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Chapter 5: Problem 78

Diethyl ether, \(\mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O}(l),\) a flammable compound that was once used as a surgical anesthetic, has the structure $$\mathrm{H}_{3} \mathrm{C}-\mathrm{CH}_{2}-\mathrm{O}-\mathrm{CH}_{2}-\mathrm{CH}_{3}$$ The complete combustion of 1 mol of \(\mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O}(l)\) to \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(l)\) yields \(\Delta H^{\circ}=-2723.7 \mathrm{kJ}\) . (a) Write a balanced equation for the combustion of 1 \(\mathrm{mol}\) of \(\mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O}(l) .\) (b) By using the information in this problem and data in Table \(5.3,\) calculate \(\Delta H_{f}^{\circ}\) for diethyl ether.

Short Answer

Expert verified
The balanced equation for the combustion of diethyl ether is: \[ \mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O}(l) + 6\mathrm{O}_{2}(g) \rightarrow 4\mathrm{CO}_{2}(g) + 5\mathrm{H}_{2} \mathrm{O}(l) \] Using Hess's Law and the given data, the standard enthalpy of formation for diethyl ether is calculated to be approximately: \[ 螖H_{f}^{\circ}(\mathrm{C}_{4}\mathrm{H}_{10}\mathrm{O}) = -279.3\,\mathrm{kJ/mol} \]

Step by step solution

01

Write a balanced equation for the combustion of diethyl ether

To write a balanced chemical equation for the combustion of diethyl ether, we need to make sure that the number of each atom in the reactants is equal to the number in the products. The general equation for the combustion of a hydrocarbon is: \[ Hydrocarbon + O_{2}(g) \rightarrow CO_{2}(g) + H_{2}O(l) \] For diethyl ether, the hydrocarbon is \(\mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O}\), so our balanced equation will be: \[ \mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O}(l) + 6\mathrm{O}_{2}(g) \rightarrow 4\mathrm{CO}_{2}(g) + 5\mathrm{H}_{2} \mathrm{O}(l) \]
02

Apply Hess's Law to calculate 螖H掳\(_f\) for diethyl ether

Now, we want to find the standard enthalpy of formation for diethyl ether. We will use the known values of 螖H掳 for the combustion reaction and the standard enthalpies of formation of the reaction products to calculate this value. According to Hess's Law: \[ 螖H_{f}^{\circ}(\mathrm{C}_{4}\mathrm{H}_{10}\mathrm{O}) = -螖H^{\circ}_{combustion} + \sum_{i=1}^{4}\left(螖H_{f}^{\circ}(\mathrm{CO}_{2})\right) + \sum_{i=1}^{5}\left(螖H_{f}^{\circ}(\mathrm{H}_{2}\mathrm{O})\right) - 6\left(螖H_{f}^{\circ}(\mathrm{O}_{2})\right) \] Note that the standard enthalpy of formation of an element in its standard state (oxygen gas in this case) is always zero, so we can simplify: \[ 螖H_{f}^{\circ}(\mathrm{C}_{4}\mathrm{H}_{10}\mathrm{O}) = -螖H^{\circ}_{combustion} + 4\left(螖H_{f}^{\circ}(\mathrm{CO}_{2})\right) + 5\left(螖H_{f}^{\circ}(\mathrm{H}_{2}\mathrm{O})\right) \] We are given 螖H掳\(_{combustion} = -2723.7\,\mathrm{kJ}\), and we can find the standard enthalpies of formation for carbon dioxide and liquid water in Table 5.3: 螖H掳\(_{f}(\mathrm{CO}_{2}) = -393.5\,\mathrm{kJ/mol}\) 螖H掳\(_{f}(\mathrm{H}_{2}\mathrm{O}) = -285.8\,\mathrm{kJ/mol}\) Now plug these values into the equation to get the standard enthalpy of formation for diethyl ether: \[ 螖H_{f}^{\circ}(\mathrm{C}_{4}\mathrm{H}_{10}\mathrm{O}) = -(-2723.7\,\mathrm{kJ}) + 4(-393.5\,\mathrm{kJ/mol}) + 5(-285.8\,\mathrm{kJ/mol}) \] \[ 螖H_{f}^{\circ}(\mathrm{C}_{4}\mathrm{H}_{10}\mathrm{O}) = 2723.7\,\mathrm{kJ} - 1574\,\mathrm{kJ} - 1429\,\mathrm{kJ} \] \[ 螖H_{f}^{\circ}(\mathrm{C}_{4}\mathrm{H}_{10}\mathrm{O}) = -279.3\,\mathrm{kJ/mol} \] So, the standard enthalpy of formation for diethyl ether is about -279.3 kJ/mol.

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Key Concepts

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

Combustion Reaction
A combustion reaction is a high-energy chemical process typically involving a fuel鈥攐ften a hydrocarbon鈥攁nd oxygen, resulting in the release of heat and the formation of water and carbon dioxide as primary products. In essence, it's what happens when something burns in the presence of oxygen. Common examples include the burning of wood in a fireplace or the combustion of gasoline in an engine.

For a combustion reaction to be complete, it needs to be balanced, meaning the number of each type of atom on the reactant side must equal the number in the products. This ensures that mass is conserved and complies with the Law of Conservation of Mass. During the combustion of diethyl ether (\rC4H10O\(l\)), the chemical reaction can be represented by:

\r\[ \mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O}(l) + 6\mathrm{O}_{2}(g) \rightarrow 4\mathrm{CO}_{2}(g) + 5\mathrm{H}_{2} \mathrm{O}(l) \]

This balanced equation indicates that for every mole of diethyl ether that combusts, six moles of oxygen are consumed, producing four moles of carbon dioxide and five moles of liquid water. Understanding combustion reactions is crucial because they are the backbone of many industrial processes and energy systems.
Hess's Law
When it comes to understanding the heat involved in chemical reactions, Hess's Law is a true cornerstone. This principle states that the total enthalpy change for a reaction is the same, regardless of the number of steps that the reaction is carried out in. Essentially, if you can take different pathways to get from reactants to products, the total energy change will be the same for each path, as long as you start and end in the same states.

This concept allows chemists to figure out the change in enthalpy (\r\(\Delta H\)) for reactions where it鈥檚 otherwise tough to measure, simply by adding up the enthalpy changes for individual steps that lead to the same final result. In our example concerning diethyl ether, Hess's Law lets us calculate the enthalpy of formation (\r\(\Delta H_{f}^{\circ}\)) by taking into account the known enthalpies for the combustion products (carbon dioxide and water) and the enthalpy of the combustion process itself.

Understanding Hess's Law is vital for students because it not only aids in solving thermodynamic problems but also provides insights into the conservation of energy within chemical reactions.
Balanced Chemical Equation
A balanced chemical equation provides a clear picture of a chemical reaction where the number of atoms for each element is conserved. In simpler terms, what goes into the reaction must come out in a different form, without any discrepancy in the quantity of atoms. It is the chemical accountant's way of ensuring that all atoms are accounted for, from reactants to products.

Creating a balanced equation involves adjusting the coefficients placed in front of chemical formulas until the number of atoms of each element on both sides of the equation matches. For example, the balanced equation for the combustion of diethyl ether includes coefficients that ensure there are equal numbers of carbon, hydrogen, and oxygen atoms on both sides:

\r\[ \mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O}(l) + 6\mathrm{O}_{2}(g) \rightarrow 4\mathrm{CO}_{2}(g) + 5\mathrm{H}_{2} \mathrm{O}(l) \]

This balance is not only a fundamental part of following the law of conservation of mass but also critical for calculations in stoichiometry, which can determine the amount of reactants needed or products formed. Mastery of balancing chemical equations is, therefore, essential for anyone studying chemistry.

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Most popular questions from this chapter

When a 6.50 -g sample of solid sodium hydroxide dissolves in 100.0 g of water in a coffee-cup calorimeter (Figure 5.18\()\) the temperature rises from 21.6 to to \(37.8^{\circ} \mathrm{C}\) . (a) Calculate the quantity of heat (in kJ) released in the reaction. (b) Using your result from part (a), calculate \(\Delta H\) (in \(\mathrm{kJ} / \mathrm{mol} \mathrm{NaOH} )\) for the solution process. Assume that the specific heat of the solution is the same as that of pure water.

The automobile fuel called \(E 85\) consists of 85\(\%\) ethanol and 15\(\%\) gasoline. E85 can be used in the so-called flex-fuel vehicles (FFVs), which can use gasoline, ethanol, or a mix as fuels. Assume that gasoline consists of a mixture of octanes (different isomers of \(\mathrm{C}_{8} \mathrm{H}_{18} ),\) that the average heat of combustion of \(\mathrm{C}_{8} \mathrm{H}_{18}(l)\) is 5400 \(\mathrm{kJ} / \mathrm{mol}\) , and that gasoline has an average density of 0.70 \(\mathrm{g} / \mathrm{mL}\) . The density of ethanol is 0.79 \(\mathrm{g} / \mathrm{mL}\) . (a) By using the information given as well as data in Appendix \(\mathrm{C},\) compare the energy produced by combustion of 1.0 L of gasoline and of 1.0 . of ethanol. (b) Assume that the density and heat of combustion of E85 can be obtained by using 85\(\%\) of the values for ethanol and 15\(\%\) of the values for gasoline. How much energy could be released by the combustion of 1.0 L of E85? (\mathbf{c} ) How many gallons of E85 would be needed to provide the same energy as 10 gal of gasoline? (d) If gasoline costs \(\$ 3.88\) per gallon in the United States, what is the break- even price per gallon of E85 if the same amount of energy is to be delivered?

Assume that the following reaction occurs at constant pressure: $$2 \mathrm{Al}(s)+3 \mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{AlCl}_{3}(s)$$ (a) If you are given \(\Delta H\) for the reaction, what additional information do you need to determine \(\Delta E\) for the process? (b) Which quantity is larger for this reaction? (c) Explain your answer to part (b).

How much work (in J) is involved in a chemical reaction if the volume decreases from 5.00 to 1.26 L against a constant pressure of 0.857 atm?

The standard enthalpies of formation of gaseous propyne \(\left(\mathrm{C}_{3} \mathrm{H}_{4}\right),\) propylene \(\left(\mathrm{C}_{3} \mathrm{H}_{6}\right),\) and propane \(\left(\mathrm{C}_{3} \mathrm{H}_{8}\right)\) are \(+185.4,+20.4,\) and \(-103.8 \mathrm{kJ} / \mathrm{mol}\) , respectively.(a) Calculate the heat evolved per mole on combustion of each substance to yield \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(g) .\) (b) Calculate the heat evolved on combustion of 1 \(\mathrm{kg}\) of each substance. (c) Which is the most efficient fuel in terms of heat evolved per unit mass?
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