Roboguru

Pertanyaan

Data hasil percobaan reaksi begin mathsize 14px style 2 A and 3 B yields 2 C end style pada volume total 100 mL adalah sebagai berikut.

Waktu yang diperlukan untuk mereaksikan A dan B dengan konsentrasi masing-masing 1,0 M pada volume total yang sama dengan percobaan di atas adalah ....space space

J. Siregar

Master Teacher

Mahasiswa/Alumni Universitas Negeri Medan

Jawaban terverifikasi

Jawaban

waktu yang diperlukan untuk mereaksikan A dan B dengan konsentrasi masing-masing 1,0 M adalah 2,5 detik.

Pembahasan

Persamaan laju reaksi dapat dicari dengan mencari orde reaksi, konstanta laju. Waktu reaksi berbanding terbalik dengan laju reaksi yang dapat dinyatakan sebagai:

begin mathsize 14px style italic t bold almost equal to bold 1 over italic v end style 

  • Penentuan orde reaksi A dan B.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell bold Orde bold space italic A bold space bold left parenthesis bold data bold space B bold space bold yang bold space bold sama bold space bold yaitu bold space bold colon bold space bold 1 bold space bold dan bold space bold 2 bold right parenthesis end cell row cell v subscript 1 over v subscript 2 end cell equals cell k subscript 1 over k subscript 2 open parentheses open square brackets A close square brackets subscript 1 over open square brackets A close square brackets subscript 2 close parentheses to the power of x open parentheses open square brackets B close square brackets subscript 1 over open square brackets B close square brackets subscript 2 close parentheses to the power of y end cell row cell fraction numerator bevelled 1 over t subscript 1 over denominator bevelled 1 over t subscript 2 end fraction end cell equals cell k subscript 1 over k subscript 2 open parentheses open square brackets A close square brackets subscript 1 over open square brackets A close square brackets subscript 2 close parentheses to the power of x open parentheses open square brackets B close square brackets subscript 1 over open square brackets B close square brackets subscript 2 close parentheses to the power of y end cell row cell fraction numerator bevelled 1 over 160 over denominator bevelled 1 over 80 end fraction end cell equals cell open parentheses fraction numerator 0 comma 25 over denominator 0 comma 50 end fraction close parentheses to the power of x open parentheses fraction numerator 0 comma 25 over denominator 0 comma 25 end fraction close parentheses to the power of y end cell row cell 80 over 160 end cell equals cell open parentheses 1 half close parentheses to the power of x end cell row cell open parentheses 1 half close parentheses to the power of x end cell equals cell 1 half end cell row x equals 1 row blank blank blank row blank blank cell bold Orde bold space B bold space bold left parenthesis bold data bold space italic A bold space bold yang bold space bold sama bold space bold yaitu bold space bold colon bold space bold 2 bold space bold dan bold space bold 3 bold right parenthesis end cell row cell v subscript 2 over v subscript 3 end cell equals cell k subscript 2 over k subscript 3 open parentheses open square brackets A close square brackets subscript 2 over open square brackets A close square brackets subscript 3 close parentheses to the power of x open parentheses open square brackets B close square brackets subscript 2 over open square brackets B close square brackets subscript 3 close parentheses to the power of y end cell row cell fraction numerator bevelled 1 over t subscript 2 over denominator bevelled 1 over t subscript 3 end fraction end cell equals cell k subscript 2 over k subscript 3 open parentheses open square brackets A close square brackets subscript 2 over open square brackets A close square brackets subscript 3 close parentheses to the power of x open parentheses open square brackets B close square brackets subscript 2 over open square brackets B close square brackets subscript 3 close parentheses to the power of y end cell row cell fraction numerator bevelled 1 over 80 over denominator bevelled 1 over 20 end fraction end cell equals cell open parentheses fraction numerator 0 comma 50 over denominator 0 comma 50 end fraction close parentheses to the power of x open parentheses fraction numerator 0 comma 25 over denominator 0 comma 50 end fraction close parentheses to the power of y end cell row cell 20 over 80 end cell equals cell open parentheses 1 half close parentheses to the power of y end cell row cell 1 fourth end cell equals cell open parentheses 1 half close parentheses to the power of y end cell row cell open parentheses 1 half close parentheses to the power of y end cell equals cell open parentheses 1 half close parentheses squared end cell row y equals 2 end table end style

  • Tentukan persamaan laju reaksi.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row v equals cell k open square brackets A close square brackets to the power of x open square brackets B close square brackets to the power of y end cell row v equals cell k open square brackets A close square brackets open square brackets B close square brackets squared end cell end table end style  

  • Tentukan nilai konstanta laju reaksi.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row k equals cell fraction numerator v subscript 1 over denominator open square brackets A close square brackets subscript 1 open square brackets B close square brackets subscript 1 superscript 2 end fraction end cell row k equals cell fraction numerator bevelled 1 over t subscript 1 over denominator open square brackets A close square brackets subscript 1 open square brackets B close square brackets subscript 1 superscript 2 end fraction end cell row k equals cell fraction numerator bevelled 1 over 160 over denominator left parenthesis 0 comma 25 right parenthesis left parenthesis 0 comma 25 right parenthesis squared end fraction end cell row k equals cell fraction numerator 0 comma 00625 over denominator 0 comma 015625 end fraction end cell row k equals cell 0 comma 4 end cell end table end style 

  • Tentukan waktu reaksi  apabila A dan B dengan konsentrasi masing-masing 1,0 M.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row v equals cell k open square brackets A close square brackets open square brackets B close square brackets squared end cell row cell 1 over t end cell equals cell k open square brackets A close square brackets open square brackets B close square brackets squared end cell row cell 1 over t end cell equals cell 0 comma 4 left parenthesis 1 comma 0 right parenthesis left parenthesis 1 comma 0 right parenthesis squared end cell row cell 1 over t end cell equals cell 0 comma 4 left parenthesis 1 comma 0 right parenthesis end cell row cell 1 over t end cell equals cell 0 comma 4 end cell row t equals cell fraction numerator 1 over denominator 0 comma 4 end fraction end cell row t equals cell 2 comma 5 space detik end cell end table end style 

Jadi, waktu yang diperlukan untuk mereaksikan A dan B dengan konsentrasi masing-masing 1,0 M adalah 2,5 detik.

2rb+

5.0 (1 rating)

Pertanyaan serupa

Untuk reaksi P (g) + Q (g)    PQ (g) diperoleh data sebagai berikut: Tentukan: a. persamaan lajunya b. orde reaksi total c. harga K

2rb+

4.0

Jawaban terverifikasi

RUANGGURU HQ

Jl. Dr. Saharjo No.161, Manggarai Selatan, Tebet, Kota Jakarta Selatan, Daerah Khusus Ibukota Jakarta 12860

Coba GRATIS Aplikasi Ruangguru

Produk Ruangguru

Produk Lainnya

Hubungi Kami

Ruangguru WhatsApp

081578200000

Email info@ruangguru.com

info@ruangguru.com

Contact 02140008000

02140008000

Ikuti Kami

©2022 Ruangguru. All Rights Reserved PT. Ruang Raya Indonesia