Atkins' Physical Chemistry
Atkins' Physical Chemistry
11th Edition
ISBN: 9780198769866
Author: ATKINS, P. W. (peter William), De Paula, Julio, Keeler, JAMES
Publisher: Oxford University Press
Question
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Chapter 19, Problem 19C.3P

(a)

Interpretation Introduction

Interpretation:

The fraction of uncovered sites under the given condition is approximately 1/αpJ has to be derived.

Concept Introduction:

According to the Langmuir isotherm for a surface-catalyzed unimolecular reaction, the rate is

  Rate =krθ=krαp1+αp

Where, α=kadkd

(a)

Expert Solution
Check Mark

Explanation of Solution

A surface-catalyzed unimolecular reaction is one in which an adsorbed molecule undergoes decomposition on a surface.  The given scenario in the question is a surface-catalyzed unimolecular reaction.  Hence, the Langmuir isotherm can be applied to the given scenario.

According to the question, when a gas J adsorbs very strongly and its pressure is pJ, the fraction of uncovered sites will be

  θ=αpJ1+αpJ1θ=11+αpJ

Where,

θ= Fraction of covered sites

1θ= Fraction of uncovered sites

For strongly adsorbed molecule, kadkd

  αpJ1

So, 1 can be neglected in comparison to αp in the denominator.  Then the fraction of uncovered sites becomes

  1θ=1αpJ

(b)

Interpretation Introduction

Interpretation:

The rate of the reaction as given in the question has to be derived.

Concept Introduction:

According to the Langmuir isotherm for a surface-catalyzed unimolecular reaction, the rate is

  Rate =krθ=krαp1+αp

Where, α=kadkd

(b)

Expert Solution
Check Mark

Explanation of Solution

According to the question, when a gas J adsorbs very strongly and its pressure is pJ, the fraction of uncovered sites will be

  θ=αpJ1+αpJ1θ=11+αpJ

Where,

θ= Fraction of covered sites

1θ= Fraction of uncovered sites

For strongly adsorbed molecule, kadkd

  αpJ1

So, 1 can be neglected in comparison to αp in the denominator.  Then the fraction of uncovered sites becomes

  1θ=1αpJ

As the reaction rate is proportional to the pressure of ammonia and the fraction of uncovered sites by the strongly adsorbed hydrogen product, the rate of the reaction can be written as

  dpNH3dt=kcpNH3(1θ)=kcpNH31+αpH2

Hydrogen is strongly adsorbed on the surface. So 1 can be neglected in the denominator. Then the above expression becomes

  dpNH3dt=kcpNH3αpH2kcpNH3pH2

(c)

Interpretation Introduction

Interpretation:

The given equation in the question has to integrated by using the initial condition that at t=0 only ammonia is present at pressure p0.

(c)

Expert Solution
Check Mark

Explanation of Solution

As the reaction rate is proportional to the pressure of ammonia and the fraction of uncovered sites by the strongly adsorbed hydrogen product, the rate of the reaction can be written as

  dpNH3dt=kcpNH3(1θ)=kcpNH31+αpH2

Hydrogen is strongly adsorbed on the surface. So 1 can be neglected in the denominator. Then the above expression becomes

  dpNH3dt=kcpNH3αpH2kcpNH3pH2

The decomposition of ammonia can be written as

  NH3(g)12N2+32H2

At t=0, pNH3(g)=p0 and at t=t, pNH3(g)=p

Then, pH2=32{p0p}

The rate of the reaction can be modified like

  dpdt=kcp32α{p0p}=kpp0p,k=2kc3α

The integration of the above equation results

  p0p(1p0p)dp=k0tdtpp0t=k+p0tlnpp0

(d)

Interpretation Introduction

Interpretation:

The form of a suitable straight-line plot to determine kc has to be derived.

(d)

Expert Solution
Check Mark

Explanation of Solution

As the reaction rate is proportional to the pressure of ammonia and the fraction of uncovered sites by the strongly adsorbed hydrogen product, the rate of the reaction can be written as

  dpNH3dt=kcpNH3(1θ)=kcpNH31+αpH2

Hydrogen is strongly adsorbed on the surface. So 1 can be neglected in the denominator. Then the above expression becomes

  dpNH3dt=kcpNH3αpH2kcpNH3pH2

The decomposition of ammonia can be written as

  NH3(g)12N2+32H2

At t=0, pNH3(g)=p0 and at t=t, pNH3(g)=p

Then, pH2=32{p0p}

The rate of the reaction can be modified like

  dpdt=kcp32α{p0p}=kpp0p,k=2kc3α

The integration of the above equation results

  p0p(1p0p)dp=k0tdtpp0t=k+p0tlnpp0

Let, A=p0tlnpp0andB=pp0t

Then, the above equation can be written in the form

  B=k+A

Hence, a plot of B against A should give a straight line with the intercept k. In other words, the difference BA should be a constant, k.

The required table for the plot:

  t/s03060100160200250p/kPa13.311.711.210.710.39.99.6B/(kPas1)0.05330.0350.0260.01880.0170.015A/(kPas1)0.05680.03810.02890.02130.01960.0173(BA)/(kPas1)0.003490.003090.002930.002500.002630.00254_

Therefore, the data fit the rate law and takin average of the values of BA, the constant came to be kc=0.00286kPas1.

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