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Chapter 36: Problem 31

Many surface reactions require the adsorption of two or more different gases. For the case of two gases, assuming that the adsorption of a gas simply limits the number of surface sites available for adsorption, derive expressions for the fractional coverage of each gas.

Short Answer

Expert verified
The expressions for the fractional coverage of two gases A and B using the Langmuir isotherm model are: Fractional coverage of gas A, \(\theta_A\) = \(\frac{K_AP_A}{1+K_AP_A+K_BP_B}\) Fractional coverage of gas B, \(\theta_B\) = \(\frac{K_BP_B}{1+K_AP_A+K_BP_B}\)

Step by step solution

01

Recall the Langmuir Isotherm model

The Langmuir isotherm model is given as: Fractional Coverage (胃) = \(\frac{KP}{1+KP}\) Where 胃 is the fractional coverage, K is the adsorption equilibrium constant, and P is the pressure of the gas.
02

Define the variables for the two gases

Let's represent the fractional coverage of gas A and gas B as \(\theta_A\) and \(\theta_B\), respectively. Let's also represent the adsorption equilibrium constants for A and B as \(K_A\) and \(K_B\), and the pressures of A and B as \(P_A\) and \(P_B\).
03

Set up the equations for fractional coverage

Using the Langmuir isotherm model, we can set up the equations for the fractional coverage for the two gases: \[\theta_A = \frac{K_AP_A}{1+K_AP_A+K_BP_B}\] \[\theta_B = \frac{K_BP_B}{1+K_AP_A+K_BP_B}\] Where we have considered the fact that the denominator accounts for the total number of surface sites.
04

Expressions for fractional coverage

Finally, we have derived the expressions for the fractional coverage of each gas in terms of their respective adsorption equilibrium constants and pressures: Fractional coverage of gas A, \(\theta_A\) = \(\frac{K_AP_A}{1+K_AP_A+K_BP_B}\) Fractional coverage of gas B, \(\theta_B\) = \(\frac{K_BP_B}{1+K_AP_A+K_BP_B}\)

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Key Concepts

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

Adsorption Equilibrium Constant
The adsorption equilibrium constant, commonly denoted as 'K' in the context of the Langmuir Isotherm model, plays a crucial role in understanding how a gas interacts with a surface. It is essentially a measure of how readily gas molecules adsorb onto a surface at a given pressure. A higher value of K indicates a stronger affinity between the gas molecules and the surface, leading to more adsorption. Conversely, a lower K value suggests that the gas molecules prefer to remain in the gaseous phase rather than being adsorbed.

When we consider two different gases competing for the same surface sites, as in our exercise, each gas will have its own unique adsorption equilibrium constant, labeled as \(K_A\) and \(K_B\) respectively. These constants will determine to what extent each gas is adsorbed when both are present.
Fractional Coverage
Fractional coverage, denoted by \(\theta\), is a term that describes the proportion of available adsorption sites on a surface that are occupied by a specific gas. This concept is integral in quantifying the extent of adsorption of gas molecules on a surface. Mathematically, fractional coverage ranges from 0 to 1, where 0 indicates no adsorption and 1 signifies that all available sites are occupied by the gas in question.

In our given problem, the Langmuir Isotherm model provides the formula to calculate the fractional coverage for individual gases, considering that each gas affects the adsorption of the other. Therefore, the availability of surface sites for gas A is influenced by the presence of gas B, and vice versa. The model accounts for this interaction in the denominator of the expressions for \(\theta_A\) and \(\theta_B\), encapsulating the dynamic balance between both gases competing for adsorption sites.
Surface Reactions
Understanding surface reactions is vital in many industrial processes, such as catalysis, where the reaction occurs on the catalyst's surface. These reactions can be influenced by the presence and adsorption of gases on the surface. In the context of the Langmuir Isotherm model, which we鈥檝e used to solve our exercise problem, adsorption is a precursor to surface reactions. The model assumes that for a reaction to proceed, reactant molecules must first adsorb onto specific sites on the surface.

This adsorption is reversible and leads to an equilibrium state described by the fractional coverage and the adsorption equilibrium constants. The interplay between the amount of gas A (\(\theta_A\)) and gas B (\(\theta_B\)) adsorbed on the surface directly affects the rate and direction of the surface reactions between these gases. For instance, if one gas dominates the surface coverage, it will likely alter the pathway and speed at which the surface reaction proceeds.

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Most popular questions from this chapter

(Challenging) Cubic autocatalytic steps are important in a reaction mechanism referred to as the "brusselator" (named in honor of the research group in Brussels that initially discovered this mechanism): $$\begin{array}{l} \mathrm{A} \stackrel{k_{1}}{\longrightarrow} \mathrm{X} \\ 2 \mathrm{X}+\mathrm{Y} \stackrel{k_{2}}{\longrightarrow} 3 \mathrm{X} \\ \mathrm{B}+\mathrm{X} \stackrel{k_{3}}{\longrightarrow} \mathrm{Y}+\mathrm{C} \\\ \mathrm{X} \stackrel{k_{4}}{\longrightarrow} \mathrm{D} \end{array}$$ If \([\mathrm{A}]\) and \([\mathrm{B}]\) are held constant, this mechanism demonstrates interesting oscillatory behavior that we will explore in this problem. a. Identify the autocatalytic species in this mechanism. b. Write down the differential rate expressions for \([\mathrm{X}]\) and \([\mathrm{Y}]\) c. Using these differential rate expressions, employ Euler's method (Section 35.6 ) to calculate \([\mathrm{X}]\) and \([\mathrm{Y}]\) versus time under the conditions \(\mathrm{k}_{1}=1.2 \mathrm{s}^{-1}, \mathrm{k}_{2}=0.5 \mathrm{M}^{-2} \mathrm{s}^{-1}, \mathrm{k}_{3}=\) \(3.0 \mathrm{M}^{-1} \mathrm{s}^{-1}, \mathrm{k}_{4}=1.2 \mathrm{s}^{-1},\) and \([\mathrm{A}]_{0}=[\mathrm{B}]_{0}=1 \mathrm{M} .\) Begin with \([\mathrm{X}]_{0}=0.5 \mathrm{M}\) and \([\mathrm{Y}]_{0}=0.1 \mathrm{M} .\) A plot of \([\mathrm{Y}]\) versus \([\mathrm{X}]\) should look like the top panel in the following figure. d. Perform a second calculation identical to that in part (c), but with \([\mathrm{X}]_{0}=3.0 \mathrm{M}\) and \([\mathrm{Y}]_{0}=3.0 \mathrm{M} .\) A plot of \([\mathrm{Y}]\) versus \([\mathrm{X}]\) should look like the bottom panel in the following figure. e. Compare the left and bottom panels in the following figure. Notice that the starting conditions for the reaction are different (indicated by the black spot). What DO the figures indicate regarding the oscillatory state the system evolves to?

In the unimolecular isomerization of cyclobutane to butylene, the following values for \(k_{u n i}\) as a function of pressure were measured at \(350 \mathrm{K}\) \\[ \begin{array}{lcccc} \boldsymbol{P}_{\mathbf{0}}(\mathbf{T o r r}) & 110 & 210 & 390 & 760 \\ \boldsymbol{k}_{\boldsymbol{u n i}}\left(\mathbf{s}^{-1}\right) & 9.58 & 10.3 & 10.8 & 11.1 \end{array} \\] Assuming that the Lindemann mechanism accurately describes this reaction, determine \(k_{1}\) and the ratio \(k_{-1} / k_{2}\)

The Rice-Herzfeld mechanism for the thermal decomposition of acetaldehyde \(\left(\mathrm{CH}_{3} \mathrm{CO}(g)\right)\) is \\[ \begin{array}{l} \mathrm{CH}_{3} \mathrm{CHO}(g) \stackrel{k_{1}}{\longrightarrow} \mathrm{CH}_{3} \cdot(g)+\mathrm{CHO} \cdot(g) \\ \mathrm{CH}_{3} \cdot(g)+\mathrm{CH}_{3} \mathrm{CHO}(g) \stackrel{k_{2}}{\longrightarrow} \mathrm{CH}_{4}(g)+\mathrm{CH}_{2} \mathrm{CHO} \cdot(g) \\ \mathrm{CH}_{2} \mathrm{CHO} \cdot(g) \stackrel{k_{3}}{\longrightarrow} \mathrm{CO}(g)+\mathrm{CH}_{3} \cdot(g) \\ \mathrm{CH}_{3} \cdot(g)+\mathrm{CH}_{3} \cdot(g) \stackrel{k_{4}}{\longrightarrow} \mathrm{C}_{2} \mathrm{H}_{6}(g) \end{array} \\] Using the steady-state approximation, determine the rate of methane \(\left(\mathrm{CH}_{4}(g)\right)\) formation.

A central issue in the design of aircraft is improving the lift of aircraft wings. To assist in the design of more efficient wings, wind-tunnel tests are performed in which the pressures at various parts of the wing are measured generally using only a few localized pressure sensors. Recently, pressure- sensitive paints have been developed to provide a more detailed view of wing pressure. In these paints, a luminescent molecule is dispersed into an oxygen- permeable paint and the aircraft wing is painted. The wing is placed into an airfoil, and luminescence from the paint is measured. The variation in \(\mathrm{O}_{2}\) pressure is measured by monitoring the luminescence intensity, with lower intensity demonstrating areas of higher \(\mathrm{O}_{2}\) pressure due to quenching. a. The use of platinum octaethylporphyrin (PtOEP) as an oxygen sensor in pressure-sensitive paints was described by Gouterman and coworkers [Review of Scientific Instruments \(61(1990): 3340] .\) In this work, the following relationship between luminescence intensity and pressure was derived: \(I_{0} / I=A+B\left(P / P_{0}\right),\) where \(I_{0}\) is the fluorescence intensity at ambient pressure \(P_{0},\) and \(I\) is the fluorescence intensity at an arbitrary pressure \(P .\) Determine coefficients \(A\) and \(B\) in the preceding expression using the Stern-Volmer equation: \(k_{\text {total}}=1 / \tau_{l}=k_{l}+k_{q}[Q] .\) In this equation \(\tau_{l}\) is the luminescence lifetime, \(k_{l}\) is the luminescent rate constant, and \(k_{q}\) is the quenching rate constant. In addition, the luminescent intensity ratio is equal to the ratio of luminescence quantum yields at ambient pressure \(\Phi_{0}\) and an arbitrary pressure \(\Phi:\) \\[ \Phi_{0} / \Phi=I_{0} / I \\] b. Using the following calibration data of the intensity ratio versus pressure observed for PtOEP, determine \(A\) and \(B\) : $$\begin{array}{cccc} I_{0} / I & P / P_{0} & I_{0} / I & P / P_{0} \\ \hline 1.0 & 1.0 & 0.65 & 0.46 \\ 0.9 & 0.86 & 0.61 & 0.40 \\ 0.87 & 0.80 & 0.55 & 0.34 \\ 0.83 & 0.75 & 0.50 & 0.28 \\ 0.77 & 0.65 & 0.46 & 0.20 \\ 0.70 & 0.53 & 0.35 & 0.10 \end{array}$$ c. \(A t\) an ambient pressure of 1 atm, \(I_{0}=50,000\) (arbitrary units \()\) and 40,000 at the front and back of the wing. The wind tunnel is turned on to a speed of Mach \(0.36,\) and the measured luminescence intensity is 65,000 and 45,000 at the respective locations. What is the pressure differential between the front and back of the wing?

Oxygen sensing is important in biological studies of many systems. The variation in oxygen content of sapwood trees was measured by del Hierro and coworkers \([J . \text { Experimental Biology } 53(2002): 559]\) by monitoring the luminescence intensity of \(\left[\operatorname{Ru}(\operatorname{dpp})_{3}\right]^{2+}\) immobilized in a sol-gel that coats the end of an optical fiber implanted into the tree. As the oxygen content of the tree increases, the luminescence from the ruthenium complex is quenched. The quenching of \(\left[\mathrm{Ru}(\mathrm{dpp})_{3}\right]^{2+}\) by \(\mathrm{O}_{2}\) was measured by Bright and coworkers [Applied Spectroscopy \(52(1998): 750]\) and the following data were obtained: $$\begin{array}{rr} I_{0} / I & \% \mathrm{O}_{2} \\ \hline 3.6 & 12 \\ 4.8 & 20 \\ 7.8 & 47 \\ 12.2 & 100 \end{array}$$ a. Construct a Stern-Volmer plot using the data supplied in the table. For \(\left[\operatorname{Ru}(\operatorname{dpp})_{3}\right]^{2+} k_{r}=1.77 \times 10^{5} \mathrm{s}^{-1},\) what is \(k_{q} ?\) b. Comparison of the Stern-Volmer prediction to the quenching data led the authors to suggest that some of the \(\left[\operatorname{Ru}(\operatorname{dpp})_{3}\right]^{2+}\) molecules are located in sol-gel environments that are not equally accessible to \(\mathrm{O}_{2}\). What led the authors to this suggestion?
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