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The decomposition of NH3 on platinum surface is zero order reaction. What are the rates of production of N2 and H2 if k = 2.5 × 10−4 mol−1 L s−1? - Chemistry

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प्रश्न

The decomposition of NH3 on platinum surface is zero order reaction. What are the rates of production of N2 and H2 if k = 2.5 × 10−4 mol−1 L s−1?

संख्यात्मक

उत्तर

The decomposition of NH3 on platinum surface is represented by the following equation.

\[\ce{2NH3_{(g)}->[Pt]N2_{(g)} +3H2_{(g)}}\]

Therefore

`"Rate" = -1/2 ("d"["NH"_3])/"dt" = ("d"["N"_2])/"dt" = 1/3("d"["H"_2])/"dt"`

However, it is given that the reaction is of zero order.

Therefore,

`-1/2 ("d"["NH"_3])/"dt" = ("d"["N"_2])/"dt" = 1/3 ("d"["H"_2])/"dt" = "k"` 

= 2.5 × 10−4 mol L1 s1

Therefore, the rate of production of N2 is

`("d"["N"_2])/"dt" = 2.5xx10^(-4)  "mol L"^(-1)  "s"^(-1)`

 And, the rate of production of H2 is

`("d"["H"_2])/"dt" = 3 xx 2.5 xx 10^(-4)  "mol L"^(-1)  "s"^(-1)`

= 7.5 × 10−4 mol L−1 s1

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अध्याय 4: Chemical Kinetics - Exercises [पृष्ठ ११७]

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एनसीईआरटी Chemistry [English] Class 12
अध्याय 4 Chemical Kinetics
Exercises | Q 3 | पृष्ठ ११७

संबंधित प्रश्न

The reaction between A and B is first order with respect to A and zero order with respect to B. Fill in the blanks in the following table:

Experiment A/mol L−1 B/mol L−1 Initial rate/mol L−1 min−1
I 0.1 0.1 2.0 × 10−2
II ______ 0.2 4.0 × 10−2
III 0.4 0.4 ______
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(iv) Further increase in pressure will change the rate of reaction.


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