Question

The heating value of combustible fuels is evaluated based on the quantities known as the higher heating value (HHV) and the lower heating value (LHV). The HHV has a higher absolute value and assumes that the water produced in the combustion reaction is formed in the liquid state. The LHV has a lower absolute value and assumes that the water produced in the combustion reaction is formed in the gaseous state. The LHV is therefore the sum of the HHV (which is negative) and the heat of vaporization of water for the number of moles of water formed in the reaction (which is positive). The table lists the enthalpy of combustion which is equivalent to the HHV-for several closely related hydrocarbons. $$\begin{array}{lc} \text { Hydrocarbon } & \Delta H_{\text {comb }}(\mathrm{kJ} / \mathrm{mol}) \\\ \mathrm{CH}_{4}(\mathrm{~g}) & -890 \\ \hline \mathrm{C}_{2} \mathrm{H}_{6}(\mathrm{~g}) & -1560 \\ \hline \mathrm{C}_{3} \mathrm{H}_{8}(\mathrm{~g}) & -2219 \\ \hline \mathrm{C}_{4} \mathrm{H}_{10}(\mathrm{~g}) & -2877 \\ \hline \mathrm{C}_{5} \mathrm{H}_{12}(I) & -3509 \\ \hline \mathrm{C}_{6} \mathrm{H}_{14}(I) & -4163 \\ \hline \mathrm{C}_{7} \mathrm{H}_{16}(I) & -4817 \\ \hline \mathrm{C}_{8} \mathrm{H}_{18}(I) & -5470 \\ \hline\end{array}$$ Use the information in the table to answer the following questions. a. Write two balanced equations for the combustion of \(\mathrm{C}_{3} \mathrm{H}_{8}\) one assuming the formation of liquid water and the other assuming the formation of gaseous water. b. Given that the heat of vaporization of water is \(44.0 \mathrm{~kJ} / \mathrm{mol}\), what is \(\Delta H_{\mathrm{rxn}}\) for each reaction in part a? Which quantity is the HHV? The LHV? c. When propane is used to cook in an outdoor grill, is the amount of heat released the HHV or the LHV? What amount of heat is released upon combustion of \(1.00 \mathrm{~kg}\) of propane in an outdoor grill? d. For each \(\mathrm{CH}_{2}\) unit added to a hydrocarbon, what is the average increase in the absolute value of \(\Delta H_{\mathrm{comb}} ?\)

Short answer

-2219 kJ/mol for HHV and -2039 kJ/mol for LHV; through outdoor cooking, LHV applies; -46330 kJ released from 1.00 kg of propane; average increase of about 654 kJ/mol per \( \text{CH}_2 \) unit added.

Step-by-step solution

- Write the Balanced Equation for Combustion Producing Liquid Water

For the combustion of propane (\text{C}_3\text{H}_8), write the balanced chemical equation assuming the formation of liquid water (H_2O(l)). The balanced equation is: \[ \text{C}_3\text{H}_8 + 5\text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O(l)} \]

- Write the Balanced Equation for Combustion Producing Gaseous Water

Now write the balanced equation for the same combustion reaction but assuming the formation of gaseous water (H_2O(g)). The balanced equation is: \[ \text{C}_3\text{H}_8 + 5\text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O(g)} \]

- Calculate \( \Delta H_{\mathrm{rxn}} \) for Combustion Producing Liquid Water (HHV)

To calculate the \( \Delta H_{\mathrm{rxn}} \) when liquid water is formed, use the given enthalpy of combustion for propane which is equivalent to the HHV (since it assumes liquid water is produced). So, \( \Delta H_{\mathrm{rxn}} = -2219 \text{ kJ/mol} \) for the formation of liquid water.

- Calculate \( \Delta H_{\mathrm{rxn}} \) for Combustion Producing Gaseous Water (LHV)

To find LHV assuming the gaseous state of water, add the heat of vaporization for the water produced to the HHV. For propane combustion, 4 moles of water are produced. So, the \( \Delta H_{\mathrm{rxn}} \) (LHV) is: \[ -2219 \text{ kJ/mol} + (4 \text{ moles} \times 44.0 \text{ kJ/mol}) = -2039 \text{ kJ/mol} \]

- Identify the HHV and LHV

The HHV is the enthalpy change when water is in liquid form which is -2219 kJ/mol and LHV is -2039 kJ/mol when water is assumed to be gaseous.

- Discuss Heat Released During Outdoor Cooking

When propane is used in an outdoor grill, the heat released is the LHV because the water produced is likely to be gaseous given the high temperatures. To find the heat released, convert 1.00 kg of propane to moles and multiply by the LHV per mole. The molar mass of propane is 44.097 g/mol. Hence, \[ 1.00 \times 10^3 \text{ g} / 44.097 \text{ g/mol} \times -2039 \text{ kJ/mol} \] gives the heat released upon combustion.

- Calculate Heat Released Upon Combustion of 1.00 kg of Propane

Using the conversion from the last step, the calculation is: \[ (1.00 \times 10^3 \text{ g}) / (44.097 \text{ g/mol}) \times (-2039 \text{ kJ/mol}) = -46330 \text{ kJ} \]

- Determine the Average Increase in \( \Delta H_{\mathrm{comb}} \) Per \( \text{CH}_2 \) Unit

Compute the differences in the \( \Delta H_{\mathrm{comb}} \) for successive hydrocarbons listed and divide by the number of \( \text{CH}_2 \) units added to find the average increase. For example, between \( \text{CH}_4 \) and \( \text{C}_2\text{H}_6 \), a \( \text{CH}_2 \) unit is added and \( \Delta H_{\mathrm{comb}} \) increases by 670 kJ/mol (from -890 to -1560 kJ/mol). Do this for all successive pairs and find the average to get the average increase per \( \text{CH}_2 \) unit.

- Calculate the Average Increase

Adding the differences and dividing by the number of instances gives the average increase. For \( \text{CH}_4 \) to \( \text{C}_8\text{H}_{18} \), there are 7 \( \text{CH}_2 \) additions and the total increase in \( \Delta H_{\mathrm{comb}} \) is 4580 kJ/mol. The average increase is thus \( 4580 \text{ kJ/mol} / 7 \approx 654 \text{ kJ per CH}_2 \text{ unit} \).

Key concepts

Understanding Higher Heating Value (HHV)

The concept of Higher Heating Value (HHV) is crucial when determining the energy potential of fuels. It's the total amount of heat energy that can be released when a fuel is completely combusted and the products have returned to a reference temperature, typically taken as 25°C. Crucially, the HHV assumes that all vapor produced during combustion condenses to liquid, which means the latent heat of vaporization of water is recovered and included in the measurement.

When performing calculations or laboratory measurements of HHV, it's essential to capture the heat content that would be lost as steam in practical situations. Hence, HHV is always greater than or equivalent to the Lower Heating Value (LHV) because it includes this additional heat.

Lower Heating Value (LHV) Explained

Conversely to HHV, the Lower Heating Value (LHV) accounts for the potential energy of a fuel disregarding the heat of condensation of the water vapor. In real-world combustions, such as in engines or furnaces, the water produced during combustion typically remains as vapor and is expelled with the exhaust gases. As a result, the energy associated with the phase change from vapor to liquid is not recovered and utilized.

For practical applications where steam is not condensed, LHV is the more relevant metric. It's important for students to grasp this distinction to correctly calculate the actual usable energy from a fuel. LHV is derived from the HHV by subtracting the heat of vaporization of the water generated during the fuel's combustion.

The Role of Enthalpy of Combustion

The enthalpy of combustion, or \( \Delta H_{\text{comb}} \), is a term that actually corresponds to the HHV of a fuel, representing the heat change when one mole of substance combusts under standard conditions. It considers all reactants and products in their stable states at 25°C, and it typically has a negative value because combustion is an exothermic reaction—energy is released.

In the context of our exercise, the enthalpy values listed for various hydrocarbons are negative, indicating that energy is liberated during combustion. For students to calculate the HHV accurately, they must understand how to interpret these enthalpy values in the context of balanced chemical equations and their stoichiometry.

Heat of Vaporization and its Impact on Fuel Values

Heat of vaporization is the energy required to convert a liquid into a vapor without a temperature change. This physical property plays a pivotal role when distinguishing between HHV and LHV, as it represents the energy not utilized when water vapor is emitted during combustion.

For instance, in the exercise, calculating the LHV for propane involves the heat of vaporization for the water produced. Subtracting the energy represented by the heat of vaporization from the HHV gives the LHV, aligning the theoretical values with the practical heat that can be utilized in systems where the water remains in the vapor phase. This reinforces the significance of considering the heat of vaporization when evaluating the effective energy output from a fuel.