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The enthalpy change for the reaction $$ \mathrm{CH}_{4}(g)+2 \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) $$ is \(-891 \mathrm{~kJ}\) for the reaction as written. a. What quantity of heat is released for each mole of water formed? b. What quantity of heat is released for each mole of oxygen reacted?

Short Answer

Expert verified
a. The heat released for each mole of water formed is -445.5 kJ/mol. b. The heat released for each mole of oxygen reacted is -445.5 kJ/mol.

Step by step solution

01

Analyze the given information

Given, the enthalpy change for the reaction: \[ \mathrm{CH}_{4}(g) + 2 \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g) + 2 \mathrm{H}_{2}\mathrm{O}(l) \] ΔH = -891 kJ
02

Determine heat released for each mole of water formed

Since two moles of water are formed in the reaction, we can calculate the heat released per mole of water. We'll divide the total heat released by the number of moles of water formed. Heat released per mole of water = \( \frac{\text{Total heat released}}{\text{Number of moles of water formed}} \) Heat released per mole of water = \( \frac{-891\,\text{kJ}}{2\,\text{moles}} \)
03

Calculate the heat released per mole of water formed

Heat released per mole of water = \( -\frac{891}{2} \,\text{kJ/mol} \) Heat released per mole of water = -445.5 kJ/mol So, for each mole of water formed, 445.5 kJ of heat is released (negative sign indicates that heat is released). Answer (a): -445.5 kJ/mol
04

Determine heat released for each mole of oxygen reacted

For every 2 moles of oxygen, the total heat released is -891 kJ. We can calculate the heat released per mole of oxygen by dividing the total heat released by the number of moles of oxygen reacted. Heat released per mole of oxygen = \( \frac{\text{Total heat released}}{\text{Number of moles of oxygen reacted}} \) Heat released per mole of oxygen = \( \frac{-891\,\text{kJ}}{2\,\text{moles}} \)
05

Calculate the heat released per mole of oxygen reacted

Heat released per mole of oxygen = \( -\frac{891}{2} \,\text{kJ/mol} \) Heat released per mole of oxygen = -445.5 kJ/mol So, for each mole of oxygen reacted, 445.5 kJ of heat is released (negative sign indicates that heat is released). Answer (b): -445.5 kJ/mol

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chemical Reaction Enthalpy
The term 'enthalpy change' in a chemical reaction refers to the amount of heat content that is either absorbed or released during the transformation of reactants into products. It is denoted by the symbol \( \Delta H \). If the process releases heat, such as in combustion reactions, it is exothermic and \( \Delta H \) is negative. Conversely, reactions that absorb heat from their surroundings are endothermic, and \( \Delta H \) is positive.

In our given exercise, the combustion of methane (\(\text{CH}_{4}(g)\)) with oxygen (\(\text{O}_{2}(g)\)) results in the production of carbon dioxide (\(\text{CO}_{2}(g)\)) and water (\(\text{H}_{2} \text{O}(l)\)), with an enthalpy change of \( -891 \text{ kJ} \). This negative \( \Delta H \) indicates that heat is released and the reaction is exothermic. Understanding how to calculate and interpret enthalpy changes allows students to predict the energy transfers within various chemical reactions.
Heat Released per Mole
The concept of 'heat released per mole' is crucial in thermochemistry, showing the amount of heat transfer associated with one mole of a reactant or product in a chemical reaction. It's a specific instance of the broader concept of molar enthalpy. The calculation typically involves dividing the total heat change by the number of moles of the specific substance being considered.

For example, the exercise demonstrates how to determine the heat released per mole of water (\(\text{H}_{2} \text{O}\)) and per mole of oxygen (\(\text{O}_{2}\)). The calculated values express the amount of energy, in kilojoules, that are released when one mole of water or oxygen takes part in the reaction. Reporting heat released per mole allows us to compare the energy profiles of different chemical substances on a per-mole basis, which is invaluable for scientists, especially when designing energy-efficient processes or understanding chemical energetics.
Thermochemistry
Thermochemistry is the branch of chemistry that studies the heat energy involved in chemical reactions and phase changes. The first law of thermodynamics, which states that energy cannot be created or destroyed, governs thermochemistry and ensures the conservation of energy within these processes.

As demonstrated in this exercise, thermochemistry involves determining the heat exchange between a system and its surroundings. In calculations, it's important to use consistent units and account for the stoichiometry of the reaction, as change in enthalpy is directly proportional to the quantity of reactants and products involved. The knowledge of thermochemistry is applied in various fields, such as predicting reaction spontaneity, engineering, and environmental studies, making it a foundational concept in both theoretical and applied chemistry.
Stoichiometry
Stoichiometry refers to the quantitative relationship between reactants and products in a chemical reaction. It's grounded on the law of conservation of mass, which states that matter is neither created nor destroyed in a chemical reaction. Hence, stoichiometry involves using the balanced chemical equation to determine the proportions of chemicals required or formed.

In the given exercise, stoichiometry helps to calculate the heat released per mole of products (e.g., water) or reactants (e.g., oxygen) based on the overall enthalpy change for the reaction. It provides a clear pathway to understand the conversion process of reactants to products and ensures that students grasp the concept of molar ratios and their practical utility in predicting the outcomes of chemical reactions, such as the exact amount of energy released per mole.

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Most popular questions from this chapter

The enthalpy of combustion of solid carbon to form carbon dioxide is \(-393.7 \mathrm{~kJ} / \mathrm{mol}\) carbon, and the enthalpy of combustion of carbon monoxide to form carbon dioxide is \(-283.3 \mathrm{~kJ} / \mathrm{mol}\) CO. Use these data to calculate \(\Delta H\) for the reaction $$ 2 \mathrm{C}(s)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}(g) $$

The heat capacity of a bomb calorimeter was determined by burning \(6.79 \mathrm{~g}\) methane (energy of combustion \(=-802 \mathrm{~kJ} / \mathrm{mol} \mathrm{CH}_{4}\) ) in the bomb. The temperature changed by \(10.8^{\circ} \mathrm{C}\). a. What is the heat capacity of the bomb? b. A \(12.6-\mathrm{g}\) sample of acetylene, \(\mathrm{C}_{2} \mathrm{H}_{2}\), produced a temperature increase of \(16.9^{\circ} \mathrm{C}\) in the same calorimeter. What is the energy of combustion of acetylene (in \(\mathrm{kJ} / \mathrm{mol}\) )?

Calculate the kinetic energy of an object with a mass of \(1.0 \times\) \(10^{-5} \mathrm{~g}\) and a velocity of \(2.0 \times 10^{5} \mathrm{~cm} / \mathrm{s}\)

Consider the reaction $$ \begin{array}{r} 2 \mathrm{HCl}(a q)+\mathrm{Ba}(\mathrm{OH})_{2}(a q) \longrightarrow \mathrm{BaCl}_{2}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) \\ \Delta H=-118 \mathrm{~kJ} \end{array} $$ Calculate the heat when \(100.0 \mathrm{~mL}\) of \(0.500 \mathrm{M} \mathrm{HCl}\) is mixed with \(300.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\). Assuming that the temperature of both solutions was initially \(25.0^{\circ} \mathrm{C}\) and that the final mixture has a mass of \(400.0 \mathrm{~g}\) and a specific heat capacity of \(4.18 \mathrm{~J} /{ }^{\circ} \mathrm{C} \cdot \mathrm{g}\), calculate the final temperature of the mixture.

Consider \(5.5 \mathrm{~L}\) of a gas at a pressure of \(3.0 \mathrm{~atm}\) in a cylinder with a movable piston. The external pressure is changed so that the volume changes to \(10.5 \mathrm{~L}\). a. Calculate the work done, and indicate the correct sign. b. Use the preceding data but consider the process to occur in two steps. At the end of the first step, the volume is \(7.0 \mathrm{~L}\). The second step results in a final volume of \(10.5\) L. Calculate the work done, and indicate the correct sign. c. Calculate the work done if after the first step the volume is \(8.0 \mathrm{~L}\) and the second step leads to a volume of \(10.5 \mathrm{~L}\). Does the work differ from that in part b? Explain.

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