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(a)
Interpretation:
The fraction of uncovered sites under the given condition is approximately 1/α pJ has to be derived.
Concept Introduction:
According to the Langmuir isotherm for a surface-catalyzed unimolecular reaction, the rate is
Rate =krθ=krα p1+α p
Where, α=kadkd
(a)
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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=α pJ1+α pJ1−θ=11+α pJ
Where,
θ= Fraction of covered sites
1−θ= Fraction of uncovered sites
For strongly adsorbed molecule, kad≫ kd
α pJ≫ 1
So, 1 can be neglected in comparison to α p in the denominator. Then the fraction of uncovered sites becomes
1−θ=1α pJ
(b)
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=krα p1+α p
Where, α=kadkd
(b)
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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=α pJ1+α pJ1−θ=11+α pJ
Where,
θ= Fraction of covered sites
1−θ= Fraction of uncovered sites
For strongly adsorbed molecule, kad≫ kd
α pJ≫ 1
So, 1 can be neglected in comparison to α p in the denominator. Then the fraction of uncovered sites becomes
1−θ=1α pJ
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(1−θ)=−kcpNH31+α pH2
Hydrogen is strongly adsorbed on the surface. So 1 can be neglected in the denominator. Then the above expression becomes
dpNH3dt=−kcpNH3α pH2≃−kcpNH3 pH2
(c)
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)
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Explanation of Solution
As the
dpNH3dt=−kcpNH3(1−θ)=−kcpNH31+α pH2
Hydrogen is strongly adsorbed on the surface. So 1 can be neglected in the denominator. Then the above expression becomes
dpNH3dt=−kcpNH3α pH2≃−kcpNH3 pH2
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{p0−p}
The rate of the reaction can be modified like
−dpdt=kcp 32α{p0−p}=kp p0−p, k=2kc3α
The integration of the above equation results
p∫p0(1−p0p) dp=kt∫0dtp−p0t=k+p0tlnpp0
(d)
Interpretation:
The form of a suitable straight-line plot to determine kc has to be derived.
(d)
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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(1−θ)=−kcpNH31+α pH2
Hydrogen is strongly adsorbed on the surface. So 1 can be neglected in the denominator. Then the above expression becomes
dpNH3dt=−kcpNH3α pH2≃−kcpNH3 pH2
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{p0−p}
The rate of the reaction can be modified like
−dpdt=kcp 32α{p0−p}=kp p0−p, k=2kc3α
The integration of the above equation results
p∫p0(1−p0p) dp=kt∫0dtp−p0t=k+p0tlnpp0
Let, A=p0tlnpp0 and B=p−p0t
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 B−A should be a constant, k.
The required table for the plot:
t/s03060100160200250p/kPa13.311.711.210.710.39.99.6B/(kPa s−1)−0.0533−0.035−0.026−0.0188−0.017−0.015A/(kPa s−1)−0.0568−0.0381−0.0289−0.0213−0.0196−0.0173(B−A)/(kPa s−1)0.003490.003090.002930.002500.002630.00254_
Therefore, the data fit the rate law and takin average of the values of B−A, the constant came to be kc=0.00286 kPa s−1.
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Chapter 19 Solutions
Atkins' Physical Chemistry
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