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Dalam bejana tertutup bervolume 5 liter, 2 mol gas N2​O5​ terurai sebagai berikut:   N2​O5​→N2​+O2​  Jika setelah 4 detik, terbentuk 2 mol gas O2​, maka laju pengurangan gas N2​O5​ adalah ...

Pertanyaan

Dalam bejana tertutup bervolume 5 liter, 2 mol gas N subscript 2 O subscript 5 terurai sebagai berikut:
 

N subscript 2 O subscript 5 yields N subscript 2 and O subscript 2 


Jika setelah 4 detik, terbentuk 2 mol gas O subscript 2, maka laju pengurangan gas N subscript 2 O subscript 5 adalah ...space space space

  1. 0,1 M/detikspace space space

  2. 0,5 M/detikspace space space

  3. 0,2 M/detikspace space space

  4. 0,01 M/detikspace space space

  5. 0,04 M/detikspace space space

I. Solichah

Master Teacher

Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah E.space space space

Pembahasan

Secara umum, untuk reaksi:

italic a A and italic b B yields italic c C and italic d D 

laju reaksi = 

negative sign 1 over italic a fraction numerator increment open square brackets A close square brackets over denominator increment italic t end fraction equals minus sign 1 over italic b fraction numerator increment open square brackets B close square brackets over denominator increment italic t end fraction equals plus 1 over italic c fraction numerator increment open square brackets C close square brackets over denominator increment italic t end fraction equals plus 1 over italic d fraction numerator increment open square brackets D close square brackets over denominator increment italic t end fraction 

Reaksi setara penguraian N subscript 2 O subscript 5 adalah sebagai berikut:

2 N subscript 2 O subscript 5 yields 2 N subscript 2 and 5 O subscript 2 

Laju reaksinya adalah: 

italic v equals minus sign 1 half fraction numerator increment open square brackets N subscript 2 O subscript 5 close square brackets over denominator increment italic t end fraction equals plus 1 half fraction numerator increment open square brackets N subscript 2 close square brackets over denominator increment italic t end fraction equals plus 1 fifth fraction numerator increment open square brackets O subscript 2 close square brackets over denominator increment italic t end fraction 

Laju pengurangan gas N subscript 2 O subscript 5 adalah:

negative sign 1 half fraction numerator increment open square brackets N subscript 2 O subscript 5 close square brackets over denominator increment italic t end fraction equals plus 1 fifth fraction numerator increment open square brackets O subscript 2 close square brackets over denominator increment italic t end fraction minus sign 1 half fraction numerator increment open square brackets N subscript 2 O subscript 5 close square brackets over denominator increment italic t end fraction equals plus 1 fifth fraction numerator open parentheses begin display style fraction numerator 2 space mol over denominator 5 space L end fraction end style minus sign 0 close parentheses over denominator open parentheses 4 minus sign 0 close parentheses space s end fraction minus sign 1 half fraction numerator increment open square brackets N subscript 2 O subscript 5 close square brackets over denominator increment italic t end fraction equals plus 1 fifth fraction numerator begin display style 2 over 5 end style space M over denominator 4 space s end fraction minus sign 1 half fraction numerator increment open square brackets N subscript 2 O subscript 5 close square brackets over denominator increment italic t end fraction equals plus 1 fifth cross times 2 over 20 space M forward slash s fraction numerator increment open square brackets N subscript 2 O subscript 5 close square brackets over denominator increment italic t end fraction equals 2 over 100 cross times 2 over 1 space M forward slash s fraction numerator increment open square brackets N subscript 2 O subscript 5 close square brackets over denominator increment italic t end fraction equals 4 over 100 space M forward slash s fraction numerator increment open square brackets N subscript 2 O subscript 5 close square brackets over denominator increment italic t end fraction equals 0 comma 04 space M forward slash s  


Jadi, jawaban yang tepat adalah E.space space space

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