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Karnataka Board PUCPUC Science 2nd PUC Class 12

Consider a first order gas phase decomposition reaction given below :A(g)⟶B(g)+C(g)The initial pressure of the system before decomposition of A was pi. After lapse of - Chemistry

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Question

Consider a first order gas phase decomposition reaction given below :
\[\ce{A(g) -> B(g) + C(g)}\]
The initial pressure of the system before decomposition of A was pi. After lapse of time ‘t’, total pressure of the system increased by x units and became ‘pt’ The rate constant k for the reaction is given as ______.

Options

  • `k = 2.303/t log  p_i/(p_i - x)`

  • `k = 2.303/t log  p_i/(2p_i - p_t)`

  • `k = 2.303/t log  p_i/(2p_i + p_t)`

  • `k = 2.303/t log  p_i/(p_i + x)`

MCQ
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Solution

The initial pressure of the system before decomposition of A was pi. After lapse of time ‘t’, total pressure of the system increased by x units and became ‘pt’ The rate constant k for the reaction is given as `k = 2.303/t log  p_i/(2p_i - p_t)`.

Explanation:

Consider a first order gas phase decomposition reaction:

\[\ce{A_{(g)} -> B_{(g)} + C_{(g)}}\]

The initial pressure of the system before decomposition of A was Pi.

After lapse of time (t), total pressure of the system increased by x units and became 'Pt'.

In other words, the pressure of A decreased by x atom.

\[\ce{A_{(g)} -> B_{(g)} + C_{(g)}}\]

Initial pressure: Pi atm

Pressure after time t: (Pi – x) atm  x atm  x atm

Pt = (Pi – x) + x + x = Pi + x atm

x = Pt – Pi

PA = Pressure of A after time t = Pi - x = Pi - (Pi - Pi) = 2Pi - P

`k = 2.303/t log  (["A"]_0)/(["A"])`

= `k = 2.303/t log  p_i/(2p_i - p_t)`

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Chapter 4: Chemical Kinetics - Exercises [Page 48]

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NCERT Exemplar Chemistry [English] Class 12
Chapter 4 Chemical Kinetics
Exercises | Q I. 5. | Page 48

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