Question

Three hydrocarbons that contain four carbons are listed here, along with their standard enthalpies of formation: $$ \begin{array}{llc} \hline \text { Hydrocarbon } & \text { Formula } & \Delta H_{f}^{0}(\mathrm{~kJ} / \mathrm{mol}) \\ \hline \text { Butane } & \mathrm{C}_{4} \mathrm{H}_{10}(g) & -125 \\ \text { 1-Butene } & \mathrm{C}_{4} \mathrm{H}_{8}(g) & -1 \\ \text { 1-Butyne } & \mathrm{C}_{4} \mathrm{H}_{6}(g) & 165 \\ \hline \end{array} $$ (a) For each of these substances, calculate the molar enthalpy of combustion to \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(l)\) (b) Calculate the fuel value, in \(\mathrm{kJ} / \mathrm{g}\), for each of these compounds. (c) For each hydrocarbon, determine the percentage of hydrogen by mass. (d) By comparing your answers for parts (b) and (c), propose a relationship between hydrogen content and fuel value in hydrocarbons.

Short answer

(a) Molar enthalpy of combustion for the three hydrocarbons: Butane: \(\Delta H_{comb}^{0} = -2877.5~kJ/mol\) 1-Butene: \(\Delta H_{comb}^{0} = -2713.6~kJ/mol\) 1-Butyne: \(\Delta H_{comb}^{0} = -2538.9~kJ/mol\) (b) Fuel values: Butane: \(49.5~kJ/g\) 1-Butene: \(48.4~kJ/g\) 1-Butyne: \(46.9~kJ/g\) (c) Percentage of hydrogen by mass: Butane: \(17.4\%\) 1-Butene: \(14.4\%\) 1-Butyne: \(11.2\%\) (d) Higher hydrogen content by mass corresponds to a higher fuel value per unit mass in hydrocarbons.

Step-by-step solution

Combustion reaction equations

Write the balanced combustion reaction equations for each hydrocarbon. Butane: \(C_4H_{10}(g) + 13/2~O_2(g) \to 4~CO_2(g) + 5~H_2O(l)\) 1-Butene: \(C_4H_{8}(g) + 6~O_2(g) \to 4~CO_2(g) + 4~H_2O(l)\) 1-Butyne: \(C_4H_{6}(g) + 11/2~O_2(g) \to 4~CO_2(g) + 3~H_2O(l)\)

Calculate the enthalpy change for the combustion reactions

Use Hess's Law, which states that the overall enthalpy change of a reaction is the sum of the enthalpy changes of the individual steps. For each reaction, consider the formation of carbon dioxide, liquid water, and the initial hydrocarbon compound. Butane: \(\Delta H_{comb}^{0} = 4 \Delta H_{f}(CO_2) + 5 \Delta H_{f}(H_2O) - \Delta H_{f}(C_4H_{10})\) 1-Butene: \(\Delta H_{comb}^{0} = 4 \Delta H_{f}(CO_2) + 4 \Delta H_{f}(H_2O) - \Delta H_{f}(C_4H_{8})\) 1-Butyne: \(\Delta H_{comb}^{0} = 4 \Delta H_{f}(CO_2) + 3 \Delta H_{f}(H_2O) - \Delta H_{f}(C_4H_{6})\) Use the provided standard enthalpies of formation and known values for \(CO_2\) and \(H_2O\) (-393.5 kJ/mol for \(CO_2\) and -285.8 kJ/mol for \(H_2O\)) to calculate the enthalpy of combustion for each hydrocarbon. (b) Calculate the fuel value for each compound in kJ/g.

Calculate the molar mass for each hydrocarbon

To find the fuel value per gram, we need to divide the molar enthalpy of combustion by the molar mass of the hydrocarbon. Butane: \(M = 4 (12.01) + 10 (1.01) \approx 58.12~g/mol\) 1-Butene: \(M = 4 (12.01) + 8 (1.01) \approx 56.10~g/mol\) 1-Butyne: \(M = 4 (12.01) + 6 (1.01) \approx 54.08~g/mol\)

Calculate the fuel value

Divide the molar enthalpy of combustion (from step 2) by the molar mass (from step 3) to obtain the fuel value in kJ/g. (c) Determine the percentage of hydrogen by mass for each hydrocarbon.

Calculate the mass of hydrogen in each hydrocarbon

For each hydrocarbon, find the mass of hydrogen atoms and divide by the molar mass, multiplying by 100 to express it as a percentage. Butane: \(\frac{10 (1.01)}{58.12} × 100\%\) 1-Butene: \(\frac{8 (1.01)}{56.10} × 100\%\) 1-Butyne: \(\frac{6 (1.01)}{54.08} × 100\%\) (d) Propose a relationship between hydrogen content and fuel value in hydrocarbons.

Compare part (b) and part (c)

Compare the results from step 4 and step 5. Observe any trends or relationships between hydrogen content and fuel value for the hydrocarbons under examination.

Key concepts

Enthalpy of Formation

When studying the enthalpy of formation, we are focusing on the amount of energy change associated with forming a compound from its elements under standard conditions. This value is crucial because it helps in calculating how much energy is absorbed or released during chemical reactions, like combusting hydrocarbons.
The enthalpy of formation is denoted by \( \Delta H_f^0 \) and is measured in kilojoules per mole (kJ/mol). For the hydrocarbons butane, 1-butene, and 1-butyne, their respective enthalpies of formation provide a starting point to analyze their combustion process. Negative values, like for butane (-125 kJ/mol), indicate temperature release when the compound forms, while positive values, such as for 1-butyne (165 kJ/mol), signify energy absorption. These values provide insights into the relative stability of each compound.

• A negative \( \Delta H_f^0 \) suggests the compound is thermodynamically stable.
• A positive \( \Delta H_f^0 \) reflects less stability, requiring energy input to form the compound.

Molar Enthalpy of Combustion

The molar enthalpy of combustion measures the heat energy released when one mole of a substance combusts completely in oxygen. This concept is fundamental to understanding the energy potential of different fuels.
By using Hess's Law, which allows the calculation of reaction enthalpies by summing known values, the combustion reactions for our hydrocarbons can be evaluated. In such reactions, each molecule breaks down into carbon dioxide and water, releasing energy in the process. For instance:

• Butane: releases significant energy when it combusts due to more hydrogen atoms that can react with oxygen.
• 1-Butene and 1-Butyne also release energy, but their values differ because of different molecular structures.

The calculations involve summing the products' enthalpies minus the reactants' enthalpies, transforming potential energy into measureable work.

Fuel Value Calculation

Calculating the fuel value of a hydrocarbon enables us to assess its energy efficiency by determining how much energy each gram can deliver. This process involves dividing the molar enthalpy of combustion by the compound's molar mass, converting the result into a practical unit: kilojoules per gram (kJ/g).
To break it down:

• Find the molar mass of the hydrocarbon, which is the sum of the atomic masses of all atoms in the molecule.
• Divide the molar enthalpy of combustion by this molar mass to get the fuel value.

This calculation can influence the selection of hydrocarbons as fuels in different applications, allowing comparisons based on energy output per gram. A higher fuel value indicates more energy can be harnessed from each gram of fuel, leading to preferences for more efficient fuels.

Hydrogen Percentage by Mass

The hydrogen percentage by mass in a hydrocarbon refers to the proportion of the molecule's mass made up by hydrogen atoms. It helps understand the fuel’s efficiency, control emissions, and predict combustion characteristics.
For each hydrocarbon:

• Calculate the total mass of hydrogen present.
• Divide by the overall molar mass and multiply by 100 to express it as a percentage.

Higher hydrogen content in hydrocarbons generally indicates higher theoretical energy output per unit mass when burned, as hydrogen combustion is highly exothermic. For our hydrocarbons, comparing this percentage with the fuel value reveals trends, often showing hydrocarbons with more hydrogen offer better fuel efficiency.
Comparisons also aid in assessing the environmental impact, as more hydrogen releases water vapor as opposed to carbon-rich emissions, impacting environmental and economic considerations.