Pertanyaan

Data percobaan untuk reaksi: 2 A ( g ) + B ( g ) + C ( g ) → hasil reaksi adalah sebagai berikut. Tentukan nilai x !

Data percobaan untuk reaksi:

$2 A (g) + B (g) + C (g) \to hasil reaksi$

adalah sebagai berikut.

Tentukan nilai x!

I. Solichah

Master Teacher

Jawaban terverifikasi

Jawaban

jawaban sesuai dengan uraian di atas.

jawaban sesuai dengan uraian di atas. space

Pembahasan

Percobaan 1 dan 2 Percobaan 2 dan 3 Percobaan 1 dan 4 Persamaan laju reaksi Konstanta laju reaksi dan satuannya Nilai x Jadi, jawaban sesuai dengan uraian di atas.

Percobaan 1 dan 2

$v subscript 1 over v subscript 2 equals fraction numerator italic k open square brackets A close square brackets subscript 1 superscript italic x open square brackets B close square brackets subscript 1 end subscript superscript italic y end superscript open square brackets C close square brackets subscript 1 superscript italic z over denominator italic k open square brackets A close square brackets subscript 2 superscript italic x open square brackets B close square brackets subscript 2 superscript italic y open square brackets C close square brackets subscript 2 superscript italic z end fraction fraction numerator 0 comma 01 over denominator 0 comma 02 end fraction equals fraction numerator up diagonal strike italic k left square bracket 0 comma 1 right square bracket subscript blank superscript italic x up diagonal strike left square bracket 0 comma 1 right square bracket subscript blank superscript italic y end strike up diagonal strike left square bracket 0 comma 1 right square bracket subscript 1 superscript italic z end strike over denominator up diagonal strike italic k left square bracket 0 comma 2 right square bracket subscript blank superscript italic x up diagonal strike left square bracket 0 comma 1 right square bracket subscript blank superscript italic y end strike up diagonal strike left square bracket 0 comma 1 right square bracket subscript 2 superscript italic z end strike end fraction fraction numerator 0 comma 01 over denominator 0 comma 02 end fraction equals open parentheses fraction numerator 0 comma 1 over denominator 0 comma 2 end fraction close parentheses to the power of x 1 half equals open parentheses 1 half close parentheses to the power of x x equals 1$

Percobaan 2 dan 3

$straight v subscript 2 over straight v subscript 3 equals fraction numerator k left square bracket straight A right square bracket subscript 2 superscript x left square bracket straight B right square bracket subscript 2 superscript y left square bracket straight C right square bracket subscript 2 superscript z over denominator k left square bracket straight A right square bracket subscript 3 superscript x left square bracket straight B right square bracket subscript 3 superscript y left square bracket straight C right square bracket subscript 3 superscript z end fraction fraction numerator 0 comma 02 over denominator 0 comma 04 end fraction equals fraction numerator up diagonal strike k up diagonal strike left square bracket 0 comma 2 right square bracket subscript blank superscript x end strike left square bracket 0 comma 1 right square bracket subscript blank superscript y up diagonal strike left square bracket 0 comma 1 right square bracket subscript 2 superscript z end strike over denominator up diagonal strike k up diagonal strike left square bracket 0 comma 2 right square bracket subscript blank superscript x end strike left square bracket 0 comma 2 right square bracket subscript blank superscript y up diagonal strike left square bracket 0 comma 1 right square bracket subscript 3 superscript z end strike end fraction fraction numerator 0 comma 02 over denominator 0 comma 04 end fraction equals open parentheses fraction numerator 0 comma 1 over denominator 0 comma 2 end fraction close parentheses to the power of straight y 1 half equals open parentheses 1 half close parentheses to the power of straight y straight y equals 1$

Percobaan 1 dan 4

$straight v subscript 1 over straight v subscript 4 equals fraction numerator k left square bracket straight A right square bracket subscript 1 superscript x left square bracket straight B right square bracket subscript 1 superscript y left square bracket straight C right square bracket subscript 1 superscript z over denominator k left square bracket straight A right square bracket subscript 4 superscript x left square bracket straight B right square bracket subscript 4 superscript y left square bracket straight C right square bracket subscript 4 superscript z end fraction fraction numerator 0 comma 01 over denominator 0 comma 09 end fraction equals fraction numerator up diagonal strike k left square bracket 0 comma 1 right square bracket subscript blank superscript italic 1 left square bracket 0 comma 1 right square bracket subscript blank superscript italic 1 left square bracket 0 comma 1 right square bracket subscript 1 superscript z over denominator up diagonal strike k left square bracket 0 comma 3 right square bracket subscript blank superscript italic 1 left square bracket 0 comma 3 right square bracket subscript blank superscript italic 1 left square bracket 0 comma 3 right square bracket subscript 4 superscript z end fraction fraction numerator 0 comma 01 over denominator 0 comma 09 end fraction equals 1 third 1 third open parentheses fraction numerator 0 comma 1 over denominator 0 comma 3 end fraction close parentheses to the power of straight z 1 over 9 equals 1 over 9 open parentheses 1 third close parentheses to the power of straight z straight z equals 0$

Persamaan laju reaksi

v equals italic k open square brackets A close square brackets open square brackets B close square brackets

Konstanta laju reaksi dan satuannya

$v subscript 1 double bond k open square brackets A close square brackets open square brackets B close square brackets 0 comma 01 space M space s to the power of negative sign 1 end exponent equals k left parenthesis 0 comma 1 space M right parenthesis left parenthesis 0 comma 1 space M right parenthesis k equals fraction numerator 0 comma 01 space M space s to the power of negative sign 1 end exponent over denominator 0 comma 01 space M squared end fraction k equals 1 space M to the power of negative sign 1 end exponent space s to the power of negative sign 1 end exponent$

Nilai x

v subscript 5 double bond k open square brackets A close square brackets open square brackets B close square brackets italic x equals 1 left parenthesis 0 comma 5 right parenthesis left parenthesis 0 comma 4 right parenthesis italic x equals 0 comma 2 space M space s to the power of negative sign 1 end exponent

Jadi, jawaban sesuai dengan uraian di atas. space

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