91Ó°ÊÓ

Methanol is formed from carbon monoxide and hydrogen in the gas-phase reaction The mole fractions of the reactive species at equilibrium satisfy the relation where \(P\) is the total pressure (atm), \(K_{c}\) the reaction equilibrium constant (atm \(^{-2}\) ), and \(T\) the temperature (K). The equilibrium constant \(K_{c}\) equals 10.5 at 373 K, and \(2.316 \times 10^{-4}\) at \(573 \mathrm{K}\). A semilog plot of \(K_{\mathrm{c}}\) (logarithmic scale) versus 1/ \(T\) (rectangular scale) is approximately linear between \(T=300 \mathrm{K}\) and \(T=600 \mathrm{K}\) (a) Derive a formula for \(K_{\mathrm{c}}(T),\) and use it to show that \(K_{\mathrm{e}}(450 \mathrm{K})=0.0548 \mathrm{atm}^{-2}\) (b) Write expressions for \(n_{A}, n_{B},\) and \(n_{C}\) (gram-moles of each species), and then \(y_{A}, y_{B},\) and \(y_{C},\) in terms of \(n_{\mathrm{A} 0}, n_{\mathrm{B} 0}, n_{\mathrm{C} 0},\) and \(\xi,\) the extent of reaction. Then derive an equation involving only \(n_{\mathrm{A} 0}, n_{\mathrm{B} 0}, n_{\mathrm{C} 0}, P, T,\) and \(\xi_{e},\) where \(\xi_{e}\) is the extent of reaction at equilibrium. (c) Suppose you begin with equimolar quantities of CO and \(\mathrm{H}_{2}\) and no \(\mathrm{CH}_{3} \mathrm{OH}\), and the reaction proceeds to equilibrium at 423 K and 2.00 atm. Calculate the molar composition of the product ( \(y_{\mathrm{A}}\), \(\left.y_{\mathrm{B}}, \text { and } y_{\mathrm{C}}\right)\) and the fractional conversion of \(\mathrm{CO}\) (d) The conversion of CO and \(\mathrm{H}_{2}\) can be enhanced by removing methanol from the reactor while leaving unreacted CO and \(\mathrm{H}_{2}\) in the vessel. Review the equations you derived in solving Part (c) and determine any physical constraints on \(\xi_{c}\) associated with \(n_{\mathrm{A} 0}=n_{\mathrm{B} 0}=1\) mol. Now suppose that 90\% of the methanol is removed from the reactor as it is produced; in other words, only 10\% of the methanol formed remains in the reactor. Estimate the fractional conversion of CO and the total gram moles of methanol produced in the modified operation. (e) Repeat Part (d), but now assume that \(n_{\mathrm{B} 0}=2\) mol. Explain the significant increase in fractional conversion of CO. (f) Write a set of equations for \(y_{\mathrm{A}}, y_{\mathrm{B}}, y_{\mathrm{C}},\) and \(f_{\mathrm{A}}\) (the fractional conversion of \(\mathrm{CO}\) ) in terms of \(y_{\mathrm{A} 0}, y_{\mathrm{B} 0}, T,\) and \(P(\) the reactor temperature and pressure at equilibrium). Enter the equations in an equation-solving program. Check the program by running it for the conditions of Part (c), then use it to determine the effects on \(f_{\mathrm{A}}\) (increase, decrease, or no effect) of separately increasing, (i) the fraction of \(\mathrm{CH}_{3} \mathrm{OH}\) in the feed, (ii) temperature, and (iii) pressure.

Step by step solution

01

Step 1. Derive \(K_{c}(T)\)

The equation for \(K_{c}(T)\) can be derived from the given fact that the semi-log plot of \(K_{c}\) against \(1/T\) is approximately linear. This implies that \(K_{c}\) versus \(1/T\) corresponds to the equation of a line. That is, \(lnK_c = -Ea/R(1/T) + lnA\), where \(Ea\) is the activation energy, \(R\) is the universal gas constant, and \(A\) is the pre-exponential factor. The values of \(lnA\) and \(-Ea/R\) can be found by considering two points on the line: (1/373, ln10.5) and (1/573, ln2.316x10^-4). Solving these, we finally get the equation to show that \(K_{e}(450K) = 0.0548 atm^{-2}\).

02

Step 2. Express Moles and Mole Fractions

We start with the balanced equation: \(A + B -> C\). Here, A stands for CO, B for H2 and C for CH3OH. At the beginning, there are \(n_{A0}\) moles of A, \(n_{B0}\) moles of B, and \(n_{C0}\) moles of C. When the volume of the container is not changed and \(ξ\) moles of A has reacted, \(n_{A} = n_{A0} - ξ, n_{B} = n_{B0} - ξ, n_{C} = n_{C0} + ξ\). Then the mole fractions become: \(y_{A} = n_{A}/(n_{A} + n_{B} + n_{C}), y_{B} = n_{B}/(n_{A} + n_{B} + n_{C}), y_{C} = n_{C}/(n_{A} + n_{B} + n_{C})\). We plug the expressions of \(n_{A}, n_{B}, n_{C}\) into the mole fractions and simplify. Then we substitute these into the given equation with mole fractions, retrieve the equation with \(ξ_e\) and simplify the result.

03

Step 3. Molar Composition and Fractional Conversion

Assuming dead state conditions for temperature and pressure, apply the equation derived in the previous step to calculate molar composition and the fractional conversion of CO.

04

Step 4. Constraints Review

Examine the revised equation from Step 2 and find any physical limits placed by the fact that the quantity of moles is always positive. Calculate the new fractional conversion of CO and the total moles of methanol produced when 90% of methanol is immediately removed from the reactor.

05

Step 5. Repeat for a Different Initial Molar Value

In this case, by changing \(n_{B0}\) to 2mol, calculate the resulting fractional conversion of CO. The sizable growth in fractional conversion is attributable to the bimolecular nature of the reaction.

06

Step 6. Expressing the Variables in a Set of Equations

The resulting equations in variable form will be used to draft a program and predict how fractional conversion is affected when certain variables are increased. The program is initially tested under the conditions of part (c).

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Key Concepts

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

Reaction Kinetics

Reaction kinetics is a branch of chemistry that deals with the rates of chemical reactions. It primarily focuses on how quickly a reaction proceeds and what factors affect this rate. For example, the activation energy, represented by \( E_a \), is a critical element in reaction kinetics. It is the energy barrier that reactants must overcome to transform into products.
Temperature also plays a significant role in reaction kinetics. Generally, as temperature increases, the reaction rate tends to increase as well. This occurs because the molecules have more kinetic energy, making it more likely for them to overcome the activation energy.
In our methanol synthesis equation, the rate of reaction is directly influenced by the concentrations of carbon monoxide (CO) and hydrogen (\( H_2 \)) as they convert into methanol (\( CH_3OH \)). The reaction rate reflects how swiftly products form from reactants in these gas-phase reactions. This rate is represented mathematically in the form of rate laws, which can be derived from the overall balanced chemical equation. By understanding reaction kinetics, we can predict how quickly equilibrium will be reached in a chemical reaction.
Moreover, in gas-phase reactions like this one, parameters such as pressure can influence both the rate and direction of the reaction. In particular, reaction kinetics can help us analyze how, at higher pressures, reactions involving gaseous components may experience increased collision frequency, potentially quickening reaction rates.

Le Chatelier's Principle

Le Chatelier's Principle is a fundamental concept in chemistry that predicts how a change in conditions affects chemical equilibrium. According to this principle, when a system at equilibrium is disturbed by a change in pressure, temperature, or concentration, the system will adjust itself to counteract the disturbance and return to a new equilibrium.
This principle is highly applicable in understanding the methanol synthesis reaction. For example:

• Pressure: An increase in pressure, due to high molar concentrations of reactants such as CO and \( H_2 \), will shift the equilibrium position towards the formation of a smaller volume of gas—in this case, methanol.
• Temperature: Because the reaction is exothermic (releases heat), increasing the temperature shifts the equilibrium to favor the reactants. This illustrates the sensitivity of the equilibrium to thermal changes.

By manipulating conditions such as temperature and pressure, chemists can optimize processes to favor the production of desired products. In the case of methanol production, controlling these factors can significantly increase the efficiency and yield of the reaction.

Gas-phase Reactions

Gas-phase reactions are chemical transformations that occur within gases at varied conditions of temperature and pressure. Understanding these reactions is crucial because gases act differently compared to liquids or solids, especially under changing conditions.
For the synthesis of methanol, factors such as pressure and temperature are critical. Higher pressures help to push the reaction towards product formation due to the reduced volume of gases when methanol is formed from CO and \( H_2 \). However, there's always a balance to be found, as gases can expand or compress, affecting reaction dynamics.

• In a closed system, the total number of moles prior to the reaction will adjust itself depending on changes in pressure and volume.
• The mole fractions of each gas component become imperative in predicting the reaction's favorability and outcome.

When examining gas-phase reactions, it’s important to consider both kinetic and equilibrium perspectives because they provide insights into how fast reactions occur and what the final composition of the system will be at equilibrium. Mathematical models and real-world applications of gas laws, like the Ideal Gas Law, combine with principles like Le Chatelier’s to predict and optimize these reactions amidst varying conditions.