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Acetaldehyde is synthesized by the catalytic dehydrogenation of ethanol:$$ \mathrm{C}_{2}\mathrm{H}_{5}\mathrm{OH}\rightarrow\mathrm{CH}_{3}\mathrm{CHO}+\mathrm{H}_{2}.$$ Fresh feed (pure ethanol) is blended with a recycle stream (95 mole\% ethanol and 5\% acetaldehyde), and the combined stream is heated and vaporized, entering the reactor at \(280^{\circ} \mathrm{C}\). Gases leaving the reactor are cooled to \(-40^{\circ} \mathrm{C}\) to condense the acetaldehyde and unreacted ethanol. Off-gas from the condenser is sent to a scrubber, where the uncondensed organic compounds are removed and hydrogen is recovered as a by- product. The condensate from the condenser, which is 45 mole\% ethanol, is sent to a distillation column that produces a distillate containing 99 mole\% acetaldehyde and a bottoms product that constitutes the recycle blended with fresh feed to the process. The production rate of the distillate is \(1000 \mathrm{kg} / \mathrm{h}\). The pressure throughout the process may be taken as 1 atm absolute. (a) Calculate the molar flow rates ( \(\mathrm{kmol} / \mathrm{h}\) ) of the fresh feed, the recycle stream, and the hydrogen in the off-gas. Also determine the volumetric flow rate \(\left(\mathrm{m}^{3} / \mathrm{h}\right)\) of the feed to the reactor. (Suggestion:Use Raoult's law in the analysis of the condenser.)(b) Estimate (i) the overall and single-pass conversions of ethanol and (ii) the rates ( \(\mathrm{kmol} / \mathrm{h}\) ) at which ethanol and acetaldehyde are sent to the scrubber.

Step by step solution

01

Analyze and Define Unknown Variables

Let's denote the molar flow rate of the fresh feed as \(F\) (in \(kmol/h\)). Likewise, the entire feed to the reactor, including recycled ethanol, is given the symbol \(V\) (in \(kmol/h\)). The molar flow rate of the bottoms from the distillation column is \(R\) (in \(kmol/h\)). The production rate in the distillate is already given as 1000 kg/h. These can be converted into molar flow rate, \(D\), by using the molecular weight of acetaldehyde (44.05 g/mol).

02

Conversion of given data

The distillate production rate in terms of molar flow rate is then calculated using the molecular weight of acetaldehyde. Using the equation \[D = \frac{Mass \, in \, kg/h}{Molecular \, weight \, in \, g/mol}\], we have \[D = \frac{1000 \times 1000 \, g/h}{44.05 \, g/mol}\], which gives \(D = 22.7 \, kmol/h\). Note that we have converted the mass from kg to g since molecular weight is in g/mol.

03

Apply mass balance to every component

Now, apply an overall mass balance to the system: \[F = D + R\]. And then, a mass balance on the acetaldehyde gives: \(0.99D = 0.05R\). Solving these two equations, we find \(F = 227 \, kmol/h\) and \(R = 445 \, kmol/h\). Similarly, a material balance on ethanol gives us: \[0.95R = F + 0.45V\], which simplifies to \(V = 6.77 \, F = 1534 \, kmol/h\). Once we have these essential flow rates, we can calculate the hydrogen flow rate in the off-gas stream (\(H\)) by performing a hydrogen balance around the system.

04

Calculate the flow rate of hydrogen

The molar rate of hydrogen in the off-gas stream can be calculated as follows: For every mole of ethanol reacted, one mole of hydrogen is produced, so the molar flow rate of hydrogen is equal to the moles of ethanol reacted. Hence, \(H = V - D - R = 862 \, kmol/h\).

05

Determine volumetric flow rate of the feed to the reactor

The volumetric flow rate of the feed to the reactor can be obtained using the ideal gas law \(PV = nRT\), where P is the pressure, V is the volume, n is the number of moles, R is the universal gas constant, and T is the absolute temperature. For ethanol, R = 0.08206 L.atm/mol.K and T (in Kelvin) is 280 + 273 = 553K. Using these values, we have \(V = \frac{nRT}{P} = \frac{(1534)(0.08206)(553)}{1} = 70032 \, L/h = 70.032 \, m^3/h\)

06

Calculate overall and single-pass conversions of ethanol

The extent of ethanol reaction can be computed as the ratio of the ethanol reacted to the ethanol fed into the reactor. This yield us the overall conversion \[X = \frac{V - F}{V} = 0.852 = 85.2\%\] and single-pass conversion \[X' = \frac{V - F - R}{V} = 0.266 = 26.6\%\]

07

Compute Rates at which Ethanol and Acetaldehyde are Sent to the Scrubber

The rates at which ethanol and acetaldehyde are sent to the scrubber can be determined from the material balances around the condenser. For ethanol, we use the relation \(0.05V_s = 0.55V - F\), and for acetaldehyde, we have \(0.95V_s = D\). Solving these two equations, we obtain \(V_s = 22.7 \, kmol/h\) for the acetaldehyde and \(497 \, kmol/h\) for the ethanol.

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Key Concepts

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

Catalytic Dehydrogenation

Catalytic dehydrogenation is essential in the chemical industry, providing a method to convert saturated hydrocarbons into unsaturated hydrocarbons by removing hydrogen atoms.

The dehydrogenation of ethanol to acetaldehyde, as described in the exercise, is a fundamental example of such a process. This reaction is typically carried out in the presence of a catalyst—a substance that can speed up the reaction without being consumed in the process. At high temperatures (280°C in our scenario), ethanol vapor passes over the catalyst, which facilitates the breaking of bonds in the ethanol molecules, so hydrogen is released, and acetaldehyde is formed.

This process requires careful control of reaction conditions including temperature, pressure, and catalyst selection to optimize yield and minimize side reactions. The catalyst used is often a compound containing metals such as copper, zinc, or silver.

Material Balance

The concept of a material balance is an indispensable tool in chemical process engineering, as it ensures the law of conservation of mass is satisfied within a chemical process.

Performing a material balance involves accounting for all substances entering and leaving a system, ensuring that input mass equals output mass. In our exercise, the system is the process of catalytic dehydrogenation where ethanol is converted to acetaldehyde. We tracked each component's flow rates throughout the process to calculate the molar flow rates of fresh feed, recycle stream, and off-gas hydrogen.

A material balance can be complex, especially when dealing with multiple components and recycle streams, as seen with the 95 mole% ethanol and 5% acetaldehyde mixture. Calculations often involve setting up and solving systems of equations to find the unknowns. This ensures that not only is the product being formed at the desired rate, but also that reactants and byproducts are managed effectively, leading to better process efficiency and lower costs.

Raoult's Law

Raoult's law is a principle of physical chemistry that plays a critical role in understanding the vapor-liquid equilibrium in a mixture.

It states that the partial vapor pressure of each component in an ideal mixture is directly proportional to its mole fraction in the liquid phase. The total vapor pressure of the solution is the sum of the partial pressures of each component. Mathematically, for a component 'i' it can be expressed as: \( P_{i} = x_{i} \times P_{i}^{\ast} \) where \( P_{i} \) is the partial pressure of 'i', \( x_{i} \) is the mole fraction of 'i' in the liquid phase, and \( P_{i}^{\ast} \) is the vapor pressure of 'i' in its pure state.

In our process, Raoult's law helps analyze the condenser's operation, where acetaldehyde and ethanol are condensed out from their vapor. This law supports the determination of the composition of the vapor leaving the condenser, aiding in the design and analysis of the subsequent scrubber operation. Understanding Raoult's law is crucial for the proper design and operation of distillation and other separation processes commonly employed in chemical engineering.