Question

The fuel in high-efficiency natural gas vehicles consists primarily of methane \(\left(\mathrm{CH}_{4}\right) .\) (a) How much heat is produced in burning 1 mol of \(\mathrm{CH}_{4}(g)\) under standard conditions if reactants and products are brought to \(298 \mathrm{~K}\) and \(\mathrm{H}_{2} \mathrm{O}(l)\) is formed? (b) What is the maximum amount of useful work that can be accomplished under standard conditions by this system?

Short answer

The heat produced in burning 1 mol of CH4(g) is 965.1 kJ/mol, and the maximum amount of useful work that can be accomplished under standard conditions by this system is 884.3 kJ/mol.

Step-by-step solution

Determine the balanced chemical equation for the combustion of CH4

Using the formula for the combustion of a hydrocarbon, we can find the balanced chemical equation for the combustion of CH4: CH4 (g) + 2 O2 (g) -> CO2 (g) + 2 H2O (l)

Find the standard enthalpy change of the reaction

We can calculate the standard enthalpy change of the reaction (∆H°) as follows: ∆H° = [n * ∆H°f(products)] - [n * ∆H°f(reactants)] Standard enthalpy of formation (∆H°f) values for the compounds are as follows: ∆H°f(CH4 (g)) = -74.8 kJ/mol ∆H°f(O2 (g)) = 0 kJ/mol (since it's an element in its standard state) ∆H°f(CO2 (g)) = -393.5 kJ/mol ∆H°f(H2O (l)) = -285.8 kJ/mol Now, we can calculate ∆H° for the reaction: ∆H° = [1(-393.5) + 2(-285.8)] - [1(-74.8) + 2(0)] ∆H° = (-393.5 - 2 * 285.8) - (-74.8) ∆H° = -965.1 kJ/mol The heat produced in burning 1 mol of CH4(g) is 965.1 kJ/mol.

Calculate the standard Gibbs free energy change of the reaction

Now we need to find the standard Gibbs free energy change of the reaction (∆G°) to determine the maximum amount of useful work that can be accomplished under standard conditions. We can do this by using the following equation: ∆G° = ∆H° - T∆S° First, we need to find the standard entropy change (∆S°) for the reaction: ∆S° = [n * S°(products)] - [n * S°(reactants)] Standard molar entropy (S°) values for the compounds are as follows: S°(CH4 (g)) = 186.3 J/mol·K S°(O2 (g)) = 205.2 J/mol·K S°(CO2 (g)) = 213.8 J/mol·K S°(H2O (l)) = 69.9 J/mol·K Now, we can calculate ∆S° for the reaction: ∆S° = [1(213.8) + 2(69.9)] - [1(186.3) + 2(205.2)] ∆S° = (213.8 + 2 * 69.9) - (186.3 + 410.4) ∆S° = -270.8 J/mol·K Now we can calculate the ∆G° for the reaction at 298 K: ∆G° = ∆H° - T∆S° ∆G° = -965.1 kJ/mol - (298 K)(-270.8 J/mol·K) / 1000 (to convert J to kJ) ∆G° = -965.1 kJ/mol + 80.8 kJ/mol ∆G° = -884.3 kJ/mol The maximum amount of useful work that can be accomplished under standard conditions by this system is 884.3 kJ/mol.

Key concepts

Standard Enthalpy Change

Standard enthalpy change, represented as \( \Delta H^\circ \), is a fundamental concept in chemical thermodynamics. It refers to the heat absorbed or released during a chemical reaction when all reactants and products are in their standard states. Values are measured under standard conditions, typically at 298 K and 1 atm pressure.

For the combustion of methane, \( \Delta H^\circ \) is calculated using the enthalpy of formation values of the involved compounds. Methane, water, and carbon dioxide each have specific enthalpy values that need to be plugged into the equation:

• \( \Delta H^\circ(\text{CH}_4 (g)) = -74.8 \text{ kJ/mol} \)
• \( \Delta H^\circ(\text{O}_2 (g)) = 0 \text{ kJ/mol} \)
• \( \Delta H^\circ(\text{CO}_2 (g)) = -393.5 \text{ kJ/mol} \)
• \( \Delta H^\circ(\text{H}_2\text{O} (l)) = -285.8 \text{ kJ/mol} \)

Using these values, we determine the overall enthalpy change as \(-965.1 \text{ kJ/mol}\). This indicates a significant release of heat energy, revealing why methane is an efficient fuel.

Standard Gibbs Free Energy

The standard Gibbs free energy change \( \Delta G^\circ \) helps us understand a reaction's spontaneity and the maximum work it can perform. A negative \( \Delta G^\circ \) indicates a spontaneous reaction at constant temperature and pressure. To calculate \( \Delta G^\circ \), use the equation:

• \( \Delta G^\circ = \Delta H^\circ - T\Delta S^\circ \)

In the methane combustion example:

• \( \Delta H^\circ = -965.1 \text{ kJ/mol} \)
• \( T = 298 \text{ K} \)
• \( \Delta S^\circ = -270.8 \text{ J/mol}\cdot\text{K} = -0.2708 \text{ kJ/mol}\cdot\text{K} \)

After calculation, we find \( \Delta G^\circ = -884.3 \text{ kJ/mol} \), confirming the reaction is highly spontaneous and could produce a significant amount of work, crucial for natural gas vehicles.

Standard Entropy Change

Standard entropy change \( \Delta S^\circ \) provides insight into the disorder or randomness in a system during a reaction. Higher entropy typically means more disorder.

In chemical thermodynamics, reactions tend toward states that increase entropy. For the methane combustion reaction, \( \Delta S^\circ \) is negative, calculated using the standard molar entropies:

• \( S^\circ(\text{CH}_4 (g)) = 186.3 \text{ J/mol}\cdot\text{K} \)
• \( S^\circ(\text{O}_2 (g)) = 205.2 \text{ J/mol}\cdot\text{K} \)
• \( S^\circ(\text{CO}_2 (g)) = 213.8 \text{ J/mol}\cdot\text{K} \)
• \( S^\circ(\text{H}_2\text{O} (l)) = 69.9 \text{ J/mol}\cdot\text{K} \)

With a calculated \( \Delta S^\circ \) of \(-270.8 \text{ J/mol}\cdot\text{K} \), it suggests the system becomes less disordered. The gases form liquid water, a more ordered state.

Natural Gas Vehicles

Natural gas vehicles (NGVs) run on compressed natural gas (CNG), primarily composed of methane. Methane's efficient combustion makes it an ideal fuel source, releasing less greenhouse gas than gasoline or diesel.

The benefits of using NGVs include:

• Lower emissions contributing to cleaner air
• Cost efficiency due to cheaper fuel prices
• Widespread availability of natural gas resources
• Silent operation and smooth driving experience

By leveraging the combustion process of methane, NGVs epitomize the practical application of chemical thermodynamics in reducing environmental impact while enhancing fuel efficiency.

Chemical Thermodynamics

Chemical thermodynamics studies energy changes, particularly heat and work, in chemical reactions. It provides a structured way to understand how and why reactions occur, combining endothermic, exothermic, and energy releases.

Key areas include:

• Understanding reaction spontaneity using Gibbs free energy
• Predicting reaction heat flow through enthalpy
• Assessing system order with entropy changes
• Applying energy principles to real-world applications, like engines and power generation

Chemical thermodynamics not only explains basic scientific principles but also frames their practical applications, such as in NGVs where efficient energy conversion is crucial for design and function.