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Dari hasil percobaan reaksi 2 NO ( g ) + 2 H 2 ​ ( g ) → N 2 ​ ( g ) + 2 H 2 ​ O ( l ) diperoleh data sebagai berikut: Tingkat reaksi untuk reaksi di atas adalah ...

Dari hasil percobaan reaksi

 

diperoleh data sebagai berikut:
 

 space 
 

Tingkat reaksi untuk reaksi di atas adalah ...space

  1. 1,0space

  2. tetapspace

  3. 2,0space

  4. 2,5space

  5. 3,0space

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A. Tri

Master Teacher

Mahasiswa/Alumni Universitas Pertamina

Jawaban terverifikasi

Jawaban

dapat disimpulkan jawaban yang tepat adalah E.

dapat disimpulkan jawaban yang tepat adalah E.space

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Pembahasan

Orde total 2+1=3. Jadi, dapat disimpulkan jawaban yang tepat adalah E.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator v 3 over denominator v 4 end fraction end cell equals cell fraction numerator italic k open square brackets N O close square brackets subscript 3 to the power of italic x open square brackets H subscript 2 close square brackets to the power of italic y over denominator italic k open square brackets N O close square brackets subscript 4 to the power of italic x open square brackets H subscript 2 close square brackets to the power of italic y end fraction end cell row cell fraction numerator italic t 4 over denominator t 3 end fraction end cell equals cell fraction numerator italic k open square brackets N O close square brackets subscript 3 to the power of italic x open square brackets H subscript 2 close square brackets to the power of italic y over denominator italic k open square brackets N O close square brackets subscript 4 to the power of italic x open square brackets H subscript 2 close square brackets to the power of italic y end fraction end cell row cell 1 fourth end cell equals cell open square brackets 0 comma 2 close square brackets to the power of x over open square brackets 0 comma 4 close square brackets to the power of x end cell row cell 1 fourth end cell equals cell 1 half to the power of x end cell row x equals 2 row blank blank blank row cell fraction numerator v 1 over denominator v 2 end fraction end cell equals cell fraction numerator italic k open square brackets N O close square brackets subscript 1 to the power of italic x open square brackets H subscript 2 close square brackets to the power of italic y over denominator italic k open square brackets N O close square brackets italic 2 to the power of italic x open square brackets H subscript 2 close square brackets to the power of italic y end fraction end cell row cell fraction numerator italic t italic 2 over denominator t 1 end fraction end cell equals cell fraction numerator italic k open square brackets N O close square brackets subscript 1 to the power of italic x open square brackets H subscript 2 close square brackets to the power of italic y over denominator italic k open square brackets N O close square brackets subscript italic 2 to the power of italic x open square brackets H subscript 2 close square brackets to the power of italic y end fraction end cell row cell fraction numerator 3 comma 2 over denominator 9 comma 6 end fraction end cell equals cell open square brackets 0 comma 1 close square brackets to the power of y over open square brackets 0 comma 3 close square brackets to the power of y end cell row cell 1 third end cell equals cell 1 third to the power of y end cell row x equals 1 row blank blank blank row blank blank blank row blank blank blank end table end style 
 

Orde total 2+1=3.

Jadi, dapat disimpulkan jawaban yang tepat adalah E.space

Latihan Bab

Pendahuluan Laju Reaksi

Laju Reaksi, Energi Aktivasi, dan Teori Tumbukan

Penentuan Laju Reaksi Single Experiment

Orde Reaksi Multiexperiment

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