Question

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 $\mathrm{\Delta }{H}^{\circ }=-2723.7\mathrm{kJ}$ . (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 $\mathrm{\Delta }{H}_f^{\circ }$ for diethyl ether.

Short answer

The balanced equation for the combustion of diethyl ether is: $${\mathrm{C}}_4{\mathrm{H}}_{10}\mathrm{O}(l)+6{\mathrm{O}}_2(g)\to 4{\mathrm{CO}}_2(g)+5{\mathrm{H}}_2\mathrm{O}(l)$$ Using Hess's Law and the given data, the standard enthalpy of formation for diethyl ether is calculated to be approximately: $$\Delta H_f^{\circ }({\mathrm{C}}_4{\mathrm{H}}_{10}\mathrm{O})=-279.3\mathrm{kJ}/\mathrm{mol}$$

Step-by-step solution

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: $$Hydrocarbon+O_2(g)\to C{O}_2(g)+H_{2}O(l)$$ For diethyl ether, the hydrocarbon is ${\mathrm{C}}_4{\mathrm{H}}_{10}\mathrm{O}$, so our balanced equation will be: $${\mathrm{C}}_4{\mathrm{H}}_{10}\mathrm{O}(l)+6{\mathrm{O}}_2(g)\to 4{\mathrm{CO}}_2(g)+5{\mathrm{H}}_2\mathrm{O}(l)$$

Apply Hess's Law to calculate ΔH°${}_f$ for diethyl ether

Now, we want to find the standard enthalpy of formation for diethyl ether. We will use the known values of ΔH° for the combustion reaction and the standard enthalpies of formation of the reaction products to calculate this value. According to Hess's Law: $$\Delta H_f^{\circ }({\mathrm{C}}_4{\mathrm{H}}_{10}\mathrm{O})=-\Delta H_{combustion}^{\circ }+\sum _{i=1}^4(\Delta H_f^{\circ }({\mathrm{CO}}_2))+\sum _{i=1}^5(\Delta H_f^{\circ }({\mathrm{H}}_2\mathrm{O}))-6(\Delta H_f^{\circ }({\mathrm{O}}_2))$$ 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: $$\Delta H_f^{\circ }({\mathrm{C}}_4{\mathrm{H}}_{10}\mathrm{O})=-\Delta H_{combustion}^{\circ }+4(\Delta H_f^{\circ }({\mathrm{CO}}_2))+5(\Delta H_f^{\circ }({\mathrm{H}}_2\mathrm{O}))$$ We are given ΔH°${}_{combustion}=-2723.7\mathrm{kJ}$, and we can find the standard enthalpies of formation for carbon dioxide and liquid water in Table 5.3: ΔH°${}_f({\mathrm{CO}}_2)=-393.5\mathrm{kJ}/\mathrm{mol}$ ΔH°${}_f({\mathrm{H}}_2\mathrm{O})=-285.8\mathrm{kJ}/\mathrm{mol}$ Now plug these values into the equation to get the standard enthalpy of formation for diethyl ether: $$\Delta H_f^{\circ }({\mathrm{C}}_4{\mathrm{H}}_{10}\mathrm{O})=-(-2723.7\mathrm{kJ})+4(-393.5\mathrm{kJ}/\mathrm{mol})+5(-285.8\mathrm{kJ}/\mathrm{mol})$$ $$\Delta H_f^{\circ }({\mathrm{C}}_4{\mathrm{H}}_{10}\mathrm{O})=2723.7\mathrm{kJ}-1574\mathrm{kJ}-1429\mathrm{kJ}$$ $$\Delta H_f^{\circ }({\mathrm{C}}_4{\mathrm{H}}_{10}\mathrm{O})=-279.3\mathrm{kJ}/\mathrm{mol}$$ So, the standard enthalpy of formation for diethyl ether is about -279.3 kJ/mol.

Key concepts

Combustion Reaction

A combustion reaction is a high-energy chemical process typically involving a fuel—often a hydrocarbon—and 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$${\mathrm{C}}_4{\mathrm{H}}_{10}\mathrm{O}(l)+6{\mathrm{O}}_2(g)\to 4{\mathrm{CO}}_2(g)+5{\mathrm{H}}_2\mathrm{O}(l)$$

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$\mathrm{\Delta }H$) for reactions where it’s 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$\mathrm{\Delta }{H}_f^{\circ }$) 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$${\mathrm{C}}_4{\mathrm{H}}_{10}\mathrm{O}(l)+6{\mathrm{O}}_2(g)\to 4{\mathrm{CO}}_2(g)+5{\mathrm{H}}_2\mathrm{O}(l)$$

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.