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Chapter 5: Problem 27

The diffusion coefficients for nickel in iron are given at two temperatures, as follows: \begin{tabular}{cc} \hline \(\boldsymbol{T}(\boldsymbol{K})\) & \(\boldsymbol{D}\left(\mathrm{m}^{2} / \mathrm{s}\right)\) \\ \hline 1473 & \(2.2 \times 10^{-15}\) \\ \hline 1673 & \(4.8 \times 10^{-14}\) \\ \hline \end{tabular} (a) Determine the values of \(D_{0}\) and the activation energy \(Q_{d^{\prime}}\) (b) What is the magnitude of \(D\) at \(1300^{\circ} \mathrm{C}(1573 \mathrm{~K}) ?\)

Short Answer

Expert verified
Question: Determine the values of \(D_{0}\) and the activation energy \(Q_{d^{\prime}}\) for the diffusion of nickel in iron using the given diffusion coefficients at two different temperatures. Also, find the magnitude of \(D\) at 1300°C (1573 K). Solution: Using the Arrhenius equation and the given diffusion coefficients at two different temperatures, we have found that: - The value of \(D_0\) is approximately \(2.14 \times 10^{-7}\,\mathrm{m^2/s}\), - The activation energy \(Q_d\) is approximately \(2.84 \times 10^5\,\mathrm{J/mol}\), - The magnitude of \(D\) at 1300°C (1573 K) is approximately \(4.06 \times 10^{-15}\,\mathrm{m^2/s}\).

Step by step solution

01

Set up the system of equations

From the Arrhenius equation, we have two equations with two unknowns, \(D_{0}\) and \(Q_{d^{\prime}}\): \[D_1 = D_0 e^{\frac{-Q_d}{R T_1}}\] \[D_2 = D_0 e^{\frac{-Q_d}{R T_2}}\] where: - \(D_1 = 2.2 \times 10^{-15}\,\mathrm{m^2/s}\) at \(T_1 = 1473\,\mathrm{K}\), - \(D_2 = 4.8 \times 10^{-14}\,\mathrm{m^2/s}\) at \(T_2 = 1673\,\mathrm{K}\).
02

Divide the equations

Divide the two equations to eliminate \(D_0\) from the system: \[\frac{D_1}{D_2} = \frac{D_0 e^{\frac{-Q_d}{R T_1}}}{D_0 e^{\frac{-Q_d}{R T_2}}}\] Simplify the equation to solve for \(Q_d\): \[\frac{D_1}{D_2} = e^{\frac{Q_d}{R}(\frac{1}{T_1}-\frac{1}{T_2})}\]
03

Solve for \(Q_d\)

Plug in the values of \(D_1, D_2, T_1,\) and \(T_2\) into the equation, then solve for \(Q_d\): \[\frac{2.2 \times 10^{-15}}{4.8 \times 10^{-14}} = e^{\frac{Q_d}{8.314}(0.000680-\frac{1}{1673})}\] To solve for \(Q_d\), take the natural logarithm of both sides and then multiply by \(RT_1 T_2\): \[Q_d = (\ln{\frac{D_1}{D_2}}) R T_1 T_2 (\frac{1}{T_1}-\frac{1}{T_2}) \approx 2.84 \times 10^5\,\mathrm{J/mol}\]
04

Solve for \(D_0\)

Now, plug in the value of \(Q_d\) into one of the original equations to solve for \(D_0\): \[D_0 = \frac{D_1}{e^{\frac{-Q_d}{R T_1}}} = \frac{2.2 \times 10^{-15}}{e^{\frac{-2.84 \times 10^{5}}{8.314\cdot 1473}}} \approx 2.14 \times 10^{-7}\,\mathrm{m^2/s}\]
05

Find the magnitude of \(D\) at 1300°C (1573 K)

Plug the values of \(D_0\) and \(Q_d\) found in part (a) into the Arrhenius equation to find the magnitude of \(D\) at 1300°C (1573 K): \[D = D_0 e^{\frac{-Q_d}{RT}} = (2.14 \times 10^{-7}) e^{\frac{-2.84 \times 10^{5}}{8.314\cdot 1573}} = 4.06 \times 10^{-15}\,\mathrm{m^2/s}\] Therefore, the magnitude of \(D\) at 1300°C (1573 K) is approximately \(4.06 \times 10^{-15}\,\mathrm{m^2/s}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Arrhenius Equation
The Arrhenius equation is a fundamental formula that describes how the rate of a chemical process, including diffusion, depends on temperature. The general form of the equation is \[ D = D_0 e^{-\frac{Q}{RT}} \] where:\
    \
  • \( D \) is the diffusion coefficient at temperature \( T \),\<\/li>\
  • \( D_0 \) is the pre-exponential factor or frequency factor, representing the maximum possible diffusion coefficient at high temperatures,\<\/li>\
  • \( Q \) is the activation energy required for diffusion,\<\/li>\
  • \( R \) is the universal gas constant (\( 8.314 \,\mathrm{J/mol\,K} \)),\<\/li>\
  • \( T \) is the absolute temperature in Kelvin.\<\/li>\<\/ul>\
    In materials science, the Arrhenius equation helps us understand how temperature affects the mobility of atoms or molecules within a material. As temperature increases, the diffusion coefficient increases, allowing for atoms to move more freely. This relationship is exponential, meaning even small temperature increases can significantly increase diffusion rates.\
    \
    When solving diffusion problems like the one from the textbook, the Arrhenius equation can be used to find unknowns such as the diffusion coefficient \( D \) at various temperatures or the activation energy \( Q \), when appropriate diffusion coefficients at known temperatures are given. Moreover, by rearranging the equation, we can compare the diffusion coefficients at two different temperatures to eliminate the pre-exponential factor \( D_0 \) and isolate \( Q \), as demonstrated in the exercise.
Activation Energy
Activation energy, denoted by \( Q \) or \( Q_d \) in the context of diffusion, is the energy barrier that must be overcome for atoms or molecules to move from one place to another within a material. This energy is essential for initiating the diffusion process.\
\
In the exercise, the activation energy is determined by using the diffusion coefficients at two different temperatures. The concept of activation energy is foundational in understanding not only diffusion but also various chemical reactions and processes.\
    \
  • It provides insights into the temperature dependence of the rate of diffusion (or reaction).\<\/li>\
  • Substances with high activation energy rates will diffuse more slowly at a given temperature compared to those with lower activation energies.\<\/li>\<\/ul>\
    Within the structure of the Arrhenius equation, the activation energy inversely relates to the temperature: as the temperature increases, the term \( e^{-\frac{Q}{RT}} \) increases, resulting in a higher diffusion rate. It indicates how sensitive a diffusion process is to temperature changes. In the provided solution, the calculated activation energy is used to further evaluate the diffusion coefficient at another temperature, showcasing its fundamental role in predicting material behaviors at the atomic level.
Diffusion Coefficients
Diffusion coefficients, symbolized as \( D \), are values that quantify the rate at which atoms or molecules can move through a material. They indicate how quickly a substance can spread out, or diffuse, from an area of high concentration to an area of low concentration. These coefficients are crucial in materials science for modeling the diffusion process and can vary with temperature, as well as the material's characteristics.\
    \
  • Higher diffusion coefficients mean faster diffusion rates.\<\/li>\
  • They depend on both intrinsic factors like atomic size and extrinsic factors such as the microstructure of the material.\<\/li>\<\/ul>\
    During the exercise, the objective was to find the magnitude of the diffusion coefficient at a new temperature using the previously determined values of \( D_0 \) and the activation energy. The calculation of \( D \) at a specific temperature is important for practical applications such as heat treatment of metals, where controlling diffusion can impact the performance and durability of the final product.\
    \
    Finding the diffusion coefficient at one temperature allows us to use the Arrhenius relation to predict the diffusion coefficients at other temperatures, giving us the ability to understand and forecast material behavior under different thermal conditions.

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Most popular questions from this chapter

The steady-state diffusion flux through a metal plate is \(7.8 \times 10^{-8} \mathrm{~kg} / \mathrm{m}^{2} \cdot \mathrm{s}\) at a temperature of \(1200^{\circ} \mathrm{C}(1473 \mathrm{~K})\) and when the concentration gradient is \(-500 \mathrm{~kg} / \mathrm{m}^{4}\). Calculate the diffusion flux at \(1000^{\circ} \mathrm{C}(1273 \mathrm{~K})\) for the same concentration gradient and assuming an activation energy for diffusion of \(145,000 \mathrm{~J} / \mathrm{mol}\).

The activation energy for the diffusion of copper in silver is \(193,000 \mathrm{~J} / \mathrm{mol}\). Calculate the diffusion coefficient at \(1200 \mathrm{~K}\left(927^{\circ} \mathrm{C}\right)\), given that \(D\) at \(1000 \mathrm{~K}\left(727^{\circ} \mathrm{C}\right)\) is \(1.0 \times 10^{-14} \mathrm{~m}^{2} / \mathrm{s}\).

Consider a diffusion couple between silver and a gold alloy that contains 10 wt \% silver. This couple is heat treated at an elevated temperature and it was found that after 850 s, the concentration of silver had increased to \(12 \mathrm{wt} \%\) at \(10 \mu \mathrm{m}\) from the interface into the Ag-Au alloy. Assuming preexponential and activation energy values of \(7.2 \times 10^{-6}\) \(\mathrm{m}^{2} / \mathrm{s}\) and \(168,000 \mathrm{~J} / \mathrm{mol}\), respectively, compute the temperature of this heat treatment. (Note: You may find Figure \(5.13\) and Equation \(5.15\) helpful.) For a steel alloy, it has been determined that a carburizing heat treatment of \(15 \mathrm{~h}\) duration will raise the carbon concentration to \(0.35\) wt \(\%\) at a point \(2.0 \mathrm{~mm}\) from the surface. Estimate the time necessary to achieve the same concentration at a \(6.0-\mathrm{mm}\) position for an identical steel and at the same carburizing temperature.

Consider the diffusion of some hypothetical metal Y into another hypothetical metal \(Z\) at \(950^{\circ} \mathrm{C}\); after \(10 \mathrm{~h}\) the concentration at the \(0.5 \mathrm{~mm}\) position (in metal \(Z\) ) is \(2.0 \mathrm{wt} \% \mathrm{Y}\). At what position will the concentration also be \(2.0 \mathrm{wt} \% \mathrm{Y}\) after a \(17.5-\mathrm{h}\) heat treatment again at \(950^{\circ} \mathrm{C}\) ? Assume preexponential and activation energy values of \(4.3 \times 10^{-4} \mathrm{~m}^{2} / \mathrm{s}\) and \(180,000 \mathrm{~J} / \mathrm{mol}\), respectively, for this diffusion system.

(a) Briefly explain the concept of a driving force. (b) What is the driving force for steady-state diffusion?
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