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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 dilute aqueous solution of sulfuric acid at \(25^{\circ} \mathrm{C}\) is used to absorb ammonia in a continuous reactor, thereby producing ammonium sulfate, a fertilizer: $$2 \mathrm{NH}_{3}(\mathrm{g})+\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq}) \rightarrow\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}(\mathrm{aq})$$ (a) If the ammonia enters the absorber at \(75^{\circ} \mathrm{C}\), the sulfuric acid enters at \(25^{\circ} \mathrm{C}\), and the product solution emerges at \(25^{\circ} \mathrm{C}\), how much heat must be withdrawn from the unit per mol of \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\) produced? (All needed physical property data may be found in Appendix B.) (b) Estimate the final temperature if the reactor of Part (a) is adiabatic and the product of the solution contains 1.00 mole \(\%\) ammonium sulfate. Take the heat capacity of the solution to be that of pure liquid water [4.184 kJ/(kg.'C)]. (c) In a real (imperfectly insulated) reactor, would the final solution temperature be less than, equal to, or greater than the value calculated in Part (b), or is there no way to tell without more information? Briefly explain your answer.

Calcium chloride is a salt used in a number of food and medicinal applications and in brine for refrigeration systems. Its most distinctive property is its affinity for water. in its anhydrous form it efficiently absorbs water vapor from gases, and from aqueous liquid solutions it can form (at different conditions) calcium chloride hydrate \(\left(\mathrm{CaCl}_{2} \cdot \mathrm{H}_{2} \mathrm{O}\right)\) dihydrate \(\left(\mathrm{CaCl}_{2} \cdot 2 \mathrm{H}_{2} \mathrm{O}\right)\) tetrahydrate \(\left(\mathrm{CaCl}_{2} \cdot 4 \mathrm{H}_{2} \mathrm{O}\right),\) and hexahydrate \(\left(\mathrm{CaCl}_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O}\right)\) You have been given the task of determining the standard heat of the reaction in which calcium chloride hexahydrate is formed from anhydrous calcium chloride: $$\mathrm{CaCl}_{2}(\mathrm{s})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{CaCl}_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O}(\mathrm{s}): \quad \Delta H_{\mathrm{r}}^{\circ}(\mathrm{k} \mathrm{J})=?$$ By definition, the desired quantity is the heat of hydration of calcium chloride hexahydrate. You cannot carry out the hydration reaction directly, so you resort to an indirect method. You first dissolve 1.00 mol of anhydrous \(\mathrm{CaCl}_{2}\) in \(10.0 \mathrm{mol}\) of water in a calorimeter and determine that \(64.85 \mathrm{kJ}\) of heat must be transferred away from the calorimeter to keep the solution temperature at \(25^{\circ} \mathrm{C}\). You next dissolve 1.00 mol of the hexahydrate salt in 4.00 mol of water and find that 32.1 kJ of heat must be transferred to the calorimeter to keep the temperature at \(25^{\circ} \mathrm{C}\). (a) Use these results to calculate the desired heat of reaction. (Suggestion: Begin by writing out the stoichiometric equations for the two dissolution processes.) (b) Calculate the standard heat of reaction in \(\mathrm{kJ}\) for \(\mathrm{Ca}(\mathrm{s}), \mathrm{Cl}_{2}(\mathrm{g})\) and \(\mathrm{H}_{2} \mathrm{O}\) reacting to form \(\mathrm{CaCl}_{2}\) (aq, \(r=10\) ). (c) Speculate about why the standard heat of reaction in forming calcium chloride hexahydrate cannot be measured directly by reacting the anhydrous salt with water in a calorimeter.

A 2.00 mole \(\%\) sulfuric acid solution is neutralized with a 5.00 mole\% sodium hydroxide solution in a continuous reactor. All reactants enter at \(25^{\circ} \mathrm{C}\). The standard heat of solution of sodium sulfate is \(-1.17 \mathrm{kJ} / \mathrm{mol} \mathrm{Na}_{2} \mathrm{SO}_{4},\) and the heat capacities of all solutions may be taken to be that of pure liquid water [4.184 kJ/(kg.'C)]. (a) How much heat (kJ/kg acid solution fed) must be transferred to or from the reactor contents (state which it is) if the product solution emerges at \(40^{\circ} \mathrm{C} ?\) (b) Estimate the product solution temperature if the reactor is adiabatic, neglecting heat transferred between the reactor contents and the reactor wall.

Methane is burned completely with 40\% excess air. The methane enters the combustion chamber at \(25^{\circ} \mathrm{C},\) the combustion air enters at \(150^{\circ} \mathrm{C},\) and the stack gas \(\left[\mathrm{CO}_{2}, \mathrm{H}_{2} \mathrm{O}(\mathrm{v}), \mathrm{O}_{2}, \mathrm{N}_{2}\right]\) exits at \(450^{\circ} \mathrm{C} .\) The chamber functions as a preheater for an air stream flowing in a pipe through the chamber to a spray dryer. The air enters the chamber at \(25^{\circ} \mathrm{C}\) at a rate of \(1.57 \times 10^{4} \mathrm{m}^{3}(\mathrm{STP}) / \mathrm{h}\) and is heated to \(181^{\circ} \mathrm{C}\). All of the heat generated by combustion is used to heat the combustion products and the air going to the spray dryer (i.e., the combustion chamber may be considered adiabatic). (a) Draw and completely label the process flow diagram and perform a degree- of-freedom analysis. (b) Calculate the required molar flow rates of methane and combustion air (kmol/h) and the volumetric flow rates \(\left(\mathrm{m}^{3} / \mathrm{h}\right)\) of the two effluent streams. State all assumptions you make. (c) When the system goes on line for the first time, environmental monitoring of the stack gas reveals a considerable quantity of CO, suggesting a problem with either the design or the operation of the combustion chamber. What changes from your calculated values would you expect to see in the temperatures and volumetric flow rates of the effluent streams [increase, decrease, cannot tell without doing the calculations]?

In a coal gasification process, carbon (the primary constituent of coal) reacts with steam to produce carbon monoxide and hydrogen (synthesis gas). The gas may either be burned or subjected to further processing to produce any of a variety of chemicals. A coal contains 10.5 wt\% moisture (water) and 22.6 wt\% noncombustible ash. The remaining fraction of the coal contains 81.2 wife \(\mathrm{C}, 13.4 \%\) O, and \(5.4 \%\) H. A coal slurry containing \(2.00 \mathrm{kg}\) coal/kg water is fed at \(25^{\circ} \mathrm{C}\) to an adiabatic gasification reactor along with a stream of pure oxygen at the same temperature. The following reactions take place in the reactor: $$\begin{array}{l}\mathrm{C}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{v}) \rightarrow \mathrm{CO}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g}): \quad \Delta H_{\mathrm{r}}^{\circ}=+131.3 \mathrm{kJ} \\\\\mathrm{C}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{CO}_{2}(\mathrm{g}): \quad \Delta H_{\mathrm{r}}^{\circ}=-393.5 \mathrm{kJ} \\ 2 \mathrm{H}(\mathrm{in} \mathrm{coal})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{v}): \quad \Delta H_{\mathrm{r}}^{\circ} \approx-242 \mathrm{kJ}\end{array}$$ Gas and slag (molten ash) leave the reactor at \(2500^{\circ} \mathrm{C}\). The gas contains \(\mathrm{CO}, \mathrm{H}_{2}, \mathrm{CO}_{2},\) and \(\mathrm{H}_{2} \mathrm{O}^{14}\) (a) Feeding oxygen to the reactor lowers the yield of synthesis gas, but no gasifier ever operates without supplementary oxygen. Why does the oxygen lower the yield? Why it is nevertheless always supplied. (Hint: All the necessary information is contained in the first two stoichiometric equations and associated heats of reaction shown above.) (b) Suppose the oxygen gas fed to the reactor and the oxygen in the coal combine with all the hydrogen in the coal (Reaction 3) and with some of the carbon (Reaction 2), and the remainder of the carbon is consumed in Reaction 1. Taking a basis of 1.00 kg coal fed to the reactor and letting \(n_{0}\) equal the moles of \(\mathrm{O}_{2}\) fed, draw and label a flowchart. Then derive expressions for the molar flow rates of the four outlet gas species in terms of \(n_{0}\). [Partial solution: \(n_{\mathrm{H}_{2}}=\left(51.3-n_{0}\right)\) mol \(\mathrm{H}_{2} . \mathrm{J}\) (c) The standard heat of combustion of the coal has been determined to be -21,400 kJ/kg, taking \(\mathrm{CO}_{2}(\mathrm{g})\) and \(\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\) to be the combustion products. Use this value and the given clemental composition of the coal to prove that the standard heat of formation of the coal is \(-1510 \mathrm{kJ} / \mathrm{kg}\). Then use an energy balance to calculate \(n_{0},\) using the following approximate heat capacities in your calculation: Take the heat of fusion of ash (the heat required to convert ash to slag) to be \(710 \mathrm{kJ} / \mathrm{kg}\).
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