Problem 93 - Chemistry The Central Science In Si Units

Question

Hydrogen gas has a higher fuel value than natural gas on a mass basis but not on a volume basis. Thus, hydrogen is not competitive with natural gas as a fuel transported long distances through pipelines. Calculate the heats of combustion of \(\mathrm{H}_{2}\) and \(\mathrm{CH}_{4}\) (the principal component of natural gas) (a) per mole of each, (b) per gram of each, \((\mathbf{c})\) per cubic meter of each at STP. Assume \(\mathrm{H}_{2} \mathrm{O}(l)\) as a product.

Step-by-Step Solution

Answer

Hydrogen releases more energy per gram but less per cubic meter compared to methane.

Step 1: Understanding Heats of Combustion

Heats of combustion are the energy released when one mole of a substance is burned completely in oxygen. For this exercise, we'll compare hydrogen (\(\text{H}_2\)) and methane (\(\text{CH}_4\)), the main component of natural gas.

Step 2: Heats of Combustion per Mole

For hydrogen:- The combustion reaction is: \[\text{2H}_2(g) + \text{O}_2(g) \rightarrow \text{2H}_2\text{O}(l)\]- The standard heat of combustion for \(\text{H}_2\) is approximately \(688 \text{ kJ/mol}\) (for 2 moles of \(\text{H}_2\), so per mole it is \(344 \text{ kJ/mol}\)).For methane:- The combustion reaction is: \[\text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(l)\]- The standard heat of combustion for \(\text{CH}_4\) is approximately \(802 \text{ kJ/mol}\).

Step 3: Heats of Combustion per Gram

For hydrogen:- Molar mass of \(\text{H}_2\) is \(2 \text{ g/mol}\).- Heat per gram: \(344 \text{ kJ/mol} \div 2 \text{ g/mol} = 172 \text{ kJ/g}\).For methane:- Molar mass of \(\text{CH}_4\) is \(16 \text{ g/mol}\).- Heat per gram: \(802 \text{ kJ/mol} \div 16 \text{ g/mol} = 50.13 \text{ kJ/g}\).

Step 4: Heats of Combustion per Cubic Meter at STP

At STP, one mole of any gas occupies approximately \(22.4 \text{ L}\) or \(0.0224 \text{ m}^3\).For hydrogen:- Heat per \(\text{m}^3\): \(344 \text{ kJ/mol} \div 0.0224 \text{ m}^3/mol \approx 15357 \text{ kJ/m}^3\).For methane:- Heat per \(\text{m}^3\): \(802 \text{ kJ/mol} \div 0.0224 \text{ m}^3/mol \approx 35804 \text{ kJ/m}^3\).

Key Concepts

Hydrogen Combustion

When we talk about hydrogen combustion, we are referring to the reaction where hydrogen gas (\(\text{H}_2\)) burns in the presence of oxygen (\(\text{O}_2\)). This process releases energy and results in water as the product. The reaction is given by: \[\text{2H}_2(g) + \text{O}_2(g) \rightarrow \text{2H}_2\text{O}(l)\]This simple reaction efficiently releases energy, generating approximately 344 kilojoules for each mole of hydrogen burned.
Hydrogen is known for its high energy yield per mass, releasing 172 kJ for every gram consumed. However, its low density means that it occupies more volume compared to other fuels, which affects its energy efficiency on a volumetric basis.

• High energy per mass: 172 kJ/g
• Reaction byproduct: Liquid water
• Key advantage: Clean energy with water as the only emission

Hydrogen's sustainability factor centers on its clean byproducts, making it a desirable contender for future fuel needs despite challenges in transport and storage.

Methane Combustion

Methane combustion illustrates how methane (\(\text{CH}_4\)), the primary component of natural gas, is utilized for energy. The combustion of methane releases carbon dioxide (\(\text{CO}_2\)) and water (\(\text{H}_2\text{O}\)) along with a significant yield of energy:\[\text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(l)\]This reaction produces approximately 802 kJ of energy per mol of methane, making it one of the most efficient fuels by mass and volume.
Despite its relatively heavier molecular weight compared to hydrogen, methane still holds an energy yield of 50.13 kJ per gram. Its effectiveness is largely due to its density, allowing it to deliver 35804 kJ per cubic meter at standard conditions.

• Heat of combustion: 50.13 kJ/g
• Byproducts: Carbon dioxide and water
• Volume efficiency: 35804 kJ/\(\text{m}^3\)

While methane's carbon emissions are of environmental concern, it remains a prevalent energy source due to its availability and comparative volumetric energy efficiency.

Standard Temperature and Pressure

In the context of combustion reactions, especially when dealing with gases, standard temperature and pressure (STP) provide a baseline for comparing energies and volumes. STP is defined as a temperature of 0°C (273.15 K) and a pressure of 1 atmosphere (101.3 kPa). At these conditions, one mole of an ideal gas occupies 22.4 liters or 0.0224 cubic meters.
Understanding the concept of STP is crucial when calculating the heats of combustion per volume. This standardization ensures accuracy and allows for meaningful comparisons of energy outputs amongst different gases.

• Standard Temperature: 0°C or 273.15 K
• Standard Pressure: 1 atm or 101.3 kPa
• Gas molar volume at STP: 22.4 L/mol or 0.0224 \(\text{m}^3\)/mol

By utilizing STP, we ascertain how much heat energy can be obtained from a specific volume of gas, be it hydrogen or methane, which is essential for evaluating their practical uses in energy calculations and applications.