Thermochemistry Calculation Heat Of Reaction For 6 Moles Of A

In the realm of chemistry, thermochemical equations serve as vital tools for understanding the energy transformations that accompany chemical reactions. These equations not only depict the chemical species involved but also provide crucial information about the heat absorbed or released during the process, quantified as the enthalpy change (ΔH). In this article, we will delve into a specific thermochemical equation to calculate the heat associated with the reaction of 6 moles of reactant A. Let's embark on this journey to unravel the intricacies of thermochemistry.

Decoding the Thermochemical Equation

Our starting point is the provided thermochemical equation:

$$2A \rightarrow AA \quad \Delta H_{rxn} = -43.0 J$$

This equation reveals that 2 moles of reactant A undergo a transformation to form product AA, accompanied by an enthalpy change (ΔHrxn) of -43.0 Joules. The negative sign preceding the ΔHrxn value signifies that this reaction is exothermic, implying that heat is released into the surroundings during the reaction. Conversely, an endothermic reaction would exhibit a positive ΔHrxn value, indicating heat absorption from the surroundings.

Stoichiometry: The Key to Unlocking the Heat

To determine the heat associated with the reaction of 6 moles of A, we must employ the principles of stoichiometry. Stoichiometry is the branch of chemistry that deals with the quantitative relationships between reactants and products in chemical reactions. The coefficients in a balanced chemical equation represent the mole ratios of the substances involved.

In our given equation, the coefficient 2 in front of A signifies that 2 moles of A react. This stoichiometric relationship serves as a conversion factor to calculate the heat associated with the reaction of any number of moles of A. We can set up a proportion to solve for the heat associated with 6 moles of A:

$$\frac{-43.0 J}{2 \text{ moles A}} = \frac{x}{6 \text{ moles A}}$$

Solving for x, we get:

$$x = \frac{-43.0 J \times 6 \text{ moles A}}{2 \text{ moles A}} = -129 J$$

Therefore, the heat associated with the reaction of 6 moles of A is -129 Joules. The negative sign reaffirms that this is an exothermic reaction, with heat being released into the surroundings.

Expressing the Answer in Three Significant Figures

The question requests the answer to be expressed in three significant figures. Significant figures are the digits in a number that carry meaningful information about its precision. In our calculated value of -129 J, all three digits are significant. Thus, the final answer, expressed in three significant figures, remains -129 J.

In summary, by meticulously analyzing the provided thermochemical equation and employing stoichiometric principles, we have successfully determined that the heat associated with the reaction of 6 moles of A is -129 Joules. This calculation underscores the importance of thermochemical equations and stoichiometry in quantifying energy changes in chemical reactions.

To further enhance our understanding, let's explore some key concepts and applications related to thermochemical equations and enthalpy changes:

Enthalpy: The Heat Content of a System

Enthalpy (H) is a thermodynamic property of a system that represents the total heat content. It is the sum of the internal energy of the system and the product of its pressure and volume. Enthalpy is a state function, meaning its value depends only on the current state of the system, not on the path taken to reach that state.

Enthalpy change (ΔH) is the change in enthalpy during a chemical reaction or physical process. It is the difference between the enthalpy of the products and the enthalpy of the reactants:

$$\Delta H = H_{products} - H_{reactants}$$

A negative ΔH indicates an exothermic process, while a positive ΔH indicates an endothermic process.

Standard Enthalpy Changes

To facilitate comparisons of enthalpy changes across different reactions, standard conditions are defined. Standard conditions are typically 298 K (25 °C) and 1 atm pressure. Enthalpy changes measured under standard conditions are called standard enthalpy changes and are denoted by the symbol ΔH°. Common types of standard enthalpy changes include:

• Standard enthalpy of formation (ΔHf°): The enthalpy change when 1 mole of a compound is formed from its elements in their standard states.
• Standard enthalpy of combustion (ΔHc°): The enthalpy change when 1 mole of a substance is completely burned in oxygen under standard conditions.
• Standard enthalpy of reaction (ΔHrxn°): The enthalpy change for a reaction carried out under standard conditions.

Hess's Law: A Powerful Tool for Calculating Enthalpy Changes

Hess's Law states that the enthalpy change for a reaction is independent of the path taken, as long as the initial and final conditions are the same. This law allows us to calculate enthalpy changes for reactions that are difficult or impossible to measure directly by combining the enthalpy changes of a series of reactions that add up to the overall reaction.

For example, consider the following reaction:

$$C(s) + O_2(g) \rightarrow CO_2(g)$$

The standard enthalpy change for this reaction can be calculated using Hess's Law by considering the following steps:

1. $$C(s) + \frac{1}{2} O_2(g) \rightarrow CO(g) \quad \Delta H_1° = -110.5 kJ$$

2. $$CO(g) + \frac{1}{2} O_2(g) \rightarrow CO_2(g) \quad \Delta H_2° = -283.0 kJ$$

Adding these two equations gives the overall reaction, and the standard enthalpy change for the overall reaction is the sum of the standard enthalpy changes for the individual steps:

$$\Delta H_{rxn}° = \Delta H_1° + \Delta H_2° = -110.5 kJ + (-283.0 kJ) = -393.5 kJ$$

Applications of Thermochemistry

Thermochemistry has numerous applications in various fields, including:

• Predicting the feasibility of reactions: Enthalpy changes can help predict whether a reaction will occur spontaneously. Exothermic reactions (negative ΔH) tend to be spontaneous, while endothermic reactions (positive ΔH) require energy input to occur.
• Designing chemical processes: Thermochemical data is essential for designing efficient chemical processes. By understanding the heat released or absorbed in a reaction, engineers can optimize reaction conditions and minimize energy consumption.
• Calculating energy content of fuels: The enthalpy of combustion is used to determine the energy content of fuels. This information is crucial for evaluating the efficiency of different fuels and for designing combustion engines.
• Understanding climate change: Greenhouse gases, such as carbon dioxide, trap heat in the atmosphere, contributing to climate change. Thermochemical principles are used to study the interactions of these gases with radiation and to model climate change scenarios.

In conclusion, thermochemical equations and enthalpy changes provide a fundamental framework for understanding energy transformations in chemical reactions. By applying stoichiometric principles, Hess's Law, and other thermochemical concepts, we can gain valuable insights into the feasibility, efficiency, and environmental impact of chemical processes.

To solidify your understanding of thermochemical calculations, let's delve into a comprehensive guide that covers various types of problems and strategies for solving them. This guide will equip you with the necessary skills to tackle a wide range of thermochemistry challenges.

Types of Thermochemical Problems

Thermochemical problems can be broadly classified into the following categories:

1. Calculating Heat Released or Absorbed in a Reaction: These problems involve determining the amount of heat released (exothermic) or absorbed (endothermic) when a specific amount of reactants undergoes a reaction. As demonstrated in the initial example, stoichiometric relationships play a crucial role in these calculations.
2. Determining Enthalpy Changes (ΔH): These problems focus on calculating the enthalpy change for a given reaction. This may involve using standard enthalpies of formation, Hess's Law, or experimental data.
3. Applying Hess's Law: These problems require the application of Hess's Law to calculate enthalpy changes for reactions that can be expressed as a sum of other reactions with known enthalpy changes.
4. Using Standard Enthalpies of Formation (ΔHf°): These problems involve calculating enthalpy changes using standard enthalpies of formation. The standard enthalpy change for a reaction can be calculated as the sum of the standard enthalpies of formation of the products, minus the sum of the standard enthalpies of formation of the reactants.
5. Relating Enthalpy Change to Bond Energies: These problems explore the relationship between enthalpy change and bond energies. Bond energy is the energy required to break one mole of a particular bond in the gas phase. Enthalpy change can be estimated by considering the difference between the energy required to break bonds in the reactants and the energy released when bonds are formed in the products.

Strategies for Solving Thermochemical Problems

To effectively tackle thermochemical problems, consider the following strategies:

1. Write a Balanced Chemical Equation: Ensure that the chemical equation is balanced to accurately represent the stoichiometric relationships between reactants and products. This is crucial for calculations involving mole ratios.

2. Identify Given Information: Carefully identify the given information in the problem, such as the amount of reactants, enthalpy changes, or standard enthalpies of formation. Organize this information to facilitate problem-solving.

3. Determine the Desired Quantity: Clearly identify the quantity you are asked to calculate, such as the heat released, enthalpy change, or standard enthalpy of formation.

4. Choose the Appropriate Method: Select the appropriate method or equation based on the given information and the desired quantity. This may involve stoichiometry, Hess's Law, or the use of standard enthalpies of formation.

5. Apply Stoichiometric Ratios: If the problem involves calculating the heat released or absorbed for a specific amount of reactants, use the stoichiometric coefficients from the balanced equation to establish mole ratios.

6. Use Hess's Law: If the reaction can be expressed as a sum of other reactions with known enthalpy changes, apply Hess's Law to calculate the overall enthalpy change.

7. Apply the Formula for Enthalpy Change Using Standard Enthalpies of Formation: If standard enthalpies of formation are provided, use the formula:

   $$\Delta H_{rxn}° = \sum \Delta H_f°(products) - \sum \Delta H_f°(reactants)$$

8. Pay Attention to Sign Conventions: Remember that exothermic reactions have negative enthalpy changes (ΔH < 0), while endothermic reactions have positive enthalpy changes (ΔH > 0). Include the correct sign in your final answer.

9. Include Units: Always include the appropriate units in your final answer. Enthalpy changes are typically expressed in Joules (J) or Kilojoules (kJ).

10. Check Your Answer: After completing the calculation, check your answer to ensure it is reasonable and consistent with the given information. Consider the magnitude and sign of the answer.

Example Problems and Solutions

To illustrate these strategies, let's consider a couple of example problems:

Example Problem 1

Calculate the heat released when 10.0 grams of methane (CH4) is completely burned in oxygen, given that the standard enthalpy of combustion of methane is -890.4 kJ/mol.

Solution

1. Balanced Chemical Equation:

   $$CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l)$$

2. Given Information:

   • Mass of methane (CH4) = 10.0 grams
   • Standard enthalpy of combustion (ΔHc°) = -890.4 kJ/mol

3. Desired Quantity:

   • Heat released

4. Method:

   • Use stoichiometry to convert grams of methane to moles of methane, then use the standard enthalpy of combustion to calculate the heat released.

5. Calculations:

   • Molar mass of methane (CH4) = 12.01 g/mol (C) + 4 * 1.01 g/mol (H) = 16.05 g/mol
   • Moles of methane = 10.0 g / 16.05 g/mol = 0.623 mol
   • Heat released = 0.623 mol * (-890.4 kJ/mol) = -554.7 kJ

6. Answer:

   • The heat released when 10.0 grams of methane is completely burned is -555 kJ (rounded to three significant figures).

Example Problem 2

Calculate the standard enthalpy change (ΔHrxn°) for the following reaction:

$$2NO(g) + O_2(g) \rightarrow 2NO_2(g)$$

Given the following standard enthalpies of formation:

• ΔHf°(NO(g)) = 90.25 kJ/mol
• ΔHf°(O2(g)) = 0 kJ/mol
• ΔHf°(NO2(g)) = 33.18 kJ/mol

Solution

1. Balanced Chemical Equation:

   $$2NO(g) + O_2(g) \rightarrow 2NO_2(g)$$

2. Given Information:

   • ΔHf°(NO(g)) = 90.25 kJ/mol
   • ΔHf°(O2(g)) = 0 kJ/mol
   • ΔHf°(NO2(g)) = 33.18 kJ/mol

3. Desired Quantity:

   • Standard enthalpy change (ΔHrxn°)

4. Method:

   • Use the formula:

   $$\Delta H_{rxn}° = \sum \Delta H_f°(products) - \sum \Delta H_f°(reactants)$$

5. Calculations:

   • ΔHrxn° = [2 * ΔHf°(NO2(g))] - [2 * ΔHf°(NO(g)) + ΔHf°(O2(g))]
   • ΔHrxn° = [2 * 33.18 kJ/mol] - [2 * 90.25 kJ/mol + 0 kJ/mol]
   • ΔHrxn° = 66.36 kJ/mol - 180.5 kJ/mol
   • ΔHrxn° = -114.14 kJ/mol

6. Answer:

   • The standard enthalpy change for the reaction is -114.14 kJ/mol.

By mastering these strategies and practicing with various types of problems, you can confidently navigate the world of thermochemical calculations. Remember to approach each problem systematically, carefully consider the given information, and apply the appropriate methods and equations.

In this comprehensive exploration of thermochemistry, we have delved into the fundamental concepts, problem-solving strategies, and practical applications of this crucial branch of chemistry. From understanding thermochemical equations and enthalpy changes to applying Hess's Law and standard enthalpies of formation, we have equipped ourselves with the tools to analyze and quantify energy transformations in chemical reactions.

Thermochemistry is not merely an academic pursuit; it is a cornerstone of various scientific and engineering disciplines. Its principles underpin the design of efficient chemical processes, the development of new energy sources, and the understanding of environmental phenomena such as climate change. By mastering thermochemistry, we gain a deeper appreciation for the intricate interplay between matter and energy and unlock the potential to solve real-world challenges.

As you continue your journey in chemistry, remember that thermochemistry is an ever-evolving field. New discoveries and advancements are constantly expanding our knowledge of energy transformations. Embrace the challenges, explore the complexities, and strive to contribute to the ongoing quest for a more sustainable and energy-efficient future.

Through dedication, practice, and a thirst for knowledge, you can master the art of thermochemistry and harness its power to shape a better world.

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What is the amount of heat associated with the reaction when 6 moles of A react, given the thermochemical equation 2A → AA with ΔHrxn = -43.0 J? Express your answer in joules to three significant figures.

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Thermochemistry Calculation Heat of Reaction for 6 Moles of A