General Chemistry
General Chemistry
4th Edition
ISBN: 9781891389603
Author: Donald A. McQuarrie, Peter A. Rock, Ethan B. Gallogly
Publisher: University Science Books
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Chapter 17, Problem 17.28P
Interpretation Introduction

Interpretation:

Rate law and rate constant of the given reaction has to be determined.

    H2C2O4(g)CO2(g)+HCHO2(g)

Concept Introduction:

Rate of a reaction depends on the species concentration that takes place in the reaction.  Chemical kinetic experiments are used in order to deduce the rate of reaction.  Rate law of a reaction can only be identified by experimental methods that are adopted.

Expert Solution & Answer
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Answer to Problem 17.28P

Rate law of the reaction is lnPH2C2O4=lnPH2C2O4okt and rate constant of the reaction is 2.9×105s1.

Explanation of Solution

Given reaction is:

    H2C2O4(g)CO2(g)+HCHO2(g)

Rate of the reaction is measured using the change in pressure.  The total pressure that is reached after 2.00×104seconds is given as follows:

PH2C2O4o/Torr(att=0)Ptotalo/Torr(att=2.00×104s)
5.07.2
7.010
8.412

By Dalton’s law of partial pressure, the total pressure at any time of the reaction can be given by the equation shown below:

  PTotal=PH2C2O4+PCO2+PHCHO2        (1)

Quantity of oxalic acid that undergoes the reaction is directly proportional to PH2C2O4oPH2C2O4.

For one mole of HCHO2 that is formed, the equation can be written as follows:

  PHCHO2=PH2C2O4oPH2C2O4

For one mole of CO2 that is formed, the equation can be written as follows:

  PCO2=PH2C2O4oPH2C2O4

Substituting these values in equation (1) as shown below:

  PTotal=PH2C2O4+PH2C2O4oPH2C2O4+PH2C2O4oPH2C2O4=2PH2C2O4oPH2C2O4

Rearranging the above equation in terms of PH2C2O4.

  PH2C2O4=2PH2C2O4oPTotal

Solving for PH2C2O4, the table obtained is:

PH2C2O4o/Torr(att=0)Ptotalo/Torr(att=2.00×104s)PH2C2O4(att=2.00×104s)
5.07.22.8
7.0104.0
8.4124.8

Considering the reaction to be first order, the rate law of the reaction can be given as follows:

    lnPH2C2O4=lnPH2C2O4okt        (2)

Substituting the values in equation (2), the rate constant for the reaction is calculated as shown below:

    ln(2.8)=ln(5.0)k(2.00×104s)k=12.00×104sln2.85.0=12.00×104s(0.5798)=0.2899×104s1=2.9×105s1

Thus the rate constant of the reaction is 2.9×105s1.

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