The activation energy for the diffusion of carbon in chromium is \(111,000 \mathrm{~J} / \mathrm{mol}\). Calculate the diffusion coefficient at \(1100 \mathrm{~K}\left(827^{\circ} \mathrm{C}\right)\), given that \(D\) at \(1400 \mathrm{~K}\left(1127^{\circ} \mathrm{C}\right)\) is \(6.25 \times\) \(10^{-11} \mathrm{~m}^{2} / \mathrm{s}\)

Short Answer

Expert verified
Answer: The diffusion coefficient of carbon in chromium at 1100K is approximately 1.79 x 10^(-12) m^2/s.

Step by step solution

01

Write down the known information

We are given the following information: - Activation energy, \(E_a = 111000 J/mol\) - Diffusion coefficient at \(1400K\), \(D_1 = 6.25\times10^{-11} m^2/s\) - Temperature at which we need to calculate the diffusion coefficient, \(T_2 = 1100K\)
02

Write down the Arrhenius equation

The Arrhenius equation is: \(D = D_0 e^{-E_a / RT}\) We need to rewrite this equation for \(D_0\) and apply it to the known values at \(T_1 = 1400K\): \(D_0 = D_1 e^{E_a / RT_1}\)
03

Calculate the value of \(D_0\)

Using the given values, we can compute \(D_0\): \(D_0 = (6.25\times10^{-11}) e^{(111000) / (8.314\cdot 1400)}\) Upon computation, we find: \(D_0 \approx 2.54\times10^{-5} m^2/s\)
04

Calculate the diffusion coefficient at \(1100K\)

Now, we can use the Arrhenius equation again to compute the diffusion coefficient at \(1100K\): \(D_2 = D_0 e^{-E_a / RT_2}\) Plug in the known values: \(D_2 = (2.54\times10^{-5}) e^{-(111000)/(8.314\cdot 1100)}\) After calculating, we find: \(D_2 \approx 1.79\times10^{-12} m^2/s\) So, the diffusion coefficient at \(1100K\) is approximately \(1.79\times10^{-12} m^2/s\).

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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 formula that connects the rate of a chemical reaction to temperature. The equation is:
\[D = D_0 e^{-E_a / RT}\]
In this context, it helps us understand how the rate of diffusion changes with temperature. Here, \(D\) is the diffusion coefficient at a given temperature, \(D_0\) is the pre-exponential factor or frequency factor, a constant that represents the diffusion coefficient at an infinite temperature, \(E_a\) is the activation energy, \(R\) is the universal gas constant, and \(T\) is the temperature in Kelvin. This relationship implies that the rate at which particles diffuse increases exponentially with an increase in temperature, provided that the activation energy remains constant. Understanding how to manipulate and apply this equation is essential for predicting material behavior in various thermal environments, such as high-temperature alloy treatments or semiconductor manufacturing processes.
When working with the Arrhenius equation, it's often necessary to rearrange the formula to solve for different variables - for instance, if the activation energy is known and you are calculating the diffusion coefficient at a new temperature. In such cases, good algebraic manipulation skills become very important to find the correct solution.
Activation Energy
Activation energy, represented as \(E_a\), is a crucial concept in materials science, particularly when dealing with diffusion processes. It is defined as the minimum energy that particles need to start a reaction, in this case, to overcome the energy barrier for diffusion. In other words, activation energy is the 'kick' necessary to initiate the movement of atoms or molecules within a material.
This energy is highly significant as it dictates how fast diffusion occurs. The higher the activation energy, the lower the rate of diffusion at a given temperature, since fewer particles will have the required energy to move. On a microscopic scale, activation energy is involved in breaking bonds or altering the structure slightly so that atoms can 'jump' from one position to another. When teaching about activation energy, it's helpful to explain it in the context of a real-life analogy, such as overcoming the effort needed to start pushing a heavy object - once that initial push is given, the object moves more easily. In our exercise, the activation energy for the diffusion of carbon in chromium helps us understand how carbon atoms move within the chromium matrix at different temperatures.
Materials Science Diffusion
Diffusion in materials science is the process by which atoms, ions, or molecules spread out and move from areas of high concentration to areas of lower concentration. This movement is driven by the atomic desire to achieve equilibrium.
Understanding diffusion is crucial as it affects properties like hardness, electrical conductivity, and corrosion resistance of materials. For instance, in metallurgy, controlling the diffusion of carbon into metal surfaces can transform material strength and weariness. The diffusion coefficient (\(D\)) is the measure of how quickly diffusion occurs. It depends on several factors, including temperature, as dictated by the Arrhenius equation.
In an educational setting, using visual aids such as animations or diagrams showing particles moving and spreading out can be a powerful way to demonstrate diffusion. It helps students visualize the sometimes abstract concept of atoms or molecules 'wandering' through a solid, liquid, or gas. Describing diffusion in a practical context, such as the hardening of steel through the introduction of carbon or the tarnishing of silver when exposed to sulfur compounds in the air, can greatly enhance comprehension and foster real-world connections to the theory.

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Most popular questions from this chapter

The preexponential and activation energy for the diffusion of iron in cobalt are \(1.1 \times 10^{-5}\) \(\mathrm{m}^{2} / \mathrm{s}\) and \(253,300 \mathrm{~J} / \mathrm{mol}\), respectively. At what temperature will the diffusion coefficient have a value of \(2.1 \times 10^{-14} \mathrm{~m}^{2} / \mathrm{s}\) ?

At approximately what temperature would a specimen of \(\gamma\)-iron have to be carburized for \(2 \mathrm{~h}\) to produce the same diffusion result as at \(900^{\circ} \mathrm{C}\) for \(15 \mathrm{~h}\) ?

When \(\alpha\)-iron is subjected to an atmosphere of hydrogen gas, the concentration of hydrogen in the iron, \(C_{\mathrm{H}}\) (in weight percent), is a function of hydrogen pressure, \(p_{\mathrm{H}_{2}}\) (in MPa), and absolute temperature \((T)\) according to $$ C_{\mathrm{H}}=1.34 \times 10^{-2} \sqrt{p_{\mathrm{H}}_{2}} \exp \left(-\frac{27.2 \mathrm{~kJ} / \mathrm{mol}}{R T}\right) $$ Furthermore, the values of \(D_{0}\) and \(Q_{d}\) for this diffusion system are \(1.4 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}\) and \(13,400 \mathrm{~J} / \mathrm{mol}\), respectively. Consider a thin iron membrane \(1 \mathrm{~mm}\) thick that is at \(250^{\circ} \mathrm{C}\). Compute the diffusion flux through this membrane if the hydrogen pressure on one side of the membrane is \(0.15 \mathrm{MPa}\) (1.48 atm), and on the other side \(7.5 \mathrm{MPa}\) (74 atm).

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

Briefly explain the concept of steady state as it applies to diffusion.

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