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Data hasil reaksi antara Na 2 S 2 O 3 dengan HCl pada berbagai konsentrasi, sebagai berikut: No [Na 2 S 2 O 3 ] (mol L -1 ) [HCl] (mol L -1 ) Laju (M/s) 1 0,20 2,0 1 2 0,10 2,0 0,5 3 0,05 2,0 0,25 4 0,05 1,5 0,25 5 0,05 1,0 0,25 Sesuai data tersebut maka orde total reaksinya adalah…

Data hasil reaksi antara Na2S2O3 dengan HCl pada berbagai konsentrasi, sebagai berikut:

No

[Na2S2O3] (mol L-1)

[HCl] (mol L-1)

Laju (M/s)

1

0,20

2,0

1

2

0,10

2,0

0,5

3

0,05

2,0

0,25

4

0,05

1,5

0,25

5

0,05

1,0

0,25

Sesuai data tersebut maka orde total reaksinya adalah…

  1. 1

  2. 2

  3. 3

  4. 4

  5. 5

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S. Hidayati

Master Teacher

Mahasiswa/Alumni Universitas Indonesia

Jawaban terverifikasi

Jawaban

orde total reaksi tersebut adalah 1 + 0 = 1

orde total reaksi tersebut adalah 1 + 0 = 1

Pembahasan

Jadi, orde total reaksi tersebut adalah 1 + 0 = 1

v subscript 2 over v subscript 1 space equals space fraction numerator begin display style open square brackets N subscript 2 S subscript 2 O subscript 3 close square brackets to the power of p space end exponent end style begin display style begin display style open square brackets H C l close square brackets end style to the power of q end style over denominator open square brackets N subscript 2 S subscript 2 O subscript 3 close square brackets to the power of p space end exponent open square brackets H C l close square brackets to the power of q end fraction fraction numerator 0 comma 5 over denominator 1 end fraction space equals space fraction numerator begin display style open square brackets 0 comma 1 close square brackets to the power of p end style begin display style space end style begin display style horizontal strike open square brackets 0 comma 2 close square brackets to the power of q end strike end style over denominator open square brackets 0 comma 1 close square brackets to the power of p space horizontal strike open square brackets 0 comma 2 close square brackets to the power of q end strike end fraction 1 half space space equals space space 1 to the power of p over 2 to the power of p p space equals space 1  fraction numerator begin display style v subscript 2 end style over denominator begin display style v subscript 1 end style end fraction space equals space fraction numerator begin display style open square brackets N subscript 2 S subscript 2 O subscript 3 close square brackets to the power of p space end exponent open square brackets H C l close square brackets to the power of q end style over denominator open square brackets N subscript 2 S subscript 2 O subscript 3 close square brackets to the power of p space end exponent open square brackets H C l close square brackets to the power of q end fraction fraction numerator 0 comma 25 over denominator 0 comma 25 end fraction space equals space fraction numerator begin display style horizontal strike open square brackets 0 comma 05 close square brackets to the power of p end strike space open square brackets 1 close square brackets to the power of q end style over denominator horizontal strike open square brackets 0 comma 05 close square brackets to the power of p end strike space open square brackets 1 comma 5 close square brackets to the power of q end fraction 1 space equals space 10 to the power of q over 15 to the power of q Q space equals space 0

Jadi, orde total reaksi tersebut adalah 1 + 0 = 1

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