Question

Today, most candles are made of paraffin wax. A typical component of paraffin wax is the hydrocarbon \(\mathrm{C}_{31} \mathrm{H}_{64}\) which is solid at room temperature. (a) Write a balanced equation for the combustion of \(\mathrm{C}_{31} \mathrm{H}_{64}(s)\) to form \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(g) .\) ( \(\mathbf{b}\) ) Without using thermochemical data, predict whether \(\Delta G^{\circ}\) for this reaction is more negative or less negative than \(\Delta H^{\circ}\).

Short answer

The balanced chemical equation for the combustion of \(\mathrm{C}_{31}\mathrm{H}_{64}(s)\) is: \[ \mathrm{C}_{31}\mathrm{H}_{64}(s) + 47 O_2(g) \rightarrow 31 CO_2(g) + 32 H_2O(g) \] We predict that \(\Delta G^{\circ}\) for this reaction is more negative than \(\Delta H^{\circ}\) without using thermochemical data, based on the exothermic nature of combustion reactions and the increase in entropy as we go from a solid hydrocarbon and oxygen gas to a mixture of carbon dioxide and water vapor gases.

Step-by-step solution

Write the balanced chemical equation for the combustion of \(\mathrm{C}_{31}\mathrm{H}_{64}\)

To write a balanced chemical equation, first write down the basic form of the equation with the reactants on the left side and the products on the right side. Then, we will balance the equation by adjusting the number of each molecule to make sure the number of each atom is the same on both sides of the equation. The basic form of the equation is: \(\mathrm{C}_{31}\mathrm{H}_{64}(s) + O_2(g) \rightarrow CO_2(g) + H_2O(g)\) Now, let's balance the equation: There are 31 carbon atoms in \(\mathrm{C}_{31}\mathrm{H}_{64}\), so we need 31 molecules of \(\mathrm{CO}_2\) on the right side of the equation: \(\mathrm{C}_{31}\mathrm{H}_{64}(sockopt) + O_2(g) \rightarrow \textbf{31} CO_2(g) + H_2O(g)\) Now, there are 64 hydrogen atoms in \(\mathrm{C}_{31}\mathrm{H}_{64}\), so we need 32 molecules of \(H_2O\) on the right side of the equation: \(\mathrm{C}_{31}\mathrm{H}_{64}(s) + O_2(g) \rightarrow 31 CO_2(g) + \textbf{32} H_2O(g)\) Finally, we have 31 molecules of carbon dioxide (31x2=62 oxygen atoms) and 32 molecules of water vapor (32x1=32 oxygen atoms), which combine for a total of 94 oxygen atoms needed on the left side of the equation. Thus, we need 47 molecules of \(O_2\): \(\mathrm{C}_{31}\mathrm{H}_{64}(s) + \textbf{47} O_2(g) \rightarrow 31 CO_2(g) + 32 H_2O(g)\) Now, the balanced chemical equation is: \[ \mathrm{C}_{31}\mathrm{H}_{64}(s) + 47 O_2(g) \rightarrow 31 CO_2(g) + 32 H_2O(g) \]

Predict if \(\Delta G^{\circ}\) is more negative or less negative than \(\Delta H^{\circ}\) without using thermochemical data

First, let's recall the Gibbs free energy equation: \[ \Delta G^{\circ} = \Delta H^{\circ} - T \Delta S^{\circ} \] Since we can't use thermochemical data, we must make a prediction based on other factors. Combustion reactions are typically exothermic, which means that they release heat. As a result, the standard enthalpy change \(\Delta H^{\circ}\) will be negative. Additionally, since we are going from a relatively ordered hydrocarbon solid and oxygen gas to a more chaotic mixture of gases, carbon dioxide and water vapor, the entropy change \(\Delta S^{\circ}\) will be positive. Considering the negative sign in the Gibbs free energy equation, the product of positive \(\Delta S^{\circ}\) and a positive temperature (T) will also be positive, which means the whole term \(-T \Delta S^{\circ}\) will be negative. Since both terms of the right side of the Gibbs free energy equation are negative, \(\Delta G^{\circ}\) will be more negative than \(\Delta H^{\circ}\). So, without using thermochemical data, we predict that \(\Delta G^{\circ}\) for this reaction is more negative than \(\Delta H^{\circ}\).

Key concepts

Combustion Reaction

A combustion reaction is a chemical process where a substance combines with oxygen to produce heat and light. The most common example of this type of reaction is the burning of hydrocarbons, like in the case of candle wax combustion.
During a combustion reaction, hydrocarbons react with oxygen to form carbon dioxide and water, usually releasing significant amounts of energy in the form of heat.
This energy release is what makes combustion reactions exothermic. Many everyday processes rely on these reactions, including engines and heating systems.

In the case of the combustion of \( \mathrm{C}_{31}\mathrm{H}_{64}(s) \), we balance the equation by ensuring the same number of carbon, hydrogen, and oxygen atoms are present on each side.

• Carbon atoms from the hydrocarbon turn into carbon dioxide.
• Hydrogen atoms convert into water vapor.
• The oxygen atoms come from the atmospheric oxygen.

These principles underpin the balancing act needed for writing a coherent combustion reaction equation.

Gibbs Free Energy

Gibbs Free Energy ( \( G \)) is a thermodynamic quantity used to predict the feasibility of a chemical reaction at constant pressure and temperature. It combines enthalpy and entropy into a single value, helping to determine whether a process will occur spontaneously.
The formula is:
\[ \Delta G = \Delta H - T \Delta S \]

• \( \Delta H \) represents the change in enthalpy or heat content.
• \( T \) is the temperature in Kelvin.
• \( \Delta S \) is the change in entropy or the disorder in the system.

In combustion reactions, the enthalpy change is often negative because these reactions release heat. Additionally, given that reactants (often a solid and gas) are converting into gaseous products, entropy increases, making \( \Delta S \) positive.
Therefore, in combustion processes:
\( -T \Delta S \) contributes negatively to \( \Delta G \), enhancing the negative nature of Gibbs Free Energy and indicating a spontaneous reaction.

Balancing Chemical Equations

Balancing chemical equations is a critical skill in chemistry, helping scientists understand reaction dynamics and product formations. When balancing, the primary goal is to have the same number of each element's atoms on both sides of the reaction equation.

• Start by writing the unbalanced equation with reactants and products.
• Adjust coefficients to balance atoms, usually beginning with the largest molecule or a singular element.
• Check and re-adjust as necessary, making sure all atoms are balanced.

For example, in the wax combustion equation, \( \mathrm{C}_{31}\mathrm{H}_{64}(s) + 47 O_2(g) \rightarrow 31 CO_2(g) + 32 H_2O(g) \), ensure each element matches:

• 31 carbon atoms in \( CO_2 \) balance with 31 carbon atoms in the hydrocarbon.
• 64 hydrogen atoms in \( H_2O \) match those in the wax.
• Oxygen is tricky here, as it involves both water and carbon dioxide products.

By carefully following these steps, you create an accurate model of what's happening in a reaction.

Hydrocarbons

Hydrocarbons are organic compounds composed entirely of hydrogen and carbon. They form a significant part of our energy sources as fuels, given their high energy release during combustion.
There are two main types of hydrocarbons based on their structure:

• Saturated hydrocarbons (alkanes): All carbon atoms form single bonds, and they are fully "saturated" with hydrogen atoms.
• Unsaturated hydrocarbons (alkenes and alkynes): Contain one or more double or triple bonds, resulting in fewer hydrogen atoms.

In the context of this exercise, paraffin wax represents a complex hydrocarbon, with many carbon atoms making it a solid at room temperature.
During a combustion reaction, hydrocarbons react with oxygen, forming carbon dioxide and water while releasing heat and light. This very reaction is foundational in maintaining fires and operating internal combustion engines.
Understanding hydrocarbons' structure and behavior in reactions is critical for both academic and practical applications in energy and environmental sciences.