square bracket over denominator left square bracket S O subscript 3 right square bracket squared end fraction equals fraction numerator left square bracket 0 comma 50 divided by 2 right square bracket squared open square brackets 0 comma 25 divided by 2 close square brackets over denominator left square bracket 0 comma 50 divided by 2 right square bracket squared end fraction$

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

141

5.0 (4 rating)

Pertanyaan serupa

Tetapan kesetimbangan bagi reaksi: X 2 ( g ) + Y 2 ( g ) ⇄ 2 XY ( g ) adalah 16 pada suhu dan tekanan tertentu. Jika X 2 , Y 2 , dan X Y masing-masing sebanyak 1 mol dicampurkan dalam...

129

5.0

Jawaban terverifikasi

Dalam ruang 4 liter terdapat reaksi kesetimbangan: NO 2 ( g ) + CO ( g ) ⇄ NO ( g ) + CO 2 ( g ) Jika pada saat setimbang terdapat gas NO 2 dan gas CO masing-masing 0,2 mol, dan gas NO ...

237

4.5

Jawaban terverifikasi

Tuliskan rumusan hukum kesetimbangan bagi reaksi-reaksi: a. NH 3 ( a q ) + H 2 O ( l ) ⇌ NH 4 + ( a q ) + OH − ( a q ) b. H 2 O ( l ) ⇌ H + ( a q ) + OH − ( a q ) c. PCl 3 ( g ) + Cl 2 ...

3.7

Jawaban terverifikasi

Pada suhu 2 7 ∘ C , sebanyak 1,2 mol gas SCl 2 dimasukkan ke dalam wadah 2 L menurut reaksi berikut. SCl 2 ( g ) + 2 C 2 H 4 ( g ) ⇌ S ( CH 2 CH 2 Cl ) 2 ( g ) Jika pada keadaan...

5.0

Jawaban terverifikasi

Tuliskan rumusan hukum kesetimbangan bagi reaksi-reaksi: d. CaO ( s ) + CO 2 ( g ) ⇌ CaCO 3 ( s ) e. Ag + ( a q ) + Fe 2 + ( a q ) ⇌ Ag ( s ) + Fe 3 + ( a q )

5.0

Jawaban terverifikasi