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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 SOsehingga 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 ...

  1. 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

  2. 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

  3. 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

  4. 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

  5. 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

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M. Robo

Master Teacher

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Pembahasan

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

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

Latihan Bab

Pendahuluan Kesetimbangan Kimia

Konsep Kesetimbangan Kimia

Konstanta Kesetimbangan (K)

Konstanta Kesetimbangan Konsentrasi (Kc)

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