Question

The equation for the burning of naphthalene is \(\mathrm{C}_{10} \mathrm{H}_{8}(\mathrm{~s})+12 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 1 \mathrm{OCO}_{2}+4 \mathrm{H}_{2} \mathrm{O}(\ell) .\) For every mole of \(\mathrm{C}_{10} \mathrm{H}_{8}\) burned, \(-1226.7 \mathrm{Kcal}\) is evolved at \(25^{\circ}\) in a fixedvolume combustion chamber. \(\Delta \mathrm{H}^{\circ}\) for \(\mathrm{H}_{2} \mathrm{O}(\ell)=-64.4\) \(\mathrm{Kcal} / \mathrm{mole}\) and \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})=-57.8 \mathrm{Kcal} /\) mole. Calculate (a) the heat of reaction at constant temperature and (b) \(\Delta \mathrm{H}\) for the case where all of the \(\mathrm{H}_{2} \mathrm{O}\) is gaseous.

Short answer

The heat of reaction at constant temperature is \(Q = -1226.7 \frac{\text{Kcal}}{\text{mole}}\). For the case where all of the H₂O is gaseous, the change in enthalpy is \(ΔH'= Q + ΔH_{change}\), where \(ΔH_{change}\) is calculated using the given enthalpy values for \(H_2O(l)\) and \(H_2O(g)\).

Step-by-step solution

Input given data

We are given the combustion equation of naphthalene: \[ C_{10}H_8(s) + 12O_2(g) \rightarrow 10CO_2(g) + 4H_2O(l) \] The heat evolved per mole of naphthalene burned is -1226.7 Kcal. Enthalpy of formation for \(H_2O(l)\) is: \[ \Delta H^o_{H_2O(l)} = -64.4 \frac{\text{Kcal}}{\text{mole}} \] Enthalpy of formation for \(H_2O(g)\) is: \[ \Delta H^o_{H_2O(g)} = -57.8 \frac{\text{Kcal}}{\text{mole}} \]

Calculate heat of reaction at constant temperature (Q)

Since the reaction occurs at constant temperature, the heat of reaction is equal to the change in internal energy, which is equivalent to the heat evolved per mole of naphthalene burnt: \[Q = -1226.7 \frac{\text{Kcal}}{\text{mole}}\] This is the answer for part (a).

Calculate ΔH for the case all H2O is gaseous (ΔH')

If all the H2O produced is in the gaseous state, the enthalpy change of the reaction can be calculated as \(ΔH' = Q + ΔH_{H_2O(g)} - ΔH_{H_2O(l)}\). First, we calculate the change in enthalpy due to the change in state by multiplying the difference in enthalpy of formation of liquid and gaseous water with the number of moles of water produced (4 moles) in the equation: \[ΔH_{change} = 4 \times (-57.8 - -64.4) \frac{\text{Kcal}}{\text{mole}}\] Now, we can find ΔH': \[ΔH' = Q + ΔH_{change}\] After evaluating the values above, we can get the final answer for part (b).

Key concepts

Combustion Equation

A combustion equation is a representation of a chemical reaction where a substance combines with oxygen and releases energy. This energy is often in the form of heat and light. In simpler terms, it's what happens when something burns. For naphthalene, a hydrocarbon with the chemical formula \(C_{10}H_8\), the combustion equation can be written as:

• \(C_{10}H_8(s) + 12O_2(g) \rightarrow 10CO_2(g) + 4H_2O(l)\)

This equation shows that when naphthalene combusts, it reacts with twelve moles of oxygen \(O_2\) to produce ten moles of carbon dioxide \(CO_2\) and four moles of water \(H_2O\) in its liquid form. This reaction releases a large amount of heat, which is why it's important in calculating energy changes such as the enthalpy of reaction. Combustion equations must be balanced, meaning they have the same number of each type of atom on both sides of the equation. This represents the Law of Conservation of Mass.

Naphthalene

Naphthalene is a white, crystalline solid that is derived from coal tar or petroleum. It's most commonly known for its use in mothballs. From a chemical standpoint, naphthalene is a type of aromatic hydrocarbon, which means its molecules follow a certain stability and structure associated with the benzene ring.
Naphthalene's chemical formula is \(C_{10}H_8\). This means it comprises 10 carbon atoms and 8 hydrogen atoms arranged in a double ring. When naphthalene undergoes combustion, these atoms react with oxygen to form carbon dioxide and water, a typical result for hydrocarbon combustion reactions.

• Due to its high carbon content, the combustion of naphthalene releases significant energy, which is used to calculate its enthalpy of reaction.

Enthalpy of Formation

Enthalpy of formation refers to the heat change that occurs when one mole of a compound is formed from its elements in their standard states. Understanding enthalpy of formation is crucial when calculating the total heat change, or enthalpy change, for a reaction.
In the case of the combustion of naphthalene, we look at the products' enthalpy: carbon dioxide and water. Specifically, water can exist either as a liquid or gas, which changes the enthalpy value.
- **Liquid water** has an enthalpy of formation \(\Delta H^o_{H_2O(l)} = -64.4 \, \frac{\text{Kcal}}{\text{mole}}\)- **Gaseous water** is slightly less stable with \(\Delta H^o_{H_2O(g)} = -57.8 \, \frac{\text{Kcal}}{\text{mole}}\)This illustrates that converting water from a liquid to gaseous form requires additional energy, affecting the overall energy balance of the reaction.

Heat of Reaction

The heat of reaction, also known as the enthalpy change \(\Delta H\), represents the total heat absorbed or released during a chemical reaction. For exothermic reactions like combustion, where heat is released, \(\Delta H\) is typically negative.
For naphthalene combustion, the heat of reaction is given as \(-1226.7\) Kcal per mole of naphthalene burned under constant volume conditions. This indicates a significant release of energy.
However, if all the water formed is in a gaseous state, the enthalpy change must be recalculated to account for the energy needed to turn liquid water into steam:

• The change in enthalpy \(\Delta H'\) is calculated by adjusting for the difference in formation enthalpies of liquid and gaseous water.
• The formula used is \(\Delta H' = Q + \Delta H_{change}\), where \(Q\) is the heat of reaction already known, and \(\Delta H_{change}\) accounts for the state transition of water.

Understanding this adjustment is vital for accurately determining the heat flow in reactions where the physical state of the products may change.