Problem 78 - Chemistry The Central Science

Question

Diethyl ether, \(\mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O}(l),\) a flammable compound that has long been used as a surgical anesthetic, has the structure \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{O}-\mathrm{CH}_{2}-\mathrm{CH}_{3}\) The complete combustion of \(1 \mathrm{~mol}\) 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) Write a balanced equation for the formation of \(\mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O}(l)\) from its elements. (c) By using the information in this problem and data in Table \(5.3,\) calculate \(\Delta H_{f}^{\circ}\) for diethyl ether.

Step-by-Step Solution

Answer

The balanced equations are: 1. Combustion: \( C_{4}H_{10}O(l) + 6\frac{1}{2} O_{2}(g) \rightarrow 4 CO_{2}(g) + 5 H_{2}O(l) \) 2. Formation: \( 4 C + 5 H_{2} + \frac{1}{2} O_{2} \rightarrow C_{4}H_{10}O \) The standard enthalpy of formation for diethyl ether is \( \Delta H_{f}^{\circ}(C_{4}H_{10}O) = -261.6 \mathrm{~kJ/mol} \).

Step 1: Write the Balanced Combustion Equation

To write the balanced equation for the combustion of 1 mol of diethyl ether (C4H10O), we need to know that: - Complete combustion means the reactants are Diethyl ether and O2; the products are CO2 and H2O. - The coefficients should be adjusted so that the atoms are conserved. The unbalanced equation looks like this: C4H10O(l) + O2(g) → CO2(g) + H2O(l) Now, balance the equation by adjusting the coefficients: C4H10O(l) + \(6\frac{1}{2}\) O2(g) → 4 CO2(g) + 5 H2O(l)

Step 2: Write the Balanced Formation Equation

To write the balanced equation for the formation of diethyl ether from its elements, we need to know that: - Formation means the reactants are the elements in their standard states; the product is diethyl ether. - The coefficients should be adjusted so that the atoms are conserved. The unbalanced equation looks like this: C + H2 + O2 → C4H10O Now, balance the equation by adjusting the coefficients: 4 C + 5 H2 + \( \frac{1}{2} \) O2 → C4H10O

Step 3: Calculate the Standard Enthalpy of Formation

To calculate the standard enthalpy of formation for diethyl ether, use the given information and the standard enthalpy of formation for the involved species from Table 5.3. We use the following equation: ∆Hf°(C4H10O) = ∆H°(products) - ∆H°(reactants) First, gather the standard enthalpy of formation values from table 5.3: ∆Hf°(C) = 0 kJ/mol (as carbon is in its standard state) ∆Hf°(H2) = 0 kJ/mol (as hydrogen is in its standard state) ∆Hf°(O2) = 0 kJ/mol (as oxygen is in its standard state) ∆Hf°(CO2) = -393.5 kJ/mol ∆Hf°(H2O) = -285.8 kJ/mol Now, we calculate the enthalpy change for the combustion of C4H10O from the given information: ∆H°(combustion) = -2723.7 kJ According to the balanced combustion equation (Step 1): ∆H°(combustion) = 4 ∆Hf°(CO2) + 5 ∆Hf°(H2O) - ∆Hf°(C4H10O) -2723.7 kJ = 4(-393.5 kJ/mol) + 5(-285.8 kJ/mol) - ∆Hf°(C4H10O) Now, solve for ∆Hf°(C4H10O): ∆Hf°(C4H10O) = 4(-393.5 kJ/mol) + 5(-285.8 kJ/mol) + 2723.7 kJ ∆Hf°(C4H10O) = - 261.6 kJ/mol The standard enthalpy of formation for diethyl ether is -261.6 kJ/mol.

Key Concepts

Enthalpy of Formation

Enthalpy of Formation is a fundamental concept in chemistry that helps us understand the energy change when 1 mole of a substance is created from its elements under standard conditions. These conditions mean a pressure of 1 atmosphere and a temperature of 25°C (298 K). The enthalpy of formation is denoted as \(\Delta H_f^{\circ}\) and usually measured in kilojoules per mole (kJ/mol).

In the specific example of diethyl ether, this value tells us how much energy is absorbed or released during its creation from carbon, hydrogen, and oxygen in their natural forms. For diethyl ether, calculated using standardized data and given reactions, the \(\Delta H_f^{\circ}\) is -261.6 kJ/mol. The negative sign indicates that its formation from the elements releases energy. This nature of energy release signifies that the compound is more stable than its individual elemental components.

Knowing \(\Delta H_f^{\circ}\) helps predict the behavior of substances during reactions and is essential for calculating enthalpy changes in various chemical processes, such as combustion or synthesis.

Balancing Chemical Equations

Balancing chemical equations is a fundamental skill for anyone studying chemistry. It ensures the conservation of mass where the number of each type of atom on the reactant side equals the number on the product side. This includes balancing all elements, even when they appear in multiple compounds.

To illustrate, consider the combustion of diethyl ether: the initial unbalanced equation is \(\text{C}_4\text{H}_{10}\text{O} + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}\). First, address the carbon atoms by ensuring there are 4 \(\text{CO}_2\) molecules for 4 carbons. Then, balance hydrogen by having 5 \(\text{H}_2\text{O}\) since there are 10 hydrogens in diethyl ether. Finally, calculate oxygen: the products need 13 oxygens, so \(6.5\) \(\text{O}_2\) will suffice. Thus, the balanced equation for combustion is \(\text{C}_4\text{H}_{10}\text{O} + 6.5 \text{O}_2 \rightarrow 4 \text{CO}_2 + 5 \text{H}_2\text{O}\).

Balancing requires practice and sometimes fractional coefficients, but it is critical for quantitatively predicting reaction outcomes. Once balanced correctly, equations can be used to find moles, enthalpy changes, and reaction stoichiometry.

Diethyl Ether

Diethyl ether, with the chemical formula \(\text{C}_4\text{H}_{10}\text{O}\), is an organic compound that belongs to the ether class. Historically, it has been used widely as a general anesthetic due to its volatility and effect on the human body. Its molecular structure \(\text{CH}_3\text{-CH}_2\text{-O-CH}_2\text{-CH}_3\) highlights the presence of two ethyl groups connected by an oxygen atom.

Diethyl ether is notably flammable, making it crucial to handle it carefully, especially around open flames or sparks. Its flammability stems from its ease of vaporization and low ignition temperature, meaning it can catch fire easily when exposed to an ignition source.

In terms of chemical reactions, diethyl ether is often involved in combustion reactions, where it burns in the presence of oxygen to produce carbon dioxide and water, releasing significant amounts of energy. This trait is characteristic of many hydrocarbons and explains its widespread investigation in energy-related and reaction stoichiometric studies.