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 \(1.0 \times 10^{-14} \mathrm{~m}^{2} / \mathrm{s}\).

Short Answer

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**Question:** Calculate the diffusion coefficient at 1200 K, given the activation energy for diffusion is 193,000 J/mol, initial temperature is 1000 K, and initial diffusion coefficient is \(1.0 \times 10^{-14}~m^2/s\). **Answer:** The diffusion coefficient at 1200 K is approximately \(3.79 \times 10^{-15}~m^2/s\).

Step by step solution

01

Convert activation energy to kJ/mol

Divide the given activation energy by 1000 to convert it to kJ/mol. $$ Q = \frac{193,000~J/mol}{1000} = 193~kJ/mol $$
02

Plug in variables

Now that we have our units converted, we'll plug in the variables into the Arrhenius equation to find the final diffusion coefficient. $$ D_2 = D_1 \times \exp { \bigg( -\frac{Q}{R} \left(\frac{1}{T_2} - \frac{1}{T_1} \right) \bigg) } $$
03

Calculate the new diffusion coefficient

Plug all the values into the equation and solve for \(D_2\): $$ D_2 = (1.0 \times 10^{-14}~m^2/s) \times \exp{ \bigg( -\frac{193 \times10^{3}J/mol}{8.314 J/(mol \cdot K)} \left(\frac{1}{1200 K} - \frac{1}{1000 K} \right) \bigg) } $$ Calculate the value inside the brackets and exponent first: $$ \frac{193 \times10^{3}J/mol}{8.314 J/(mol \cdot K)} \left(\frac{1}{1200 K} - \frac{1}{1000 K} \right) \approx 3.275 $$ Now, calculate the exponential value and multiply it with the initial diffusion coefficient: $$ D_2 = (1.0 \times 10^{-14}~m^2/s) \times \exp{(-3.275)} \approx 3.79 \times 10^{-15}~m^2/s $$ So, the diffusion coefficient at 1200 K (927°C) is approximately \(3.79 \times 10^{-15}~m^2/s\).

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Key Concepts

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

Arrhenius Equation
Understanding the basics of the Arrhenius equation is crucial for students tackling problems related to reaction rates in chemistry and diffusion processes in materials science. At its core, it provides a mathematical relationship between the rate of a chemical reaction or a diffusion process and temperature.
The equation is expressed as: \[\begin{equation}D = D_0 \times \text{exp}\bigg(-\frac{Q}{RT}\bigg)\end{equation}\]Here, \(D\) is the diffusion coefficient, \(D_0\) is the pre-exponential factor, \(Q\) represents the activation energy, \(R\) is the universal gas constant, and \(T\) is the absolute temperature in Kelvin. In simple terms, this equation suggests that as the temperature increases, the diffusion coefficient also increases, meaning diffusion occurs more rapidly.
It's vital to understand the exponential function in this equation, as it indicates that small changes in temperature can lead to significant changes in the diffusion rate. This concept is widely applied in materials science to predict how materials will behave under different thermal conditions. In the given exercise, the Arrhenius equation helps calculate the diffusion coefficient of copper in silver at various temperatures.
Activation Energy
Activation energy, often symbolized as \(Q\) in the Arrhenius equation, is a fundamental concept in both chemistry and materials science, representing the minimum energy barrier that must be overcome for a chemical reaction or a diffusion process to occur.

Role of Activation Energy

Activation energy can be thought of as the 'gatekeeper' of chemical reactions and diffusion processes. A higher activation energy means that reactant molecules or diffusing atoms need more energy to start the process. In the context of the exercise, the activation energy of 193,000 J/mol is quite significant, indicating that copper atoms require a considerable amount of energy to diffuse through silver.

Temperature Influence

Temperature plays a vital role in overcoming this energy barrier. As temperature increases, the kinetic energy of atoms or molecules also increases, enabling them to overcome the activation energy more easily. This is why the exercise asks for the diffusion coefficient at a higher temperature to understand how the diffusion process accelerates with increased thermal energy.
Understanding this concept is vital for developing and optimizing materials for specific uses, as different applications may require materials that exhibit certain diffusion characteristics at various temperature ranges.
Materials Science
Materials science is an interdisciplinary field that studies the properties of matter and its applications in various areas, including engineering, biology, and physics. This discipline involves the investigation of the relationship between the structure of materials at atomic or molecular scales and their macroscopic properties.

Diffusion in Materials Science

One of the key phenomena studied within materials science is diffusion - the process by which atoms or molecules spread out as a result of random motion from regions of high concentration to regions of low concentration. Diffusion is crucial in materials processing, such as alloy production, semiconductor fabrication, and the heat treatment of metals.
The exercise provided involves calculating the diffusion coefficient, a quantity that gives insights into how quickly atoms, like copper in silver, can move through a solid matrix. This coefficient determines how a material will respond under thermal treatment, which is essential for tailoring its properties for specific applications.

Importance in Everyday Materials

From the creation of durable building materials to the development of efficient electronic devices, understanding the science behind material properties is key. The exercise not only serves as a practical application of the Arrhenius equation and concepts of activation energy but also demonstrates the importance of materials science in developing advanced materials with precise behaviors and functionalities.

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Most popular questions from this chapter

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.

For a steel alloy, it has been determined that a carburizing heat treatment of \(15 \mathrm{~h}\) duration will raise the carbon concentration to \(0.35 \mathrm{wt} \%\) at a point \(2.0 \mathrm{~mm}\) from the surface. Estimate the time necessary to achieve the same concentration at a 6.0-mm position for an identical steel and at the same carburizing temperature.

A sheet of BCC iron \(2-\mathrm{mm}\) thick was exposed to a carburizing gas atmosphere on one side and a decarburizing atmosphere on the other side at \(675^{\circ} \mathrm{C}\). After reaching steady state, the iron was quickly cooled to room temperature. The carbon concentrations at the two surfaces of the sheet were determined to be \(0.015\) and \(0.0068\) wt \(\%\), respectively. Compute the diffusion coefficient ifthe diffusion flux is \(7.36 \times 10^{-9} \mathrm{~kg} / \mathrm{m}^{2}+\mathrm{s}\). Hint: Use Equation \(4.9\) to convert the concentrations from weight percent to kilograms of carbon per cubic meter of iron.

For the predeposition heat treatment of a semiconducting device, gallium atoms are to be diffused into silicon at a temperature of \(1150^{\circ} \mathrm{C}\) for \(2.5 \mathrm{~h}\). If the required concentration of \(\mathrm{Ga}\) at a position \(2 \mu \mathrm{m}\) below the surface is \(8 \times 10^{23}\) atoms \(/ \mathrm{m}^{3}\), compute the required surface concentration of \(\mathrm{Ga}\). Assume the following: (i) The surface concentration remains constant (ii) The background concentration is \(2 \times 10^{19} \mathrm{Ga}\) atoms \(/ \mathrm{m}^{3}\) (iii) Preexponential and activation energy values are \(3.74 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\) and \(3.39 \mathrm{eV} /\) atom, respectively.

The purification of hydrogen gas by diffusion through a palladium sheet was discussed in Section 5.3. Compute the number of kilograms of hydrogen that pass per hour through a \(6-\mathrm{mm}\) thick sheet of palladium having an area of \(0.25 \mathrm{~m}^{2}\) at \(600^{\circ} \mathrm{C}\). Assume a diffusion coefficient of \(1.7 \times 10^{-8} \mathrm{~m}^{2} / \mathrm{s}\), that the respective concentrations at the high- and low-pressure sides of the plate are \(2.0\) and \(0.4 \mathrm{~kg}\). of hydrogen per cubic meter of palladium, and that steady-state conditions have been attained.

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