/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 26 A common procedure for surface m... [FREE SOLUTION] | 91Ó°ÊÓ

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A common procedure for surface micromachining involves chemical etching of sacrificial layers. In one procedure, phosphosilicate glass is etched by hydrofluoric acid, with an overall reaction of $$ 6 \mathrm{HF}+\text { Si } \mathrm{O}_{2} \rightarrow \mathrm{H}_{2} \mathrm{Si} \mathrm{F}_{6}+2 \mathrm{H}_{2} \mathrm{O} $$ When the etching depth is small, the reaction rate is rate-controlled; but for deep etching, as required to machine channels, diffusion of acid through the solution also plays a role. Assuming quasi-steady diffusion and a one- dimensional model, obtain an expression for the etching depth as a function of time. The reaction is approximately first-order in HF molar concentration with rate constant \(k\) ". Hence determine the time required to etch a channel \(100 \mu \mathrm{m}\) deep if \(k^{\prime \prime}=2 \times 10^{-6} \mathrm{~m} / \mathrm{s}\), \(\mathscr{D}_{1 m}=2 \times 10^{-9} \mathrm{~m}^{2} / \mathrm{s}\), and the bulk concentration of HF in an aqueous solution is \(7.0 \mathrm{kmol} / \mathrm{m}^{3}\).

Short Answer

Expert verified
The time required to etch a 100 µm deep channel is approximately 0.0002 seconds.

Step by step solution

01

Understand the Reaction

The chemical reaction given is \(6 \mathrm{HF} + \mathrm{SiO}_2 \rightarrow \mathrm{H}_{2}\mathrm{SiF}_6 + 2 \mathrm{H}_2\mathrm{O}\). This reaction is first-order with respect to HF, meaning that the rate of reaction is proportional to the concentration of HF.
02

Define Variables for Diffusion and Reaction

For quasi-steady-state diffusion in a one-dimensional model, use Fick's first law of diffusion. The diffusion coefficient \(\mathscr{D}\) is given as \(2 \times 10^{-9} \mathrm{~m}^2/\mathrm{s}\), and the reaction rate constant for surface reaction \(k''\) is given as \(2 \times 10^{-6} \mathrm{~m}/\mathrm{s}\). The bulk concentration of HF \(C_b\) is \(7.0 \mathrm{kmol}/\mathrm{m}^3\).
03

Establish the Rate Equation

The rate of etching can be expressed through a balance of surface reaction and diffusion, given by the formula: \( R = \frac{k'' C_s}{1 + \frac{k''}{\mathscr{D}} \cdot H}\), where \(C_s\) is the concentration at the surface and \(H\) is the etching depth.
04

Solve for Surface Concentration \(C_s\)

Utilize the relationship that at equilibrium, diffusion rate equals reaction rate, and solve for \(C_s\) : \( C_s = \frac{C_b}{1 + \frac{k''}{\mathscr{D}} \cdot H} \). This establishes the dependency of surface concentration on etching depth and initial concentration.
05

Determine Depth-Time Relationship

Differentiate the depth-time relationship keeping in mind the previous substitution for \(C_s\). You'll arrive at: \( H(t) = \frac{\mathscr{D}}{k''} \ln \left( \frac{C_b}{C_{s,in}} \cdot \exp \left( \frac{k''}{\mathscr{D}}t \right) \right) \). Here, \(C_{s,in}\) is the initial surface concentration.
06

Calculate Time for \(100 \mu m\) Depth

Given that \(H = 100 \mu \mathrm{m} \), convert this to \(0.1 \mathrm{mm} \) or \(0.0001 \mathrm{m} \). Substitute \(H\), \(C_b\), and given constants into the depth-time relationship and solve for \(t\). The required time \(t\) can be calculated using: \( H = \frac{\mathscr{D}}{k''} \ln \left( \frac{C_b}{C_s} \right) \).
07

Final Calculation

Substitute values: \( H = 0.0001 \mathrm{m} \), \( \mathscr{D} = 2 \times 10^{-9} \mathrm{m}^2/\mathrm{s} \), \( k'' = 2 \times 10^{-6} \mathrm{m}/\mathrm{s} \), \( C_b = 7.0 \mathrm{kmol/m}^3 \). Calculate \( t \) by rearranging the formula: \( t = \frac{\mathscr{D}}{k''} \ln \left( 1 + \frac{k''}{\mathscr{D}} \cdot H \right) \approx 0.0002 \, \text{seconds}\).

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Key Concepts

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

Surface Micromachining
Surface micromachining is a technique used to create miniature structures on a substrate by selectively removing material. This process is essential in industries like microelectronics and MEMS (Micro-Electro-Mechanical Systems). The goal is to create detailed patterns or structures by adding and removing thin layers.
In this particular exercise, phosphosilicate glass undergoes chemical etching, a common aspect of surface micromachining. Here, hydrofluoric acid (HF) is used to dissolve sacrificial layers, resulting in cavities or channels. The efficiency and precision of this process rely on understanding how various factors affect the removal rate of these layers. Chemical etching is central because it allows for intricate designs and is generally less costly than other alternatives like photolithography.
Ultrafine surfaces require careful coordination between reaction rates and diffusion processes. Ensuring the correct balance results in precise dimensions, which are crucial for devices that need exact specifications.
Diffusion
Diffusion is the movement of molecules from an area of higher concentration to one of lower concentration. It plays a vital role in many chemical processes, including the etching of microscale structures.
In the context of the exercise, diffusion determines how fast the hydrofluoric acid reaches the surface being etched. The diffusion coefficient, \(\mathscr{D} = 2 \times 10^{-9} \, \mathrm{m}^2/\mathrm{s}\), gives insight into how quickly this happens. The lower the coefficient, the slower the movement, impacting how fast the etching proceeds.
Understanding diffusion helps predict how experimental controls or environmental conditions might accelerate or slow down the process. Especially when dealing with toxic or volatile chemicals like HF, optimizing diffusion parameters is crucial for both safety and efficiency. Knowing how diffusion interacts with other factors, like concentration gradients, allows for more refined control over the etching depth and time.
Reaction Rate
Reaction rate is a measure of how quickly a chemical reaction occurs. In this micromachining context, it primarily refers to how quickly hydrofluoric acid reacts with the material being etched.
The rate of reaction for HF and SiO2 etching is first-order concerning HF concentration. In simple terms, this means the reaction rate depends linearly on the concentration of HF present. The reaction rate constant, \(k'' = 2 \times 10^{-6} \, \mathrm{m}/\mathrm{s}\), further defines this relationship.
Knowing the reaction rate is crucial for predicting how long it will take to etch specific depths. As the exercise illustrates, when the reaction rate is known, it's possible to calculate the time required to achieve a desired etching depth. This calculation helps engineers plan and control micromachining processes, ensuring they're efficient and produce the desired outcomes.
Quasi-Steady Diffusion
Quasi-steady diffusion describes a state where, over a certain period, the concentration gradient driving diffusion reaches a point where it remains relatively stable. This assumes that any changes in the system over short periods are slow enough to allow predicting future behavior without continuous recalculation.
This concept simplifies managing the chemical etching process. By assuming a quasi-steady state, engineers can use mathematical models to reasonably predict the etching depth over time without needing constant recalibration. The exercise applies this by using a one-dimensional diffusion model, combining both diffusion and reaction rates into one expression.
Understanding quasi-steady diffusion allows precise control over the etching rate. With this knowledge, engineers can optimize the balance between the diffusion and reaction rates necessary for other processes. It provides clarity on the interactions within a complex chemical system, which is crucial for accurate and efficient surface micromachining.

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Most popular questions from this chapter

A laboratory experiment requires a supply of ammonia gas. In order to maintain the supply pressure close to 1 atm, it is proposed to provide a \(3 \mathrm{~mm}\)-I.D., 10 m-long vent line to connect the supply pipe with ambient air outside the laboratory. To evaluate the performance of the venting process, your supervisor has asked you to determine the rate at which ammonia gas diffuses through the vent line. (i) If the temperature and pressure are \(300 \mathrm{~K}\) and 1 atm, respectively, calculate the rate at which \(\mathrm{NH}_{3}\) is lost by diffusion in g/day. Perform your analysis on a molar basis to take advantage of the almost constant temperature and pressure along the tube. (ii) The diffusion flow rate calculated in part (i) is very small. Calculate the pressure difference required to give a bulk flow rate of \(\mathrm{NH}_{3}\) equal to the diffusion flow. Since this pressure difference is extremely small, discuss the relevance of the diffusion calculation to the practical venting problem.

An automobile catalytic converter has a ceramic matrix with \(1 \mathrm{~mm}\)-square passages and \(0.2 \mathrm{~mm}\)-thick walls. The walls are coated with a \(20 \mu \mathrm{m}\)-thick layer of porous alumina that is impregnated with a platinum alloy catalyst (the washcoat). The purpose of the porous washcoat can be viewed as giving an effective reaction rate constant \(k_{\text {eff }}^{\prime \prime}\) based on \(s\)-surface area that is much larger than if the catalyst were simply coated onto a solid wall. (i) If the actual rate constant for \(\mathrm{CO}\) oxidation is \(k^{\prime \prime}=7 \times 10^{-3} \mathrm{~m} / \mathrm{s}, \mathscr{D}_{1, \text { eff }}\) in the washcoat is \(1.6 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\), and the washcoat area-to-volume ratio is \(a_{p}=5 \times 10^{6} \mathrm{~cm}^{2} / \mathrm{cm}^{3}\), determine \(k_{\text {eff }}^{\prime \prime}\) (ii) If the exhaust is at \(800 \mathrm{~K}, 1.1\) bar, is the removal of \(\mathrm{CO}\) controlled by diffusion of CO from the gas stream or by reaction within the washcoat?

A closed cylindrical vessel containing stagnant air stands with its axis vertical. Each end is maintained at a uniform temperature with the base colder than the upper end; the cylindrical wall is insulated. Neglecting the variation in hydrostatic pressure, ascertain whether there are vertical gradients of (i) partial density of oxygen. (ii) partial pressure of oxygen. (iii) mass fraction of oxygen.

Methane is burned with \(20 \%\) excess air. At \(1250 \mathrm{~K}\) the equilibrium composition of the product is: \(\begin{array}{rlcccc}\text { Species } i: & \mathrm{CO}_{2} & \mathrm{H}_{2} \mathrm{O} & \mathrm{O}_{2} & \mathrm{~N}_{2} & \mathrm{NO} \\ x_{i}: & 0.0803 & 0.160 & 0.0325 & 0.727 & 0.000118\end{array}\) Determine the mean molecular weight \(M\) and gas constant \(R\) of the mixture, and the mass fraction of the pollutant nitric oxide in parts per million.

A new type of cooling tower packing has been tested in a small-scale test rig containing a \(28 \mathrm{~cm} \times 28 \mathrm{~cm}\) square cross-section, \(0.9 \mathrm{~m}-\mathrm{high}\) packing element. The following data were measured: \(\dot{m}_{G, \text { in }}=0.178 \mathrm{~kg} / \mathrm{s} ; \dot{m}_{L, \text { in }}=0.550 \mathrm{~kg} / \mathrm{s}\); \(T_{\mathrm{DB}, \text { in }}=24.3^{\circ} \mathrm{C}, T_{\mathrm{WB}, \text { in }}=14.5^{\circ} \mathrm{C} ; T_{\mathrm{WB}, \text { out }}=25.2^{\circ} \mathrm{C} ; T_{L, \text { in }}=28.90^{\circ} \mathrm{C} ;\) \(T_{L, \text { out }}=26.03^{\circ} \mathrm{C}\). Also, \(P=1\) atm. (i) Perform an energy balance to assess the accuracy of the data. (ii) Determine the NTU obtained in the test. (iii) The manufacturer has proposed the following correlation for the mass transfer conductance times specific area product: $$ g_{m} a=1.46 L^{0.39} G^{0.5} \mathrm{~kg} / \mathrm{m}^{3} \mathrm{~s} $$ Compare the measured \(g_{m} a\) with the manufacturer's value.

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