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Pada reaksi kesetimbangan CO +4mu(g) + 3 H2​ (g) ⇄ CH4​ (g) + H2​O (g) diperoleh data sebagai berikut: Harga Kc dari reaksi tersebut..

Pertanyaan

P a d a space r e a k s i space k e s e t i m b a n g a n space C O thin space left parenthesis g right parenthesis space plus space 3 space H subscript 2 space left parenthesis g right parenthesis space rightwards arrow over leftwards arrow space C H subscript 4 space left parenthesis g right parenthesis space plus space H subscript 2 O space left parenthesis g right parenthesis space d i p e r o l e h space d a t a space s e b a g a i space b e r i k u t colon

H a r g a space K c space d a r i space r e a k s i space t e r s e b u t..

  1. K c space equals space fraction numerator left square bracket 0 comma 05 right square bracket space left square bracket 0 comma 05 right square bracket over denominator left square bracket 0 comma 15 right square bracket space left square bracket 0 comma 15 right square bracket end fraction

  2. K c space equals space fraction numerator left square bracket 0 comma 05 right square bracket space left square bracket 0 comma 05 right square bracket over denominator left square bracket 0 comma 20 right square bracket space left square bracket 0 comma 30 right square bracket end fraction

  3. K c space equals space fraction numerator left square bracket 0 comma 05 right square bracket space left square bracket 0 comma 05 right square bracket over denominator left square bracket 0 comma 20 right square bracket space left square bracket 0 comma 30 right square bracket cubed end fraction

  4. K c space equals space fraction numerator left square bracket 0 comma 05 right square bracket space left square bracket 0 comma 05 right square bracket over denominator left square bracket 0 comma 15 right square bracket space left square bracket 0 comma 15 right square bracket cubed end fraction

  5. K c space equals space fraction numerator left square bracket 0 comma 15 right square bracket space left square bracket 0 comma 15 right square bracket cubed over denominator left square bracket 0 comma 05 right square bracket space left square bracket 0 comma 05 right square bracket end fraction

J. Siregar

Master Teacher

Mahasiswa/Alumni Universitas Negeri Medan

Jawaban terverifikasi

Pembahasan

P e n e n t u a n space h a r g a space t e t a p a n space k e s e t i m b a n g a n space k i m i a space s u a t u space r e a k s i space d i h i t u n g space s a a t space k e a d a a n space s e t i m b a n g space left parenthesis s e t e l a h space r e a k s i space s e l e s a i right parenthesis. space  S e h i n g g a space p e r s a m a a n y a space colon  K c space equals space fraction numerator left square bracket C H subscript 4 right square bracket space left square bracket space H subscript 2 0 right square bracket over denominator left square bracket C O right square bracket space left square bracket H subscript 2 right square bracket squared end fraction space equals space fraction numerator left square bracket 0 comma 05 right square bracket space left square bracket space 0 comma 05 right square bracket over denominator left square bracket 0 comma 15 right square bracket space left square bracket 0 comma 15 right square bracket cubed end fraction

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