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Chapter 9: Problem 3

The standard heat of the combustion reaction of liquid \(n\) -hexane to form \(\mathrm{CO}_{2}(\mathrm{g})\) and \(\mathrm{H}_{2} \mathrm{O}(\mathrm{l}),\) with all reactants and products at \(77^{\circ} \mathrm{F}\) and 1 atm, is \(\Delta H_{\mathrm{r}}^{\prime}=-1.791 \times 10^{6} \mathrm{Btu} .\) The heat of vaporization of hexane at \(77^{\circ} \mathrm{F}\) is \(13,550 \mathrm{Btu} / \mathrm{b}\) -mole and that of water is \(18.934 \mathrm{Btu} / \mathrm{h}\) -mole. (a) Is the reaction exothermic or endothermic at \(77^{\circ} \mathrm{F}\) ? Would you have to heat or cool the reactor to keep the temperature constant? What would the temperature do if the reactor ran adiabatically? What can you infer about the energy required to break the molecular bonds of the reactants and that released when the product bonds form? (b) Use the given data to calculate \(\Delta H_{\mathrm{r}}^{\mathrm{r}}\) (Btu) for the combustion of \(n\) -hexane vapor to form \(\mathrm{CO}_{2}(\mathrm{g})\) and \(\overline{\mathrm{H}}_{2} \mathrm{O}(\mathrm{g})\) (c) If \(\dot{Q}=\Delta \dot{H},\) at what rate in \(\mathrm{B}_{\text {tu } / \mathrm{s}}\) is heat absorbed or released (state which) if \(120 \mathrm{lb}_{\mathrm{n}} / \mathrm{s}\) of \(\mathrm{O}_{2}\) is consumed in the combustion of hexane vapor, water vapor is the product, and the reactants and products are all at \(77^{\circ} \mathrm{F} ?\) (d) If the reaction were carried out in a real reactor, the actual value of \(\dot{Q}\) would be greater (less negative) than the value calculated in Part (c). Explain why.

Short Answer

Expert verified
(a) The reaction is exothermic at \(77^{\circ} \mathrm{F}\) where one would have to cool the reactor in order to keep the temperature constant. If the reactor ran adiabatically, temperature would rise. More energy is released in formation of product bonds than is required to break the molecular bonds of the reactants. (b) \(\Delta H_{\mathrm{r}}^{\mathrm{r}} = 鈥�1.760 \times 10^{6} \text{Btu}\). (c) Heat is released at a rate that is obtained from the formula \(\dot{Q} = \(\Delta \dot{H}\ = \Delta H_{\mathrm{r}}^{\mathrm{r}} \) \times \mathrm{Rate\ of\ O_{2}\ consumption\). (d) In a real reactor, the actual value of \(\dot{Q}\) would be greater (less negative) due to possible heat losses and incomplete combustion.

Step by step solution

01

Step 1(a): Identify The Type of The Reaction

The standard heat of combustion of liquid n-hexane is given as \(\Delta H_{\mathrm{r}}^{\prime}=-1.791 \times 10^{6} \mathrm{Btu}\). The negative sign indicates that the reaction is exothermic i.e., heat is released during the process.
02

Step 2(a): Temperature Effect on Reactor

To keep the temperature constant in an exothermic reaction, heat needs to be removed. If the reactor was adiabatically insulated (no heat transfer), the temperature inside the reactor would increase because of the heat being produced in this exothermic combustion reaction.
03

Step 3(a): Energy Requirement for Bond Breaking

In an exothermic reaction, the energy released when product bonds are formed is more than the energy required to break the bonds of the reactants.
04

Step 1(b): Calculate The Enthalpy Change

The overall enthalpy change \(\Delta H_{\mathrm{r}}^{\mathrm{r}}\) includes the heat of combustion plus the heat of vaporization of hexane, and the heat of vaporization of water, i.e., \(\Delta H_{\mathrm{r}}^{\mathrm{r}} = \Delta H_{\mathrm{r}}^{\prime} + \Delta_{\mathrm{vap}} H_{\mathrm{Hexane}} + \Delta_{\mathrm{vap}} H_{\mathrm{Water}} = 鈥�1.791 \times 10^6 \text{Btu} + 13,550 \text{Btu} - 18,934 \text{Btu}\)
05

Step 1(c): Rate of Heat Absorption or Release

Since it is an exothermic reaction, heat is released. Also, the heat released or absorbed per second \(\dot{Q}\) can be calculated using the rate of consumption of O2, and the overall enthalpy change of the reaction. Hence, \(\dot{Q} = \(\Delta \dot{H}\) = \(\Delta H_{\mathrm{r}}^{\mathrm{r}} \) \times \mathrm{Rate\ of\ O_{2}\ consumption\).
06

Step 1(d): Actual Value of Heat Transfer

In reality, the actual heat transfer \(\dot{Q}\) might be less negative than calculated because not all heat might be effectively transferred due to losses in real conditions. Also, it might be due to incomplete combustion.

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Key Concepts

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

Exothermic and Endothermic Reactions
Chemical reactions can either release or absorb energy, known as exothermic and endothermic reactions, respectively. When a reaction is exothermic, like the combustion of n-hexane, it releases heat. This is evident from the negative sign of the enthalpy change e.g., \(\Delta H_{\mathrm{r}}^{\prime}=-1.791 \times 10^{6} \ \mathrm{Btu}\). In such reactions, the energy produced from forming the product bonds exceeds the energy required to break the bonds of the reactants.
If the reaction proceeds without any heat exchange with the surroundings (adiabatically), the temperature of the system will increase. To maintain a constant temperature, the heat must be actively removed from the system. This concept highlights the intrinsic nature of exothermic reactions and the balance between bond-breaking and bond-making energies.
Heat of Combustion
The heat of combustion is the energy released when a substance burns completely in oxygen. It is a critical measure to understand energy changes in combustion reactions. Typically measured for standard conditions, it tells us how much energy is obtained from a fuel. For n-hexane, this value \(-1.791 \times 10^6\ \mathrm{Btu}\) indicates a significant release of energy. To analyze a reaction properly, it is vital to consider additional factors like the state of all reactants and products.
  • The heat of vaporization is crucial for calculations involving phase changes, such as liquid to vapor. For example, when liquid hexane vaporizes, it requires energy (\(13,550 \ \mathrm{Btu/b} \text{-mole} \) for n-hexane), which must be factored in to accurately determine the total heat change during its combustion to vapor products.
Enthalpy Change Calculations
Enthalpy change calculations quantify the heat exchange in chemical reactions. Enthalpy, denoted as \(\Delta H\), encompasses different factors like bond energies and phase changes. For a combustion reaction, the overall enthalpy change can include contributions from both the combustion heat and any phase change enthalpies. The formula to find the overall enthalpy change is:\[\Delta H_{\mathrm{r}}^{\mathrm{r}} = \Delta H_{\mathrm{r}}^{\prime} + \Delta_{\mathrm{vap}} H_{\mathrm{Hexane}} + \Delta_{\mathrm{vap}} H_{\mathrm{Water}}\]This equation helps us account for energy differences due to vaporization. By adding the heat of vaporization for both hexane and water, you can determine the total energy change when the reactants and products are in their gaseous states. This calculation is essential, especially in industrial applications where precise energy accounting affects processes and efficiency.

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Most popular questions from this chapter

A 12.0-molar solution of sodium hydroxide ( \(\mathrm{SG}=1.37\) ) is neutralized with \(75.0 \mathrm{mL}\) of a \(4.0 \mathrm{molar}\) solution of sulfuric acid ( \(\mathrm{SG}=1.23\) ) in a well-insulated container. (a) Estimate the volume of the sodium hydroxide solution and the final solution temperature if both feed solutions are at \(25^{\circ} \mathrm{C}\). The heat capacity of the product solution may be taken to be that of pure liquid water, the standard heat of solution of sodium sulfate is \(-1.17 \mathrm{kJ} / \mathrm{mol},\) and the energy balance reduces to \(Q=\Delta H\) for this constant-pressure batch process. (b) List several additional assumptions you made to arrive at your estimated volume and temperature.

You are checking the performance of a reactor in which acetylene is produced from methane in the reaction $$2 \mathrm{CH}_{4}(\mathrm{g}) \rightarrow \mathrm{C}_{2} \mathrm{H}_{2}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g})$$ An undesired side reaction is the decomposition of acetylene: $$\mathrm{C}_{2} \mathrm{H}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{C}(\mathrm{s})+\mathrm{H}_{2}(\mathrm{g})$$ Methane is fed to the reactor at \(1500^{\circ} \mathrm{C}\) at a rate of \(10.0 \mathrm{mol} \mathrm{CH}_{4} / \mathrm{s}\). Heat is transferred to the reactor at a rate of \(975 \mathrm{kW}\). The product temperature is \(1500^{\circ} \mathrm{C}\) and the fractional conversion of methane is 0.600 . A flowchart of the process and an enthalpy table are shown below. (a) Using the heat capacitics given below for enthalpy calculations, write and solve material balances and an energy balance to determine the product component flow rates and the yield of acctylene (mol \(\mathbf{C}_{2} \mathbf{H}_{2}\) produced/mol \(\mathbf{C H}_{4}\) consumed). $$\begin{aligned}\mathrm{CH}_{4}(\mathrm{g}): & C_{p} \approx 0.079 \mathrm{kJ} /\left(\mathrm{mol} \cdot^{\circ} \mathrm{C}\right) \\ \mathrm{C}_{2} \mathrm{H}_{2}(\mathrm{g}): & C_{p} \approx 0.052 \mathrm{kJ} /\left(\mathrm{mol} \cdot^{\circ} \mathrm{C}\right) \\ \mathrm{H}_{2}(\mathrm{g}): & C_{p} \approx 0.031 \mathrm{kJ} /\left(\mathrm{mol} \cdot^{\circ} \mathrm{C}\right) \\ \mathrm{C}(\mathrm{s}): & C_{p} \approx 0.022 \mathrm{kJ} /\left(\mathrm{mol} \cdot^{\circ} \mathrm{C}\right)\end{aligned}$$ For example, the specific enthalpy of methane at \(1500^{\circ} \mathrm{C}\) relative to methane at \(25^{\circ} \mathrm{C}\) is \(\left[0.079 \mathrm{kJ} /\left(\mathrm{mol} \cdot^{\circ} \mathrm{C}\right)\right]\left(1500^{\circ} \mathrm{C}-25^{\circ} \mathrm{C}\right)=116.5 \mathrm{kJ} / \mathrm{mol}\) (b) The reactor efficiency may be defined as the ratio (actual acetylene yield/acetylene yield with no side reaction). What is the reactor efficiency for this process? (c) The mean residence time in the reactor \([\tau(\mathrm{s})]\) is the average time gas molecules spend in the reactor in going from inlet to outlet. The more \(\tau\) increases, the greater the extent of reaction for every reaction occurring in the process. For a given feed rate, \(\tau\) is proportional to the reactor volume and inversely proportional to the feed stream flow rate. (i) If the mean residence time increases to infinity, what would you expect to find in the product stream? Explain. (ii) Someone proposes running the process with a much greater feed rate than the one used in Part (a), separating the products from the unconsumed reactants, and recycling the reactants. Why would you expect that process design to increase the reactor efficiency? What else would you need to know to determine whether the new design would be cost-effective?

The equilibrium constant for the ethane dehydrogenation reaction, $$\mathrm{C}_{2} \mathrm{H}_{6}(\mathrm{g}) \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g})$$ is defined as $$K_{p}(\mathrm{atm})=\frac{y_{\mathrm{C}_{2} \mathrm{H}_{4}} y_{\mathrm{H}_{2}}}{y_{\mathrm{C}_{2} \mathrm{H}_{6}}} P$$ where \(P(\text { atm })\) is the total pressure and \(y_{i}\) is the mole fraction of the ith substance in an equilibrium mixture. The equilibrium constant has been found experimentally to vary with temperature according to the formula $$K_{p}(T)=7.28 \times 10^{6} \exp [-17,000 / T(\mathrm{K})]$$ The heat of reaction at \(1273 \mathrm{K}\) is \(+145.6 \mathrm{kJ}\), and the heat capacities of the reactive species may be approximated by the formulas $$\left.\begin{array}{rl}\left(C_{p}\right)_{\mathrm{C}_{2} \mathrm{H}_{4}} & =9.419+0.1147 T(\mathrm{K}) \\\\\left(C_{p}\right)_{\mathrm{H}_{2}} & =26.90+4.167 \times 10^{-3} T(\mathrm{K}) \\ \left(C_{p}\right)_{\mathrm{C}_{2} \mathrm{H}_{6}} & =11.35+0.1392 T(\mathrm{K}) \end{array}\right\\}[\mathrm{J} /(\mathrm{mol} \cdot \mathrm{K})]$$ Suppose pure cthane is fed to a continuous constant-pressure adiabatic reactor at \(1273 \mathrm{K}\) and pressure \(P(\text { atm }),\) the products emerge at \(T_{\mathrm{f}}(\mathrm{K})\) and \(P(\mathrm{atm}),\) and the residence time of the reaction mixture in the reactor is large enough for the outlet stream to be considered an equilibrium mixture of ethane, ethylene, and hydrogen. (a) Prove that the fractional conversion of ethane in the reactor is $$f=\left(\frac{K_{p}}{P+K_{p}}\right)^{1 / 2}$$ (b) Write an energy balance on the reactor, and use it to prove that $$f=\frac{1}{1+\phi\left(T_{\mathrm{f}}\right)}$$ where Finally, substitute for \(\Delta H_{\mathrm{r}}\) and the heat capacities in Equation 4 to derive an explicit expression for \(\phi\left(T_{\mathrm{f}}\right)\) (c) We now have two expressions for the fractional conversion \(f\) : Equation 2 and Equation 3 . If these expressions are equated, \(K_{p}\) is replaced by the expression of Equation \(1,\) and \(\phi\left(T_{\mathrm{f}}\right)\) is replaced by the expression derived in Part (b), the result is one equation in one unknown, \(T_{\mathrm{f}}\). Derive this equation, and transpose the right side to obtain an expression of the form $$\psi\left(T_{\mathrm{f}}\right)=0$$ (d) Prepare a spreadsheet to take \(P\) as input, solve Equation 5 for \(T_{\mathrm{f}}\) (use Goal Seek or Solver), and determine the final fractional conversion, \(\left.f \text { . (Suggestion: Set up columns for } P, T_{\mathrm{f}}, f, K_{p}, \phi, \text { and } \psi .\right)\) Run the program for \(P(\text { atm })=0.01,0.05,0.10,0.50,1.0,5.0,\) and \(10.0 .\) Plot \(T_{\mathrm{f}}\) versus \(P\) and \(f\) versus \(P,\) using a logarithmic coordinate scale for \(P\).

Sulfur dioxide is oxidized to sulfur trioxide in a small pilot-plant reactor. SO \(_{2}\) and \(100 \%\) excess air are fed to the reactor at \(450^{\circ} \mathrm{C}\). The reaction proceeds to a \(65 \% \mathrm{SO}_{2}\) conversion, and the products emerge from the reactor at \(550^{\circ} \mathrm{C}\). The production rate of \(\mathrm{SO}_{3}\) is \(1.00 \times 10^{2} \mathrm{kg} / \mathrm{min}\). The reactor is surrounded by a water jacket into which water at \(25^{\circ} \mathrm{C}\) is fed. (a) Calculate the feed rates (standard cubic meters per second) of the \(\mathrm{SO}_{2}\) and air feed streams and the extent of reaction, \(\xi\) (b) Calculate the standard heat of the SO_ oxidation reaction, \(\Delta H_{\mathrm{t}}^{\mathrm{r}}(\mathrm{kJ}) .\) Then, taking molecular species at \(25^{\circ} \mathrm{C}\) as references, prepare and fill in an inlet-outlet enthalpy table and write an energy balance to calculate the necessary rate of heat transfer ( \(\mathrm{kW}\) ) from the reactor to the cooling water. (c) Calculate the minimum flow rate of the cooling water if its temperature rise is to be kept below \(15^{\circ} \mathrm{C}\) (d) Briefly state what would have been different in your calculations and results if you had taken elemental species as references in Part (b).

Cumene \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{C}_{3} \mathrm{H}_{7}\right)\) is produced by reacting benzene with propylene \(\left[\Delta H_{\mathrm{r}}\left(77^{\circ} \mathrm{F}\right)=-39,520 \mathrm{Btu}\right]\) A liquid feed containing 75 mole \(\%\) propylene and \(25 \%\) n-butane and a second liquid stream containing essentially pure benzene are fed to the reactor. Fresh benzene and recycled benzene, both at \(77^{\circ} \mathrm{F},\) are mixed in a 1: 3 ratio \((1 \text { mole fresh feed } / 3\) moles recycle) and passed through a heat exchanger, where they are heated by the reactor effluent before being fed to the reactor. The reactor effluent enters the exchanger at \(400^{\circ} \mathrm{F}\) and leaves at \(200^{\circ} \mathrm{F}\). The pressure in the reactor is sufficient to maintain the effluent stream as a liquid. After being cooled in the heat exchanger, the reactor effluent is fed to a distillation column (T1). All of the butane and unreacted propylene are removed as overhead product from the column, and the cumene and unreacted benzene are removed as bottoms product and fed to a second distillation column (T2) where they are scparated. The benzenc leaving the top of the sccond column is the recycle that is mixed with the fresh benzene feed. Of the propylene fed to the process, \(20 \%\) does not react and leaves in the overhead product from the first distillation column. The production rate of cumene is \(1200 \mathrm{lb}_{\mathrm{m}} / \mathrm{h}\). (a) Calculate the mass flow rates of the streams fed to the reactor, the molar flow rate and composition of the reactor effluent, and the molar flow rate and composition of the overhead product from the first distillation column, T1. (b) Calculate the temperature of the benzene stream fed to the reactor and the required rate of heat addition to or removal from the reactor. Use the following approximate heat capacities in your calculations: \(C_{p}\left[\operatorname{Btu} /\left(\operatorname{lb}_{m} \cdot F\right)\right]=0.57\) for propylene, 0.55 for butane, 0.45 for benzene, and 0.40 for cumene. (c) Most people unfamiliar with the chemical process industry imagine that chemical engineers are people who deal mainly with chemical reactions carried out on a large scale. In fact, in most industrial processes, a visitor to the plant would have trouble finding the reactor in a maze of towers and tanks and pipes that were added to the process design to improve the profitability of the process. Briefly explain how the heat exchanger, the two distillation columns, and the recycle stream in the cumene process serve that function.
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