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Chapter 9: Problem 4

The standard heat of the reaction $$\mathrm{CaC}_{2}(\mathrm{s})+5 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{CaO}(\mathrm{s})+2 \mathrm{CO}_{2}(\mathrm{g})+5 \mathrm{H}_{2}(\mathrm{g})$$ is \(\Delta H_{\mathrm{t}}^{\circ}=+69.36 \mathrm{kJ}\). (a) Is the reaction exothermic or endothermic at \(25^{\circ} \mathrm{C}\) ? Would you have to heat or cool the reactor to kecp 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) Calculate \(\Delta U_{\mathrm{r}}^{\circ}\) for this reaction. (See Example \(9.1-2 .\) ) Briefly explain the physical significance of your calculated value. (c) Suppose you charge \(150.0 \mathrm{g}\) of \(\mathrm{CaC}_{2}\) and liquid water into a rigid container at \(25^{\circ} \mathrm{C}\), heat the container until the calcium carbide reacts completely, and cool the products back down to \(25^{\circ} \mathrm{C}\). condensing essentially all the unconsumed water. Write and simplify the energy balance equation for this closed constant-volume system and use it to determine the net amount of heat (kJ) that must be transferred to or from the reactor (state which). (d) If in Part (c) the term "rigid container" were replaced with "container at a constant pressure of 1 atm," the calculated value of \(Q\) would be slightly in error. Explain why. (e) If you placed 1 mol of solid calcium carbide and 5 mol of liquid water in a container at \(25^{\circ} \mathrm{C}\) and left them there for several days, upon returning you would not find 1 mol of solid calcium oxide, 2 mol of carbon dioxide, and 5 mol of hydrogen gas. Explain why not.

Short Answer

Expert verified
a) The reaction is endothermic, so we would have to heat the reactor to keep the temperature constant. In an adiabatic reactor, the temperature would decrease. b) \(\Delta U_{\mathrm{r}}^{\circ}=+69.36 kJ\), the positive value signifies that the energy absorbed to break the bonds in the reactants is more than that released during the formation of product bonds. c) The net amount of heat is positive, indicating that heat must be transferred into the reactor. d) If we were to use a container with constant pressure, the value of Q would differ because of possible volume change. e) Not finding the expected amount of products is due to the reaction being in a dynamic equilibrium at 25掳C.

Step by step solution

01

Determine the Nature of the Reaction

From the given heat of reaction, \(\Delta H_{\mathrm{t}}^{\circ}\), and that it's a positive value (+69.36 kJ), we can determine that the reaction is endothermic as it absorbs heat from the surroundings. So, the reaction would require heat to keep the temperature constant. If the reaction was carried in an adiabatic reactor, the temperature would decrease because the reaction absorbs heat.
02

Calculate Internal Energy Change

The formula for calculating the change in internal energy \(\Delta U_{\mathrm{r}}^{\circ}\) is: \(\Delta U_{\mathrm{r}}^{\circ} = \Delta H_{\mathrm{t}}^{\circ} - RT\Delta n\). Here, \(R\) is the ideal gas constant (8.314 J/mol.K), \(T\) is the temperature in Kelvin (298K), and \(\Delta n\) is the change in number of moles of gas (product moles - reactant moles = 5 - 5 = 0). After substituting these values into the formula you get: \(\Delta U_{\mathrm{r}}^{\circ}=+69.36 kJ - 0 = +69.36 kJ\)
03

Determine Net Amount of Heat Transferred

To calculate the net amount of heat that must be transferred, we first need to calculate the number of moles of calcium carbide reacting, by dividing the mass by the molar mass. Once we have the number of moles (n), we can calculate the heat (Q) using the formula Q = n\(\Delta H_{\mathrm{t}}^{\circ}\). In this case, heat is absorbed, so it's a positive value.
04

Explain Influence of Pressure

If a container at constant pressure were used, the calculated value of Q would be slightly in error. This is because constant pressure allows the system volume to change, which can affect the amount of work done by the system and therefore alter the amount of heat transferred.
05

Explain why the Reaction does not Fully Proceed

Even if sufficient quantities of reactants are present, reactions do not always proceed to completion because of limitations imposed by equilibrium. At 25掳C, it is likely that the reaction is in a state of dynamic equilibrium, where the forward and reverse reactions occur at the same rate, resulting in stable concentration of reactants and products over time.

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Key Concepts

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

Endothermic Reactions
In the study of chemistry, endothermic reactions are processes that absorb energy from their surroundings, usually in the form of heat. This is in contrast to exothermic reactions which release heat.

For a reaction where calcium carbide reacts with water to produce calcium oxide, carbon dioxide, and hydrogen gas, the heat of the reaction, denoted as \(\Delta H_{\mathrm{t}}^{\circ}\), is a positive value (+69.36 kJ). This positive sign indicates the endothermic nature of the reaction, meaning it requires an input of energy.

In practical terms, maintaining the reaction temperature constant would necessitate the addition of heat to the system. If the reaction ran adiabatically, which means without the exchange of heat with the surroundings, the temperature within the reactor would drop as the reaction absorbs heat. This characteristic is central to understanding energy changes during chemical processes.
Internal Energy Change (\(\Delta U\))
The internal energy change, expressed as \(\Delta U\), is a fundamental concept in thermodynamics which portrays the total change in energy within a system. In the context of chemical reactions, \(\Delta U\) represents the energy difference between the reactants and the products.

When calculating \(\Delta U\) for a reaction, the relationship between \(\Delta U\) and the enthalpy change \(\Delta H\) is often used, taking into account the work done by the system due to changes in volume and pressure. This is given by the equation \(\Delta U = \Delta H - P\Delta V\), and when dealing with gases at constant pressure, it simplifies to \(\Delta U_{\mathrm{r}}^{\circ} = \Delta H_{\mathrm{t}}^{\circ} - RT\Delta n\), where \(R\) is the gas constant, \(T\) the temperature, and \(\Delta n\) the change in the number of moles of gases.

In the exercise, since \(\Delta n\) is zero, the internal energy change is equal to the heat of reaction, revealing that the energy stored within the chemical system does not change due to volume work on or by the system under these conditions.
Heat Transfer in Reactions
Understanding heat transfer in reactions is crucial for controlling and optimizing chemical processes. The amount of heat that must be transferred to or from a system to drive a reaction is calculated based on the reaction's \(\Delta H\) and the amount of reactant.

In the textbook exercise, we are presented with a scenario where calcium carbide reacts with water at constant volume, necessitating a heat calculation to maintain the temperature. For systems at constant volume, the heat transfer (\(Q\)) can directly be calculated using the amount of reactant (moles) and the heat of reaction (\(\Delta H_{\mathrm{t}}^{\circ}\)). The formula for this calculation is \(Q = n\Delta H_{\mathrm{t}}^{\circ}\), with the sign indicating the direction of heat flow.

This is essential because whether heat is added or removed influences the reaction鈥檚 direction and extent, showcasing the intimate connection between thermodynamics and chemical kinetics.
Reaction Equilibrium
Lastly, reaction equilibrium is a dynamic state where the rate of the forward reaction equals the rate of the reverse reaction, resulting in no net change in the concentration of reactants and products over time.

In the given textbook example, even with sufficient reactants, the reaction does not progress to full completion because it reaches an equilibrium state. At 25掳C, the reactants and products have reached a state of balance that is governed by the reaction's equilibrium constant.

This concept is pivotal when considering the feasibility and yield of a chemical reaction under specific conditions. It explains why, under certain conditions, a mixture of reactants and products is found even after an extended period, highlighting the influence of thermodynamic principles in predicting the outcome of chemical processes.

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Most popular questions from this chapter

Metallic iron is produced in the reaction between ferrous oxide and carbon monoxide: $$\mathrm{FeO}(\mathrm{s})+\mathrm{CO}(\mathrm{g}) \rightarrow \mathrm{Fe}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g}), \quad \Delta H_{\mathrm{r}}=-16.480 \mathrm{kJ}$$ The flowchart shown below depicts this process for a basis of 1 mol \(\mathrm{FeO}\) fed at \(298 \mathrm{K}\). (a) We wish to explore the effects of the variables \(n_{0}\) (the molar feed ratio of CO to \(\mathrm{FeO}\) ), \(T_{0}\) (the feed temperature of the carbon monoxide), \(X\) (the fractional conversion of \(\mathrm{FeO}\) ), and \(T\) (the product temperature) on \(Q\) (the heat duty on the reactor). Without doing any calculations, sketch the shapes of the curves you would expect to obtain for the following plots: (i) Let \(n_{0}=1\) mol \(\mathrm{CO}\) fed/mol \(\mathrm{FeO}\) fed, \(T_{0}=400 \mathrm{K},\) and \(X=1 .\) Vary \(T\) from \(298 \mathrm{K}\) to \(1000 \mathrm{K}\) calculate \(Q\) for each \(T,\) and plot \(Q\) versus \(T\). (ii) Let \(n_{0}=1\) mol CO fed/mol FeO fed, \(T=700\) K, and \(X=1 .\) Vary \(T_{0}\) from 298 K to 1000 K. calculate \(Q\) for each \(T_{0}\), and plot \(Q\) versus \(T_{0}\). (iii) Let \(n_{0}=1\) mol CO fed/mol \(\mathrm{FeO}\) fed, \(T_{0}=400 \mathrm{K},\) and \(T=500 \mathrm{K}\). Vary \(X\) from 0 to 1 calculate \(Q\) for each \(X,\) and plot \(Q\) versus \(X\) (iv) Let \(X=0.5, T_{0}=400 \mathrm{K},\) and \(T=400 \mathrm{K}\). Vary \(n_{0}\) from 0.5 to \(2 \mathrm{mol}\) CO fed/mol FeO fed, calculate \(Q\) for each \(n_{0},\) and plot \(Q\) versus \(n_{0}\) (b) Following is an inlet-outlet enthalpy table for the process: Write an expression for the heat duty on the reactor, \(Q(\mathrm{kJ})\), in terms of the \(n\) s and \(\hat{H}\) s in the table, the standard heat of the given reaction, and the extent of reaction, \(\xi\). Then derive expressions for the quantities \(\xi, n_{1}, n_{2}, n_{3},\) and \(n_{4}\) interms of the variables \(n_{0}\) and \(X\). Finally, derive expressions for \(\hat{H}_{0}\) as a function of \(T_{0}\) and for \(\hat{H}_{1}, \hat{H}_{2}, \hat{H}_{3},\) and \(\hat{H}_{4}\) as functions of \(T\). In the latter derivations, use the following formulas for \(C_{p}[\mathrm{kJ} /(\mathrm{mol} \cdot \mathrm{K})]\) in terms of \(T(\mathrm{K})\) adapted from Table \(2-151\) of Perry's Chemical Engineers' Handbook (sce Footnote 2): $$\begin{aligned}&\mathrm{FeO}(\mathrm{s}): \quad C_{p}=0.05280+6.243 \times 10^{-6} T-3.188 \times 10^{2} T^{-2}\\\&\mathrm{Fe}(\mathrm{s}): \quad C_{p}=0.01728+2.67 \times 10^{-5} T\\\ &\mathrm{CO}(\mathrm{g}): \quad C_{p}=0.02761+5.02 \times 10^{-6} T\\\&\mathrm{CO}_{2}(\mathrm{g}): \quad C_{p}=0.04326+1.146 \times 10^{-5} T-8.180 \times 10^{2} T^{-2}\end{aligned}$$ (c) Calculate the heat duty, \(Q(\mathrm{kJ}),\) for \(n_{0}=2.0 \mathrm{mol} \mathrm{CO}, T_{0}=350 \mathrm{K}, T=550 \mathrm{K},\) and \(X=0.700 \mathrm{mol}\) FeO reacted/mol FeO fed. (d) Prepare a spreadsheet that has the following format (a partial solution is given for one set of process variables): of reaction. Use the spreadsheet to generate the four plots described in Part (a). If the shapes of the plots do not match your predictions, explain why.

Biodiesel fuel - a sustainable alternative to petroleum diesel as a transportation fuel- -is produced via the transesterification of triglyceride molecules derived from vegetable oils or animal fats. For every \(9 \mathrm{kg}\) of biodiesel produced in this process, \(1 \mathrm{kg}\) of glycerol, \(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}_{3},\) is produced as a byproduct. Finding a market for the glycerol is important for biodiesel manufacturing to be economically viable. A process for converting glycerol to the industrially important specialty chemical intermediates acrolein, \(C_{3} \mathrm{H}_{4} \mathrm{O},\) and hydroxyacetone (acetol), \(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}_{2},\) has been proposed. $$\begin{array}{l}\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}_{3} \rightarrow \mathrm{C}_{3} \mathrm{H}_{4} \mathrm{O}+2 \mathrm{H}_{2} \mathrm{O} \\ \mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}_{3} \rightarrow \mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}_{2}+\mathrm{H}_{2} \mathrm{O} \end{array}$$ The reactions take place in the vapor phase at \(325^{\circ} \mathrm{C}\) in a fixed bed reactor over an acid catalyst. The feed to the reactor is a vapor stream at \(325^{\circ} \mathrm{C}\) containing 25 mol\% glycerol, \(25 \%\) water, and the balance nitrogen. All of the glycerol is consumed in the reactor, and the product stream contains acrolein and hydroxyacctone in a 9: 1 mole ratio. Data for the process species are shown below. $$\begin{array}{|l|c|c|}\hline \text { Species } & \Delta \hat{H}_{\mathrm{f}}(\mathrm{kJ} / \mathrm{mol}) & C_{p}\left[\mathrm{kJ} /\left(\mathrm{mol} \cdot^{\circ} \mathrm{C}\right)\right] \\ \hline \text { glycerol(v) } & -620 & 0.1745 \\ \hline \text { acrolein(v) } & -65 & 0.0762 \\\\\hline \text { hydroxyacetone(v) } & -372 & 0.1096 \\ \hline \text { water(v) } & -242 & 0.0340 \\\\\hline \text { nitrogen(g) } & 0 & 0.0291 \\ \hline\end{array}$$ (a) Assume a basis of 100 mol fed to the reactor, and draw and completely label a flowchart. Carry out a degree-of-freedom analysis assuming that you will use extents of reaction for the material balances. Then calculate the molar amounts of all product species. (b) Calculate the total heat added or removed from the reactor (state which it is), using the constant heat capacities given in the above table. (c) Assuming this process is implemented along with biodiesel production, how would you determine whether the biodiesel is an cconomically viable alternative to petroleum diesel? (d) If you do a degree-of-freedom analysis based on atomic species balances, you are likely to count one more equation than you have unknowns, and yet you know the system has zero degrees of freedom. Guess what the problem is, and then prove it.

A city with a population of 200,000 people operates a \(20.0 \times 10^{6}\) gal/day wastewater treatment plant. For every million gallons of wastewater treated, \(1875 \mathrm{lb}_{\mathrm{m}}\) of solids are generated, with \(75 \%\) of the solids being classified as "volatile" (meaning they are converted to gases during digestion). Solids generated in the treatment plant are fed to an anaerobic digester in which microorganisms break down biodegradable material in the absence of oxygen. The feed to the digester contains 45,000 mg solids/L at \(55^{\circ} \mathrm{F}\) and has a density approximately that of water. The digester operates at \(95^{\circ} \mathrm{F},\) converts \(50 \%\) of the volatile solids to a biogas containing 65 vol\% \(\mathrm{CH}_{4}\) and \(35 \%\) \(\mathrm{CO}_{2}\) at a rate of \(15 \mathrm{SCF}\) (standard cubic feet) gas per \(\mathrm{Ib}_{\mathrm{m}}\) of solids converted, and loses approximately \(250,000 \mathrm{Btu} / \mathrm{h}\) of heat to the surroundings. To supply the heat needed to raise the feed temperature from \(55^{\circ} \mathrm{F}\) to \(95^{\circ} \mathrm{F}\) and to make up for the heat loss, a stream of sludge is pumped from the digester through a heat exchanger in which it comes into thermal contact with a stream of hot water. The heated sludge is returned to the digester. The digester biogas is fed to a furnace in which a fraction of it is burned to heat the water for the heat exchanger from \(160^{\circ} \mathrm{F}\) to \(180^{\circ} \mathrm{F}\). A schematic of part of the process is shown below. (a) Calculate the rate (SCF/h) at which biogas is produced in the digester and the total heating value (Btu/h) of the gas (= fuel flow rate \(\times\) lower heating value). (b) Calculate the rate of heat transfer (Btu/h) between the hot water and sludge and the volumetric flow rate (ft \(^{3} / \mathrm{h}\) ) of the water passing through the heat exchanger. Assume the heat of reaction of the anaerobic digestion process is negligible. (c) If the biogas is burned in a boiler with \(80 \%\) efficiency (that is, \(80 \%\) of the heating value of the fuel goes to produce hot water for the heat exchanger), what fraction of the digester gas must be burned to heat the water from \(160^{\circ} \mathrm{F}\) to \(180^{\circ} \mathrm{F}\) ? What happens to the other \(20 \%\) of the heating value? (d) If there is excess digester gas available after meeting the process-water heating demand, what are its potential uses?

The standard heat of combustion of liquid \(n\) -octane to form \(\mathrm{CO}_{2}\) and liquid water at \(25^{\circ} \mathrm{C}\) and \(1 \mathrm{atm}\) is \(\Delta \hat{H}_{\mathrm{c}}=-5471 \mathrm{kJ} / \mathrm{mol}\) (a) Briefly explain what that means. Your explanation may take the form "When ___ (specify quantities of reactant species and their physical states) react to form ___ (quantities of product species and their physical states), the change in enthalpy is ___. (b) Is the reaction exothermic or endothermic at \(25^{\circ} \mathrm{C}\) ? 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? (c) If \(25.0 \mathrm{mol} / \mathrm{s}\) of liquid octane is consumed and the reactants and products are all at \(25^{\circ} \mathrm{C}\), estimate the required rate of heat input or output (state which) in kilowatts, assuming that \(Q=\Delta H\) for the process. What have you also assumed about the reactor pressure in your calculation? (You don't have to assume that it equals 1 atm.) (d) The standard heat of combustion of \(n\) -octane vapor is \(\Delta \hat{H}_{\mathrm{c}}=-5528 \mathrm{kJ} / \mathrm{mol}\). What is the physical significance of the \(57 \mathrm{kJ} / \mathrm{mol}\) difference between this heat of combustion and the one given previously? (e) The value of \(\Delta \hat{H}_{c}\) given in Part (d) applies to \(n\) -octane vapor at \(25^{\circ} \mathrm{C}\) and 1 atm, and yet the normal boiling point of \(n\) -octane is \(125.5^{\circ} \mathrm{C}\). Can \(n\) -octane exist as a vapor at \(25^{\circ} \mathrm{C}\) and a total pressure of 1 atm? Explain your answer.

Use Hess's law to calculate the standard heat of the water-gas shift reaction $$\mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{v}) \rightarrow \mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g})$$ from each of the two sets of data given here. (a) \(\mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g}): \quad \Delta H_{\mathrm{r}}^{\circ}=+1226 \mathrm{Btu}\) $$\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{v}): \quad \Delta \hat{H}_{\mathrm{v}}=+18,935 \mathrm{Btu} / \mathrm{lb}-\mathrm{mole}$$ $$\begin{aligned}&\text { (b) } \mathrm{CO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{CO}_{2}(\mathrm{g}): \quad \Delta H_{\mathrm{r}}^{\circ}=-121,740 \mathrm{Btu}\\\&\mathrm{H}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{v}): \quad \Delta H_{\mathrm{r}}^{\circ}=-104,040 \mathrm{Btu} \end{aligned}$$
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