Problem 108 - Chemistry The Central Science

Question

Three common hydrocarbons that contain four carbons are listed here, along with their standard enthalpies of formation: (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. \((\mathrm{c})\) 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.

Step-by-Step Solution

Answer

The molar enthalpy of combustion for the three hydrocarbons are: Butane: \(-2880\,\mathrm{kJ/mol}\), Butene: \(-2823\,\mathrm{kJ/mol}\), and Butyne: \(-2776\,\mathrm{kJ/mol}\). Their fuel values are: Butane: \(-49.5\,\mathrm{kJ/g}\), Butene: \(-50.3\,\mathrm{kJ/g}\), and Butyne: \(-51.3\,\mathrm{kJ/g}\). The percentage of hydrogen by mass for each hydrocarbon is: Butane: \(17.4\%\), Butene: \(14.4\%\), and Butyne: \(11.2\%\). Comparing the fuel values and hydrogen content, we can propose an inverse relationship between hydrogen content and fuel value in hydrocarbons.

Step 1: Part (a): Molar Enthalpy of Combustion

To find the molar enthalpy of combustion for these hydrocarbons to \(\mathrm{CO_2}\) and \(\mathrm{H_2O}\), we will need the balanced chemical equations for their combustion. 1. Butane: \(\mathrm{C_4H_{10}(g) + \frac{13}{2}O_{2}(g) \rightarrow 4CO_{2}(g) + 5H_2O(l)}\) Using Hess's Law, the enthalpy change of the reaction is: \(\Delta H^\circ_\mathrm{comb} = 4\Delta H^\circ_\mathrm{f}[\mathrm{CO_2}] + 5\Delta H^\circ_\mathrm{f}[\mathrm{H_2O}] - \Delta H^\circ_\mathrm{f}[\mathrm{C_4H_{10}}]\) By plugging in the values for the standard enthalpies of formation, we get \(\Delta H^\circ_\mathrm{comb}= -2880\,\mathrm{kJ/mol}\) for Butane 2. Butene: \(\mathrm{C_4H_8(g) + 6O_2(g) \rightarrow 4CO_2(g) + 4H_2O(l)}\) Using Hess's Law, the enthalpy change of the reaction is: \(\Delta H^\circ_\mathrm{comb} = 4\Delta H^\circ_\mathrm{f}[\mathrm{CO_2}] + 4\Delta H^\circ_\mathrm{f}[\mathrm{H_2O}] - \Delta H^\circ_\mathrm{f}[\mathrm{C_4H_8}]\) By plugging in the values for the standard enthalpies of formation we get \(\Delta H^\circ_\mathrm{comb}=-2823\,\mathrm{kJ/mol}\) for Butene 3. Butyne: \(\mathrm{C_4H_6(g) + \frac{11}{2}O_2(g) \rightarrow 4CO_2(g) + 3H_2O(l)}\) Using Hess's Law, the enthalpy change of the reaction is: \(\Delta H^\circ_\mathrm{comb} = 4\Delta H^\circ_\mathrm{f}[\mathrm{CO_2}] + 3\Delta H^\circ_\mathrm{f}[\mathrm{H_2O}] - \Delta H^\circ_\mathrm{f}[\mathrm{C_4H_6}]\) By plugging in the values for the standard enthalpies of formation, we get \(\Delta H^\circ_\mathrm{comb}=-2776\,\mathrm{kJ/mol}\) for Butyne

Step 2: Part (b): Fuel Value in kJ/g

To find the fuel value in kJ/g, we need to divide the molar enthalpy of combustion (in kJ/mol) by the molar mass of the hydrocarbon (in g/mol). 1. Butane: Fuel value = \(\frac{-2880\,\mathrm{kJ/mol}}{58.12\,\mathrm{g/mol}} = -49.5\,\mathrm{kJ/g}\) 2. Butene: Fuel value = \(\frac{-2823\,\mathrm{kJ/mol}}{56.11\,\mathrm{g/mol}} = -50.3\,\mathrm{kJ/g}\) 3. Butyne: Fuel value = \(\frac{-2776\,\mathrm{kJ/mol}}{54.10\,\mathrm{g/mol}} = -51.3\,\mathrm{kJ/g}\)

Step 3: Part (c): Percentage of Hydrogen by Mass

To find the percentage of hydrogen by mass, we need to divide the mass of hydrogen in the molecule by the total mass of the molecule. 1. Butane: Percentage of hydrogen = \(\frac{10 \times 1.01\,\mathrm{g/mol}}{58.12\,\mathrm{g/mol}} \times 100 = 17.4\%\) 2. Butene: Percentage of hydrogen = \(\frac{8 \times 1.01\,\mathrm{g/mol}}{56.11\,\mathrm{g/mol}} \times 100 = 14.4\%\) 3. Butyne: Percentage of hydrogen = \(\frac{6 \times 1.01\,\mathrm{g/mol}}{54.10\,\mathrm{g/mol}} \times 100 = 11.2\%\)

Step 4: Part (d): Relationship Between Hydrogen Content and Fuel Value

Comparing the results of parts (b) and (c), we can see that as the percentage of hydrogen by mass in a hydrocarbon increases, the fuel value (kJ/g) decreases. Therefore, we can propose that there is an inverse relationship between hydrogen content and fuel value in hydrocarbons.

Key Concepts

Hess's Law

Hess's Law is a principle in chemistry that states that the total enthalpy change for a chemical reaction is the same, regardless of the pathway the reaction takes, as long as the initial and final conditions are identical. This is incredibly useful when trying to calculate difficult-to-measure enthalpy changes, as it allows us to use known enthalpy changes of other steps.
This law helps to solve complex problems by breaking them down into simpler steps.
When applying Hess's Law, you can perform calculations using enthalpy changes:

• Formulate the desired chemical reaction and identify known reactions that can be added to equal the desired reaction.
• Use the known enthalpy changes of these reactions to calculate the unknown enthalpy of your desired reaction.

In our example, we use Hess's Law to calculate the molar enthalpy of combustion of different hydrocarbons (butane, butene, butyne) into carbon dioxide and water. We take standard enthalpies of formation of products and reactants to make these calculations. This concept is crucial for energy considerations in chemistry, particularly in combustion reactions.

Standard Enthalpy of Formation

The Standard Enthalpy of Formation (\( \Delta H^ ext{°}_ ext{f} \)) is the change in enthalpy when one mole of a compound is formed from its elements in their standard states. This concept is foundational to understanding energy changes in chemical reactions. The standard state refers to the most stable form of a substance at a pressure of 1 atmosphere and a specified temperature, usually 25°C (298 K).
This value provides a reference point that allows chemists to determine reaction energetics and predict thermochemical behaviors of compounds.

• Elements in their standard state have a standard enthalpy of formation of zero, like\( \ce{O2(g), N2(g), H2(g)} \).
• Compounds formed from elements can have positive or negative\( \Delta H^ ext{°}_ ext{f} \) values, indicating endothermic (energy absorbed, positive) or exothermic (energy released, negative) processes.

During calculations in the exercise, we use the given standard enthalpies of formation to find the enthalpy changes when hydrocarbons are combusting to form CO2 and H2O. Remembering that these values are given for specific thermodynamic conditions makes our calculations precise and reproducible.

Hydrocarbons

Hydrocarbons are organic compounds composed entirely of hydrogen and carbon atoms. They are the primary building blocks of many chemical processes and are abundant in fuels, including natural gas, oil, and coal.
In the exercise, we explore three hydrocarbons - butane, butene, and butyne. Each has the formula\( \ce{C4H10} \),\( \ce{C4H8} \), and\( \ce{C4H6} \) respectively. These molecules represent different types of hydrocarbons based on the bonds between carbon atoms:

• Butane is an alkane with only single bonds, making it saturated.
• Butene is an alkene with at least one double bond, unsaturated.
• Butyne is an alkyne, containing one triple bond, also unsaturated.

Understanding the structure is important, as it affects a compound's chemical reactivity and physical properties.
When combusted, hydrocarbons release energy, and the different structures suggest varying efficiencies and energy outputs, which are calculated in terms of enthalpy of combustion. Hydrocarbons are crucial to our understanding of chemical energy and drive much of our exploration into alternative energy sources.