Problem 61 - Chemistry The Central Science In Si Units

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 \(\Delta H^{\circ}\).

Step-by-Step Solution

Answer

The balanced equation is \(\mathrm{C}_{31} \mathrm{H}_{64}(s) + 47\mathrm{O}_2(g) \rightarrow 31\mathrm{CO}_2(g) + 32\mathrm{H}_2 \mathrm{O}(g)\). \(\Delta G^{\circ}\) is less negative than \(\Delta H^{\circ}\) because of the increase in entropy.

Step 1: Write the Unbalanced Equation

To begin, write out the unbalanced chemical equation for the combustion of \(\mathrm{C}_{31} \mathrm{H}_{64}(s)\). This process involves adding oxygen \(\mathrm{O}_2(g)\), and producing carbon dioxide \(\mathrm{CO}_2(g)\) and water \(\mathrm{H}_2 \mathrm{O}(g)\):\[\mathrm{C}_{31} \mathrm{H}_{64}(s) + \mathrm{O}_2(g) \rightarrow \mathrm{CO}_2(g) + \mathrm{H}_2 \mathrm{O}(g)\]

Step 2: Balance the Carbon Atoms

Balance the carbon atoms by ensuring that there are 31 carbon atoms on both sides of the equation. Thus, the equation becomes:\[\mathrm{C}_{31} \mathrm{H}_{64}(s) + \mathrm{O}_2(g) \rightarrow 31\mathrm{CO}_2(g) + \mathrm{H}_2 \mathrm{O}(g)\]

Step 3: Balance the Hydrogen Atoms

Balance the hydrogen atoms by making sure there are 64 hydrogen atoms on each side. Since water has two hydrogen atoms per molecule, you need 32 water molecules:\[\mathrm{C}_{31} \mathrm{H}_{64}(s) + \mathrm{O}_2(g) \rightarrow 31\mathrm{CO}_2(g) + 32\mathrm{H}_2 \mathrm{O}(g)\]

Step 4: Balance the Oxygen Atoms

Now, balance the oxygen atoms. There are a total of \(31 \times 2 = 62\) oxygen atoms in the carbon dioxide and \(32 \times 1 = 32\) in the water, giving a total of 94 oxygen atoms needed. Thus,\(47\) molecules of \(\mathrm{O}_2\) are required:\[\mathrm{C}_{31} \mathrm{H}_{64}(s) + 47\mathrm{O}_2(g) \rightarrow 31\mathrm{CO}_2(g) + 32\mathrm{H}_2 \mathrm{O}(g)\]

Step 5: Compare \(\Delta G^{\circ}\) and \(\Delta H^{\circ}\)

In an exothermic reaction such as combustion, \(\Delta H^{\circ}\) is negative, indicating heat is released. Without specific data, typically \(\Delta G^{\circ}\) also tends to be negative but not as much so as \(\Delta H^{\circ}\), because entropy increases significantly (more moles of gas produced), reducing the negativity of \(\Delta G^{\circ}\) due to the \(T\Delta S\) term in \(\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}\).

Key Concepts

Combustion

Combustion is a type of chemical reaction where a substance reacts rapidly with oxygen (usually from the air) and releases energy in the form of heat and light. This is why we often associate it with burning. Most combustion reactions involve hydrocarbons, compounds made up of hydrogen and carbon atoms. When hydrocarbons combust, the carbon and hydrogen atoms combine with oxygen to form carbon dioxide (4CO_24) and water (4H_2O4).

For instance, when you're burning a candle made from paraffin wax, which contains hydrocarbons like 4C_{31}H_{64}4, the combustion process occurs as:

• The wax melts and evaporates.
• Its vapors are ignited, combining with the oxygen in the air.
• This chemical reaction releases heat and light.

Reactions like these are critically important because they transform energy stored in chemical bonds into heat and light, which can be used to warm our homes or illuminate spaces.

Thermochemistry

Thermochemistry is the branch of chemistry that deals with the energy changes that accompany chemical reactions, especially heat energy. Every chemical reaction is either exothermic or endothermic. In an exothermic reaction, such as combustion, energy is released in the form of heat, resulting in a negative enthalpy change (4ΔH^{0}4).

When predicting if 4ΔG^{0}4, the change in free energy, will be more or less negative than 4ΔH^{0}4 in a reaction, it's vital to consider the entropy change (4ΔS^{0}4), which accounts for the disorder or randomness. The formula is:
\[ΔG^{0} = ΔH^{0} - TΔS^{0}\]
Where:

• 4T4 is the temperature in Kelvin.
• 4ΔS^{0}4 reflects the change in disorder.

During combustion, more gas molecules (4CO_24 and 4H_2O4) are produced than are consumed, causing 4ΔS^{0}4 to be positive. Thus, although both 4ΔH^{0}4 and 4ΔG^{0}4 are usually negative (indicating spontaneous and exothermic reactions), 4ΔG^{0}4 tends not to be as negative as 4ΔH^{0}4.

Chemical Equation Balancing

Balancing chemical equations ensures that the same number of each type of atom is present on both sides of the equation, following the law of conservation of mass. In the combustion of 4C_{31}H_{64}4,

• First, you ensure all carbon atoms in 4C_{31}H_{64}4 become part of 4CO_24, giving 431 4 4CO_24 molecules.
• Next, the hydrogen atoms convert to 4H_2O4, requiring 432 4 4H_2O4 molecules per the balance with 464 4 hydrogens.
• Finally, balance oxygen, combining the total needed for 431 4 4CO_24 and 432 4 4H_2O4, to obtain 447 4 4O_24 molecules.

It's like a chemical puzzle, ensuring that every piece is perfectly placed to reflect the true transformation occurring in reactions. This systematic approach helps in visualizing where atoms rearrange during a reaction, critical for both theoretical calculations and practical lab work.