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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) \(\mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g}): \quad \Delta H_{\mathrm{r}}^{\circ}=+1226 \mathrm{Btu}\) $$\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{v}): \quad \Delta \hat{H}_{\mathrm{v}}=+18,935 \mathrm{Btu} / \mathrm{lb}-\mathrm{mole}$$ $$\begin{aligned}&\text { (b) } \mathrm{CO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{CO}_{2}(\mathrm{g}): \quad \Delta H_{\mathrm{r}}^{\circ}=-121,740 \mathrm{Btu}\\\&\mathrm{H}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{v}): \quad \Delta H_{\mathrm{r}}^{\circ}=-104,040 \mathrm{Btu} \end{aligned}$$

Short Answer

Expert verified
The standard heat of the water-gas shift reaction calculated from set (a) of the given data is +1226 Btu + 18935 Btu/lb-mole. From set (b) of the data, it is -121740 Btu + 104040 Btu.

Step by step solution

01

Determine Reaction Segments

Firstly, let's focus on the target reaction which is \(\mathrm{CO(g)}+\mathrm{H}_{2}\mathrm{O(v)}=\mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g})\). We need to split this reaction into segments that match the given reactions. For the set (a), this can be done as follows: \(\mathrm{CO(g)}+\mathrm{H}_{2}\mathrm{O(l)}=\mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g})\) and \(\mathrm{H}_{2} \mathrm{O(l)} = \mathrm{H}_{2}\mathrm{O(v)}\). Then, we can add these reactions together to get the target equation.
02

Calculate Heat of Reaction for Step 1 (Set A)

For the first reaction, the heat of reaction \(\Delta H_{\mathrm{r}}^{\circ}\) is given as +1226 Btu. The second reaction has the heat of vaporization \(\Delta \hat{H}_{\mathrm{v}}\) equal to +18935 Btu/lb-mole. Adding these two together, we get the total enthalpy change for the first set of data - \(\Delta H_{\mathrm{total}}^{\circ} = +1226 Btu + 18935 Btu/lb-mole\).
03

Determine Reaction Segments for Set B

Moving on to set (b), we have two more reactions. These are: \(\mathrm{CO(g)}+\frac{1}{2}\mathrm{O}_{2}(\mathrm{g})=\mathrm{CO}_{2}(\mathrm{g})\) and \(\mathrm{H}_{2}(\mathrm{g})+\frac{1}{2}\mathrm{O}_{2}(\mathrm{g})=\mathrm{H}_{2}\mathrm{O(v)}\). We can see that adding these reactions together doesn't directly give the target equation. We'll need to reverse the second reaction and then add them.
04

Calculate Heat of Reaction for Step 2 (Set B)

The heat of reaction for the two reactions are given as -121740 Btu and -104040 Btu respectively. However, since we are reversing the second equation, the sign of the heat of reaction for it should be changed. The total enthalpy change for the second set of data is thus \(\Delta H_{\mathrm{total}}^{\circ} = -121740 Btu + 104040 Btu\).

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Key Concepts

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

Thermodynamics
Thermodynamics is the study of energy, heat, and work. It's a core subject in chemistry and physics, helping us understand how energy is transferred in systems. When we talk about chemical reactions, thermodynamics allows us to predict if a reaction will be spontaneous or not. This prediction is crucial in determining if a reaction will occur naturally without added energy.
Thermodynamic principles are essential in calculating changes in different types of energy, such as enthalpy, entropy, and free energy. In this exercise, thermodynamics helps us use Hess’s Law—a key principle—to find the enthalpy change of the water-gas shift reaction. By understanding thermodynamic laws, we can fully grasp how energy moves and changes within chemical reactions, making it a fascinating and practical field.
Enthalpy Change
Enthalpy change is a measure of the heat change during a chemical or physical process at constant pressure. It represents the total energy change in a system and includes internal energy and the energy required to make room for it by displacing its environment.
When bonds are broken and new ones are formed in a chemical reaction, energy is either absorbed or released. This is known as the enthalpy change (\( \Delta H \)). For exothermic reactions, energy is released, and \( \Delta H \) is negative. For endothermic reactions, energy is absorbed, and \( \Delta H \) is positive.
  • In uses of Hess's Law, enthalpy change for two or more reactions can be added or subtracted to find the enthalpy change of a target reaction.
  • In the original exercise, Hess's Law helps us combine reactions and calculate the enthalpy change, using the provided reactions and their \( \Delta H \) values.
By understanding and calculating enthalpy change, we understand how energy exchanges in chemical reactions impact the environment and the universe.
Chemical Reactions
Chemical reactions are processes where reactants are transformed into products through breaking and forming chemical bonds. These changes are fundamental in chemistry as they rearrange elements into new compounds.
In a balanced chemical reaction, the number of atoms for each element is the same on the reactants side and on the products side, reflecting the law of conservation of mass. Understanding the nature of reactants and products, their structures, and the energy involved in breaking or forming bonds is key to predicting the behavior of chemical reactions.
  • The original exercise breaks down the water-gas shift reaction into smaller steps and reactions: each step is part of the process of rearranging reactant molecules into product molecules.
  • By analyzing these steps, we better understand the sequence of reactions needed to achieve the desired chemical transformation.
Learning about chemical reactions enhances our understanding of how molecules interact, facilitating fields like pharmaceuticals, energy, and materials science.
Water-Gas Shift Reaction
The water-gas shift reaction is a chemical reaction where carbon monoxide and water vapor react to form carbon dioxide and hydrogen gas: \( \text{CO(g)} + \text{H}_2\text{O(v)} \rightarrow \text{CO}_2\text{(g)} + \text{H}_2\text{(g)} \).
This reaction is useful for producing hydrogen gas and is often part of industrial processes. The equation is crucial in energy production, particularly in processes like hydrocarbon reforming and biomass gasification. It is a means to increase hydrogen yield while minimizing carbon monoxide emissions.
  • By utilizing Hess's Law, we can determine the energy changes involved in this reaction using standard enthalpy changes from other known reactions.
  • This method helps in designing industrial processes where controlling energy flow is essential for efficiency.
Understanding the specifics of the water-gas shift reaction aids in improving the sustainability and efficiency of energy systems, making it highly relevant in our pursuit of cleaner energy solutions.

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Most popular questions from this chapter

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.

Ammonia scrubbing is one of many processes for removing sulfur dioxide from flue gases. The gases are bubbled through an aqueous solution of ammonium sulfite, and the SO_reacts to form ammonium bisulfite: $$\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{3}(\mathrm{aq})+\mathrm{SO}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow 2 \mathrm{NH}_{4} \mathrm{HSO}_{3}(\mathrm{aq})$$ Subsequent process steps yield concentrated SO \(_{2}\) and regenerate ammonium sulfite, which is recycled to the scrubber. The sulfur dioxide is either oxidized and absorbed in water to form sulfuric acid or reduced to elemental sulfur. Flue gas from a power-plant boiler containing \(0.30 \% \mathrm{SO}_{2}\) by volume enters a scrubber at a rate of \(50,000 \mathrm{mol} / \mathrm{h}\) at \(50^{\circ} \mathrm{C} .\) The gas is bubbled through an aqueous solution containing \(10.0 \mathrm{mole} \%\) ammonium sulfite that enters the scrubber at \(25^{\circ} \mathrm{C}\). The gas and liquid effluents from the scrubber both emerge at \(35^{\circ} \mathrm{C}\). The scrubber removes \(90 \%\) of the \(S O_{2}\) entering with the flue gas. The effluent liquid is analyzed and is found to contain 1.5 moles \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{3}\) per mole of \(\mathrm{NH}_{4} \mathrm{HSO}_{3}\). The heat of formation of \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{3}(\mathrm{aq})\) at \(25^{\circ} \mathrm{C}\) is \(-890.0 \mathrm{kJ} / \mathrm{mol},\) and that of \(\mathrm{NH}_{4} \mathrm{HSO}_{3}(\mathrm{aq})\) is \(-760 \mathrm{kJ} / \mathrm{mol} .\) The heat capacities of all liquid solutions may be taken to be \(4.0 \mathrm{J} /\left(\mathrm{g} \cdot^{\circ} \mathrm{C}\right)\) and that of the flue gas may be taken to be that of nitrogen. Evaporation of water may be neglected. Calculate the required rate of heat transfer to or from the scrubber ( \(\mathrm{kW}\) ).

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.

The synthesis of cthyl chloride is accomplished by reacting ethylene with hydrogen chloride in the presence of an aluminum chloride catalyst: $$\mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{g})+\mathrm{HCl}(\mathrm{g}) \stackrel{\text { catallyst }}{\longrightarrow} \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}(\mathrm{g}) ; \quad \Delta H_{\mathrm{r}}\left(0^{\circ} \mathrm{C}\right)=-64.5 \mathrm{kJ}$$ Process data and a simplified schematic flowchart are given here. Data Reactor: adiabatic, outlet temperature \(=50^{\circ} \mathrm{C}\) Feed A: \(100 \% \mathrm{HCl}(\mathrm{g}), 0^{\circ} \mathrm{C}\) Feed \(\mathrm{B}: 93\) mole \(\% \mathrm{C}_{2} \mathrm{H}_{4}, 7 \% \mathrm{C}_{2} \mathrm{H}_{6}, 0^{\circ} \mathrm{C}\) Reactor: adiabatic, outlet temperature \(=50^{\circ} \mathrm{C}\) Feed A: 100\% HCl(g), 0"C Feed B: 93 mole\% C_H_4, 7\% C_H_0, 0"C Product C: Consists of 1.5\% of the HCl, 1.5\% of the C_2 \(\mathrm{H}_{4}\), and all of the \(\mathrm{C}_{2} \mathrm{H}_{6}\) that enter the reactor Product D: \(1600 \mathrm{kg} \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}(\mathrm{l}) / \mathrm{h}, 0^{\circ} \mathrm{C}\) Recycle to reactor: \(\mathbf{C}_{2} \mathrm{H}_{5} \mathrm{Cl}(\mathrm{l}), 0^{\circ} \mathrm{C}\) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}: \Delta \hat{H}_{\mathrm{y}}=24.7 \mathrm{kJ} / \mathrm{mol}\) (assume independent of \(T\) ) \(\left(C_{p}\right)_{C_{2} H_{3} C(v)}\left[\mathrm{kJ} /\left(\mathrm{mol} \cdot^{\circ} \mathrm{C}\right)\right]=0.052+8.7 \times 10^{-5} T\left(^{\circ} \mathrm{C}\right)\) The reaction is exothermic, and if the heat of reaction is not removed in some way, the reactor temperature could increase to an undesirably high level. To avoid this occurrence, the reaction is carried out with the catalyst suspended in liquid cthyl chloride. As the reaction proceeds, most of the heat liberated goes to vaporize the liquid, making it possible to keep the reaction temperature at or below 50^'C. The stream leaving the reactor contains cthyl chloride formed by reaction and that vaporized in the reactor. This stream passes through a heat exchanger where it is cooled to \(0^{\circ} \mathrm{C},\) condensing essentially all of the cthyl chloride and leaving only unreacted \(\mathrm{C}_{2} \mathrm{H}_{4}, \mathrm{HCl}\), and \(\mathrm{C}_{2} \mathrm{H}_{6}\) in the gas phase. A portion of the liquid condensate is recycled to the reactor at a rate equal to the rate at which ethyl chloride is vaporized, and the rest is taken off as product. At the process conditions, heats of mixing and the influence of pressure on enthalpy may be neglected. (a) At what rates (kmol/h) do the two feed streams enter the process? (b) Calculate the composition (component mole fractions) and molar flow rate of product stream \(\mathrm{C}\). (c) Write an energy balance around the reactor and use it to determine the rate at which ethyl chloride must be recycled. (d) A number of simplifying assumptions were made in the process description and the analysis of this process system, so the results obtained using a more realistic simulation would differ considerably from those you should have obtained in Parts (a)-(c). List as many of these assumptions as you can think of.

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.

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