The activation enthalpy and pre-exponential constant for the diffusion of oxygen ( O 18 ) into ZnO .

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Answer 1

Question

Chapter 4, Problem 4.20P

To determine

The activation enthalpy and pre-exponential constant for the diffusion of oxygen (O18) into ZnO.

Expert Solution & Answer

Answer to Problem 4.20P

The activation enthalpy is 3.83×105 J/mole of O atoms and the pre exponential constant is 0.40×10−3 m2/s.

Explanation of Solution

Formula used:

The diffusion coefficient of point 1 is given by,

D0ZnO1=D0ZnO0exp(−ΔH D 0ZnO kT1) ....... (I)

Here, D0Zn01 is the diffusion coefficient of point 1, D0ZnO is the pre-exponential constant for oxygen diffusion into ZnO, k is the Boltzmann’s constant, T1 is the temperature at point 1 and ΔHD0ZnO is the activation enthalpy for oxygen into ZnO.

The diffusion coefficient of point 2 is given by,

D0ZnO2=D0ZnO0exp(−ΔH D 0ZnO kT2) ....... (II)

Here, D0Zn02 is the diffusion coefficient of point 1 and T2 is the temperature at point 2.

The ratio of diffusion coefficient of points 1 and 2 is given by,

D 0 ZnO 1 D 0 ZnO 2 =D 0 ZnO 0 exp( − Δ H D 0ZnO k T 1 )D 0 ZnO 0 exp( − Δ H D 0ZnO k T 2 )=exp( − Δ H D 0ZnO k T 1 )exp( − Δ H D 0ZnO k T 2 )=exp[−( Δ H D 0ZnO k)( 1 T 1 − 1 T 2 )] ....... (III)

Calculation:

The activation enthalpy is calculated as,

Substitute 1×10−12 cm2/s for D0ZnO1, 1×10−18 cm2/s for D0ZnO2, 8.62×10−5 eV/atom⋅K for k, 1587 K for T1 and 1075 K for T2 in equation (III).

1× 10 −12 cm 2/s1× 10 −18 cm 2/s=exp[−( Δ H D 0ZnO 8.62× 10 −5 eV/ atom ⋅K)( 1 1587 K− 1 1075 K)]106=exp[−ΔH D 0ZnO(−3.48 atom/ eV)]ln( 106)=ΔHD 0ZnO(3.48 atom/eV)ΔHD 0ZnO=ln( 10 6 )3.48 atom/eV

Solve further,

ΔHD 0ZnO=13.823.48 atom/eV=(3.97 eV/atom× 96.521× 10 3 J/ mole of O atoms 1 eV/ atom )=3.83×105 J/mole of O atoms

The diffusion coefficient is calculated as,

Substitute 1×10−12 cm2/s for D0ZnO1, 8.62 ×10−5 eV/atom⋅K for k, 1587 K for T1 and 3.97 eV/atom for ΔHD0ZnO in equation (I).

1×10−12 cm2/s=D0 ZnO0exp(− 3.97 eV/ atom ( 8.62× 10 −5 eV/ atom ⋅K )( 1587 K ))D0 ZnO0=( 1× 10 −12 cm 2 /s × 10 −4 m 2 1 cm 2 )exp( −29.02)=1× 10 −16 m 2/s2.32 × 10 −13=0.40×10−3 m2/s

Conclusion:

Therefore, the activation enthalpy is 3.83×105 J/mole of O atoms and the pre exponential constant is 0.40×10−3 m2/s.

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Answer 2

Question

Chapter 4, Problem 4.20P

To determine

The activation enthalpy and pre-exponential constant for the diffusion of oxygen (O18) into ZnO.

Expert Solution & Answer

Answer to Problem 4.20P

The activation enthalpy is 3.83×105 J/mole of O atoms and the pre exponential constant is 0.40×10−3 m2/s.

Explanation of Solution

Formula used:

The diffusion coefficient of point 1 is given by,

D0ZnO1=D0ZnO0exp(−ΔH D 0ZnO kT1) ....... (I)

Here, D0Zn01 is the diffusion coefficient of point 1, D0ZnO is the pre-exponential constant for oxygen diffusion into ZnO, k is the Boltzmann’s constant, T1 is the temperature at point 1 and ΔHD0ZnO is the activation enthalpy for oxygen into ZnO.

The diffusion coefficient of point 2 is given by,

D0ZnO2=D0ZnO0exp(−ΔH D 0ZnO kT2) ....... (II)

Here, D0Zn02 is the diffusion coefficient of point 1 and T2 is the temperature at point 2.

The ratio of diffusion coefficient of points 1 and 2 is given by,

D 0 ZnO 1 D 0 ZnO 2 =D 0 ZnO 0 exp( − Δ H D 0ZnO k T 1 )D 0 ZnO 0 exp( − Δ H D 0ZnO k T 2 )=exp( − Δ H D 0ZnO k T 1 )exp( − Δ H D 0ZnO k T 2 )=exp[−( Δ H D 0ZnO k)( 1 T 1 − 1 T 2 )] ....... (III)

Calculation:

The activation enthalpy is calculated as,

Substitute 1×10−12 cm2/s for D0ZnO1, 1×10−18 cm2/s for D0ZnO2, 8.62×10−5 eV/atom⋅K for k, 1587 K for T1 and 1075 K for T2 in equation (III).

1× 10 −12 cm 2/s1× 10 −18 cm 2/s=exp[−( Δ H D 0ZnO 8.62× 10 −5 eV/ atom ⋅K)( 1 1587 K− 1 1075 K)]106=exp[−ΔH D 0ZnO(−3.48 atom/ eV)]ln( 106)=ΔHD 0ZnO(3.48 atom/eV)ΔHD 0ZnO=ln( 10 6 )3.48 atom/eV

Solve further,

ΔHD 0ZnO=13.823.48 atom/eV=(3.97 eV/atom× 96.521× 10 3 J/ mole of O atoms 1 eV/ atom )=3.83×105 J/mole of O atoms

The diffusion coefficient is calculated as,

Substitute 1×10−12 cm2/s for D0ZnO1, 8.62 ×10−5 eV/atom⋅K for k, 1587 K for T1 and 3.97 eV/atom for ΔHD0ZnO in equation (I).

1×10−12 cm2/s=D0 ZnO0exp(− 3.97 eV/ atom ( 8.62× 10 −5 eV/ atom ⋅K )( 1587 K ))D0 ZnO0=( 1× 10 −12 cm 2 /s × 10 −4 m 2 1 cm 2 )exp( −29.02)=1× 10 −16 m 2/s2.32 × 10 −13=0.40×10−3 m2/s

Conclusion:

Therefore, the activation enthalpy is 3.83×105 J/mole of O atoms and the pre exponential constant is 0.40×10−3 m2/s.

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Answer 3

Question

Chapter 4, Problem 4.20P

To determine

The activation enthalpy and pre-exponential constant for the diffusion of oxygen (O18) into ZnO.

Expert Solution & Answer

Answer to Problem 4.20P

The activation enthalpy is 3.83×105 J/mole of O atoms and the pre exponential constant is 0.40×10−3 m2/s.

Explanation of Solution

Formula used:

The diffusion coefficient of point 1 is given by,

D0ZnO1=D0ZnO0exp(−ΔH D 0ZnO kT1) ....... (I)

Here, D0Zn01 is the diffusion coefficient of point 1, D0ZnO is the pre-exponential constant for oxygen diffusion into ZnO, k is the Boltzmann’s constant, T1 is the temperature at point 1 and ΔHD0ZnO is the activation enthalpy for oxygen into ZnO.

The diffusion coefficient of point 2 is given by,

D0ZnO2=D0ZnO0exp(−ΔH D 0ZnO kT2) ....... (II)

Here, D0Zn02 is the diffusion coefficient of point 1 and T2 is the temperature at point 2.

The ratio of diffusion coefficient of points 1 and 2 is given by,

D 0 ZnO 1 D 0 ZnO 2 =D 0 ZnO 0 exp( − Δ H D 0ZnO k T 1 )D 0 ZnO 0 exp( − Δ H D 0ZnO k T 2 )=exp( − Δ H D 0ZnO k T 1 )exp( − Δ H D 0ZnO k T 2 )=exp[−( Δ H D 0ZnO k)( 1 T 1 − 1 T 2 )] ....... (III)

Calculation:

The activation enthalpy is calculated as,

Substitute 1×10−12 cm2/s for D0ZnO1, 1×10−18 cm2/s for D0ZnO2, 8.62×10−5 eV/atom⋅K for k, 1587 K for T1 and 1075 K for T2 in equation (III).

1× 10 −12 cm 2/s1× 10 −18 cm 2/s=exp[−( Δ H D 0ZnO 8.62× 10 −5 eV/ atom ⋅K)( 1 1587 K− 1 1075 K)]106=exp[−ΔH D 0ZnO(−3.48 atom/ eV)]ln( 106)=ΔHD 0ZnO(3.48 atom/eV)ΔHD 0ZnO=ln( 10 6 )3.48 atom/eV

Solve further,

ΔHD 0ZnO=13.823.48 atom/eV=(3.97 eV/atom× 96.521× 10 3 J/ mole of O atoms 1 eV/ atom )=3.83×105 J/mole of O atoms

The diffusion coefficient is calculated as,

Substitute 1×10−12 cm2/s for D0ZnO1, 8.62 ×10−5 eV/atom⋅K for k, 1587 K for T1 and 3.97 eV/atom for ΔHD0ZnO in equation (I).

1×10−12 cm2/s=D0 ZnO0exp(− 3.97 eV/ atom ( 8.62× 10 −5 eV/ atom ⋅K )( 1587 K ))D0 ZnO0=( 1× 10 −12 cm 2 /s × 10 −4 m 2 1 cm 2 )exp( −29.02)=1× 10 −16 m 2/s2.32 × 10 −13=0.40×10−3 m2/s

Conclusion:

Therefore, the activation enthalpy is 3.83×105 J/mole of O atoms and the pre exponential constant is 0.40×10−3 m2/s.

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Answer 4

Question

Chapter 4, Problem 4.20P

To determine

The activation enthalpy and pre-exponential constant for the diffusion of oxygen (O18) into ZnO.

Expert Solution & Answer

Answer to Problem 4.20P

The activation enthalpy is 3.83×105 J/mole of O atoms and the pre exponential constant is 0.40×10−3 m2/s.

Explanation of Solution

Formula used:

The diffusion coefficient of point 1 is given by,

D0ZnO1=D0ZnO0exp(−ΔH D 0ZnO kT1) ....... (I)

Here, D0Zn01 is the diffusion coefficient of point 1, D0ZnO is the pre-exponential constant for oxygen diffusion into ZnO, k is the Boltzmann’s constant, T1 is the temperature at point 1 and ΔHD0ZnO is the activation enthalpy for oxygen into ZnO.

The diffusion coefficient of point 2 is given by,

D0ZnO2=D0ZnO0exp(−ΔH D 0ZnO kT2) ....... (II)

Here, D0Zn02 is the diffusion coefficient of point 1 and T2 is the temperature at point 2.

The ratio of diffusion coefficient of points 1 and 2 is given by,

D 0 ZnO 1 D 0 ZnO 2 =D 0 ZnO 0 exp( − Δ H D 0ZnO k T 1 )D 0 ZnO 0 exp( − Δ H D 0ZnO k T 2 )=exp( − Δ H D 0ZnO k T 1 )exp( − Δ H D 0ZnO k T 2 )=exp[−( Δ H D 0ZnO k)( 1 T 1 − 1 T 2 )] ....... (III)

Calculation:

The activation enthalpy is calculated as,

Substitute 1×10−12 cm2/s for D0ZnO1, 1×10−18 cm2/s for D0ZnO2, 8.62×10−5 eV/atom⋅K for k, 1587 K for T1 and 1075 K for T2 in equation (III).

1× 10 −12 cm 2/s1× 10 −18 cm 2/s=exp[−( Δ H D 0ZnO 8.62× 10 −5 eV/ atom ⋅K)( 1 1587 K− 1 1075 K)]106=exp[−ΔH D 0ZnO(−3.48 atom/ eV)]ln( 106)=ΔHD 0ZnO(3.48 atom/eV)ΔHD 0ZnO=ln( 10 6 )3.48 atom/eV

Solve further,

ΔHD 0ZnO=13.823.48 atom/eV=(3.97 eV/atom× 96.521× 10 3 J/ mole of O atoms 1 eV/ atom )=3.83×105 J/mole of O atoms

The diffusion coefficient is calculated as,

Substitute 1×10−12 cm2/s for D0ZnO1, 8.62 ×10−5 eV/atom⋅K for k, 1587 K for T1 and 3.97 eV/atom for ΔHD0ZnO in equation (I).

1×10−12 cm2/s=D0 ZnO0exp(− 3.97 eV/ atom ( 8.62× 10 −5 eV/ atom ⋅K )( 1587 K ))D0 ZnO0=( 1× 10 −12 cm 2 /s × 10 −4 m 2 1 cm 2 )exp( −29.02)=1× 10 −16 m 2/s2.32 × 10 −13=0.40×10−3 m2/s

Conclusion:

Therefore, the activation enthalpy is 3.83×105 J/mole of O atoms and the pre exponential constant is 0.40×10−3 m2/s.

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Answer 5

Chapter 4, Problem 4.20P

To determine

The activation enthalpy and pre-exponential constant for the diffusion of oxygen (O18) into ZnO.

Expert Solution & Answer

Answer to Problem 4.20P

The activation enthalpy is 3.83×105 J/mole of O atoms and the pre exponential constant is 0.40×10−3 m2/s.

Explanation of Solution

Formula used:

The diffusion coefficient of point 1 is given by,

D0ZnO1=D0ZnO0exp(−ΔH D 0ZnO kT1) ....... (I)

Here, D0Zn01 is the diffusion coefficient of point 1, D0ZnO is the pre-exponential constant for oxygen diffusion into ZnO, k is the Boltzmann’s constant, T1 is the temperature at point 1 and ΔHD0ZnO is the activation enthalpy for oxygen into ZnO.

The diffusion coefficient of point 2 is given by,

D0ZnO2=D0ZnO0exp(−ΔH D 0ZnO kT2) ....... (II)

Here, D0Zn02 is the diffusion coefficient of point 1 and T2 is the temperature at point 2.

The ratio of diffusion coefficient of points 1 and 2 is given by,

D 0 ZnO 1 D 0 ZnO 2 =D 0 ZnO 0 exp( − Δ H D 0ZnO k T 1 )D 0 ZnO 0 exp( − Δ H D 0ZnO k T 2 )=exp( − Δ H D 0ZnO k T 1 )exp( − Δ H D 0ZnO k T 2 )=exp[−( Δ H D 0ZnO k)( 1 T 1 − 1 T 2 )] ....... (III)

Calculation:

The activation enthalpy is calculated as,

Substitute 1×10−12 cm2/s for D0ZnO1, 1×10−18 cm2/s for D0ZnO2, 8.62×10−5 eV/atom⋅K for k, 1587 K for T1 and 1075 K for T2 in equation (III).

1× 10 −12 cm 2/s1× 10 −18 cm 2/s=exp[−( Δ H D 0ZnO 8.62× 10 −5 eV/ atom ⋅K)( 1 1587 K− 1 1075 K)]106=exp[−ΔH D 0ZnO(−3.48 atom/ eV)]ln( 106)=ΔHD 0ZnO(3.48 atom/eV)ΔHD 0ZnO=ln( 10 6 )3.48 atom/eV

Solve further,

ΔHD 0ZnO=13.823.48 atom/eV=(3.97 eV/atom× 96.521× 10 3 J/ mole of O atoms 1 eV/ atom )=3.83×105 J/mole of O atoms

The diffusion coefficient is calculated as,

Substitute 1×10−12 cm2/s for D0ZnO1, 8.62 ×10−5 eV/atom⋅K for k, 1587 K for T1 and 3.97 eV/atom for ΔHD0ZnO in equation (I).

1×10−12 cm2/s=D0 ZnO0exp(− 3.97 eV/ atom ( 8.62× 10 −5 eV/ atom ⋅K )( 1587 K ))D0 ZnO0=( 1× 10 −12 cm 2 /s × 10 −4 m 2 1 cm 2 )exp( −29.02)=1× 10 −16 m 2/s2.32 × 10 −13=0.40×10−3 m2/s

Conclusion:

Therefore, the activation enthalpy is 3.83×105 J/mole of O atoms and the pre exponential constant is 0.40×10−3 m2/s.

Answer 6

Chapter 4, Problem 4.20P

To determine

The activation enthalpy and pre-exponential constant for the diffusion of oxygen (O18) into ZnO.

Expert Solution & Answer

Answer to Problem 4.20P

The activation enthalpy is 3.83×105 J/mole of O atoms and the pre exponential constant is 0.40×10−3 m2/s.

Explanation of Solution

Formula used:

The diffusion coefficient of point 1 is given by,

D0ZnO1=D0ZnO0exp(−ΔH D 0ZnO kT1) ....... (I)

Here, D0Zn01 is the diffusion coefficient of point 1, D0ZnO is the pre-exponential constant for oxygen diffusion into ZnO, k is the Boltzmann’s constant, T1 is the temperature at point 1 and ΔHD0ZnO is the activation enthalpy for oxygen into ZnO.

The diffusion coefficient of point 2 is given by,

D0ZnO2=D0ZnO0exp(−ΔH D 0ZnO kT2) ....... (II)

Here, D0Zn02 is the diffusion coefficient of point 1 and T2 is the temperature at point 2.

The ratio of diffusion coefficient of points 1 and 2 is given by,

D 0 ZnO 1 D 0 ZnO 2 =D 0 ZnO 0 exp( − Δ H D 0ZnO k T 1 )D 0 ZnO 0 exp( − Δ H D 0ZnO k T 2 )=exp( − Δ H D 0ZnO k T 1 )exp( − Δ H D 0ZnO k T 2 )=exp[−( Δ H D 0ZnO k)( 1 T 1 − 1 T 2 )] ....... (III)

Calculation:

The activation enthalpy is calculated as,

Substitute 1×10−12 cm2/s for D0ZnO1, 1×10−18 cm2/s for D0ZnO2, 8.62×10−5 eV/atom⋅K for k, 1587 K for T1 and 1075 K for T2 in equation (III).

1× 10 −12 cm 2/s1× 10 −18 cm 2/s=exp[−( Δ H D 0ZnO 8.62× 10 −5 eV/ atom ⋅K)( 1 1587 K− 1 1075 K)]106=exp[−ΔH D 0ZnO(−3.48 atom/ eV)]ln( 106)=ΔHD 0ZnO(3.48 atom/eV)ΔHD 0ZnO=ln( 10 6 )3.48 atom/eV

Solve further,

ΔHD 0ZnO=13.823.48 atom/eV=(3.97 eV/atom× 96.521× 10 3 J/ mole of O atoms 1 eV/ atom )=3.83×105 J/mole of O atoms

The diffusion coefficient is calculated as,

Substitute 1×10−12 cm2/s for D0ZnO1, 8.62 ×10−5 eV/atom⋅K for k, 1587 K for T1 and 3.97 eV/atom for ΔHD0ZnO in equation (I).

1×10−12 cm2/s=D0 ZnO0exp(− 3.97 eV/ atom ( 8.62× 10 −5 eV/ atom ⋅K )( 1587 K ))D0 ZnO0=( 1× 10 −12 cm 2 /s × 10 −4 m 2 1 cm 2 )exp( −29.02)=1× 10 −16 m 2/s2.32 × 10 −13=0.40×10−3 m2/s

Conclusion:

Therefore, the activation enthalpy is 3.83×105 J/mole of O atoms and the pre exponential constant is 0.40×10−3 m2/s.

Answer 7

Chapter 4, Problem 4.20P

To determine

The activation enthalpy and pre-exponential constant for the diffusion of oxygen (O18) into ZnO.

Answer 8

Expert Solution & Answer

Answer to Problem 4.20P

The activation enthalpy is 3.83×105 J/mole of O atoms and the pre exponential constant is 0.40×10−3 m2/s.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Formula used:

The diffusion coefficient of point 1 is given by,

D0ZnO1=D0ZnO0exp(−ΔH D 0ZnO kT1) ....... (I)

Here, D0Zn01 is the diffusion coefficient of point 1, D0ZnO is the pre-exponential constant for oxygen diffusion into ZnO, k is the Boltzmann’s constant, T1 is the temperature at point 1 and ΔHD0ZnO is the activation enthalpy for oxygen into ZnO.

The diffusion coefficient of point 2 is given by,

D0ZnO2=D0ZnO0exp(−ΔH D 0ZnO kT2) ....... (II)

Here, D0Zn02 is the diffusion coefficient of point 1 and T2 is the temperature at point 2.

The ratio of diffusion coefficient of points 1 and 2 is given by,

D 0 ZnO 1 D 0 ZnO 2 =D 0 ZnO 0 exp( − Δ H D 0ZnO k T 1 )D 0 ZnO 0 exp( − Δ H D 0ZnO k T 2 )=exp( − Δ H D 0ZnO k T 1 )exp( − Δ H D 0ZnO k T 2 )=exp[−( Δ H D 0ZnO k)( 1 T 1 − 1 T 2 )] ....... (III)

Calculation:

The activation enthalpy is calculated as,

Substitute 1×10−12 cm2/s for D0ZnO1, 1×10−18 cm2/s for D0ZnO2, 8.62×10−5 eV/atom⋅K for k, 1587 K for T1 and 1075 K for T2 in equation (III).

1× 10 −12 cm 2/s1× 10 −18 cm 2/s=exp[−( Δ H D 0ZnO 8.62× 10 −5 eV/ atom ⋅K)( 1 1587 K− 1 1075 K)]106=exp[−ΔH D 0ZnO(−3.48 atom/ eV)]ln( 106)=ΔHD 0ZnO(3.48 atom/eV)ΔHD 0ZnO=ln( 10 6 )3.48 atom/eV

Solve further,

ΔHD 0ZnO=13.823.48 atom/eV=(3.97 eV/atom× 96.521× 10 3 J/ mole of O atoms 1 eV/ atom )=3.83×105 J/mole of O atoms

The diffusion coefficient is calculated as,

Substitute 1×10−12 cm2/s for D0ZnO1, 8.62 ×10−5 eV/atom⋅K for k, 1587 K for T1 and 3.97 eV/atom for ΔHD0ZnO in equation (I).

1×10−12 cm2/s=D0 ZnO0exp(− 3.97 eV/ atom ( 8.62× 10 −5 eV/ atom ⋅K )( 1587 K ))D0 ZnO0=( 1× 10 −12 cm 2 /s × 10 −4 m 2 1 cm 2 )exp( −29.02)=1× 10 −16 m 2/s2.32 × 10 −13=0.40×10−3 m2/s

Conclusion:

Therefore, the activation enthalpy is 3.83×105 J/mole of O atoms and the pre exponential constant is 0.40×10−3 m2/s.

Answer 12

Ch. 4 - Prob. 1CQ Ch. 4 - Prob. 2CQ Ch. 4 - Prob. 3CQ Ch. 4 - Prob. 4CQ Ch. 4 - Prob. 5CQ Ch. 4 - Prob. 6CQ Ch. 4 - Prob. 7CQ Ch. 4 - Prob. 8CQ Ch. 4 - Prob. 9CQ Ch. 4 - Prob. 10CQ

The activation enthalpy and pre-exponential constant for the diffusion of oxygen ( O 18 ) into ZnO .

Concept explainers

The activation enthalpy and pre-exponential constant for the diffusion of oxygen (O18) into ZnO.

Answer to Problem 4.20P

Explanation of Solution

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Chapter 4 Solutions