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Calculate \(\Delta \mathrm{H}^{\circ} \mathrm{r}\) for the combustion of methane, \(\mathrm{CH}_{4}\). The balanced reaction is \(\mathrm{CH}_{4}(\mathrm{~g})+2 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(\ell)\) \(\Delta \mathrm{H}^{\circ} \mathrm{f}\) in \(\mathrm{Kcal} / \mathrm{mole}=-17.89,0,-94.05,2(-68.32)\)

Short Answer

Expert verified
The standard enthalpy change of reaction (\(\Delta H^{\circ}_r\)) for the combustion of methane is calculated using the formula: \(\Delta H^{\circ}_r = \sum \Delta H^{\circ}_f \text{(products)} - \sum \Delta H^{\circ}_f\text{(reactants)}\). By substituting the provided values, we find that \(\Delta H^{\circ}_r\) for the combustion of methane is \(-212.8 \text{ kcal/mole}\).

Step by step solution

01

Balanced Equation

The combustion of methane is given by the balanced reaction: \[CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l)\] Step 2: Identify standard enthalpy of formation values
02

Standard Enthalpy of Formation Values

The values of standard enthalpy of formation (\(\Delta H^{\circ}_f\)) are given in kcal/mole: $CH_4(g): -17.89 \\ O_2(g): 0 \\ CO_2(g): -94.05 \\ H_2O(l): -68.32$ Step 3: Apply the formula to determine \(\Delta H^{\circ}_r\)
03

Calculate \(\Delta H^{\circ}_r\)

Using the formula: \(\Delta H^{\circ}_r = \sum \Delta H^{\circ}_f\text{(products)} - \sum \Delta H^{\circ}_f\text{(reactants)}\), we can calculate the standard enthalpy change of reaction: \(\Delta H^{\circ}_r = [1(-94.05) + 2(-68.32)] - [1(-17.89) + 2(0)]\) Step 4: Solve for \(\Delta H^{\circ}_r\)
04

Solve the Equation

Now, perform the calculations to find the value of \(\Delta H^{\circ}_r\) for the combustion of methane. \(\Delta H^{\circ}_r = (-94.05 - 136.64) - (-17.89) = -230.69 + 17.89\) \(\Delta H^{\circ}_r = -212.8 \text{ kcal/mole}\) So, the standard enthalpy change of reaction for the combustion of methane is -212.8 kcal/mole.

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Key Concepts

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

Standard Enthalpy of Formation
The standard enthalpy of formation, symbolized as \( \Delta H^\circ_f \), is a measure of the energy change when one mole of a substance is formed from its elements in their standard states. This concept is critical in chemical thermodynamics as it allows us to understand the energy involved in forming different chemical compounds. A standard state refers to the pure form of the substance at 1 atmosphere pressure and at a specified temperature, usually 25°C (298 K).

For example, when considering the combustion of methane, knowing the enthalpy of formation for each reactant and product enables us to calculate the overall energy change during the reaction. Elements in their most stable form, like \( O_2(g) \) in our exercise, have a standard enthalpy of formation of zero because no energy is required to form a pure element from itself.

To improve understanding, it's useful to remember that exothermic reactions, where energy is released, have a negative \( \Delta H^\circ_f \), indicating that the formation of the product is energetically favorable. Conversely, endothermic reactions absorb energy and have a positive \( \Delta H^\circ_f \).
Combustion of Methane
The combustion of methane is an exothermic chemical reaction where methane gas is oxidized by oxygen to produce carbon dioxide, water, and energy in the form of heat. Represented by the balanced chemical equation \( CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l) \) , it is a common example of a combustion reaction involving a hydrocarbon.

In the realm of energy production, methane combustion is a fundamental process since methane is the primary component of natural gas. When methane burns, it is converted into \( CO_2 \) and \( H_2O \) , and the energy released can be harnessed for heating, electricity generation, and various industrial processes.

For students, visualizing the reaction can be helpful. Picture methane molecules colliding with oxygen molecules in a fiery dance that breaks old chemical bonds and forms new ones, releasing energy as a captivating finale. This reaction is also a vital tutorial of the conservational aspect of matter, as the number of atoms of each element remains constant before and after the reaction, adhering to the Law of Conservation of Mass.
Chemical Thermodynamics
Chemical thermodynamics is an area of chemistry that involves the study of the interrelation of heat and work with chemical reactions or with physical changes of state within the confines of the laws of thermodynamics. It provides a framework for understanding how energy is transformed and conserved in chemical processes and gives us tools like the concept of enthalpies of formation and reaction to quantify these changes.

It hinges on key principles, like the First Law of Thermodynamics, which asserts that energy cannot be created or destroyed, only transformed. This is seen in exercises like the combustion of methane, where chemical bonds are broken and formed, and energy is released but conserved overall. Students may find it useful to think of thermodynamics as the accountants for energy, maintaining a precise balance sheet for every reaction to ensure that the energy books always balance out.

By applying these principles, we can predict the feasibility and extent of chemical reactions, design processes that make efficient use of energy, and even assess the environmental impact of chemical processes by examining the heat they release or absorb.
Balanced Chemical Equation
A balanced chemical equation is the cornerstone of reaction stoichiometry, representing the principle of the conservation of mass. It shows the reactants transforming into products with the number of atoms of each element balanced on both sides of the equation. For example, in the combustion of methane, \( CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l) \) is the balanced chemical equation illustrating that one methane molecule reacts with two oxygen molecules to form one carbon dioxide molecule and two water molecules.

Balancing an equation is not just about numbers; it offers a visual and quantitative representation of the reaction that is essential for calculating reactant and product amounts, and for understanding the energy changes as described in the concept of enthalpy. To enhance comprehension, students should practice balancing equations regularly and conceptually link the atoms' rearrangement with energy changes occurring throughout the reaction. This skill is fundamental for any student who wishes to excel in chemistry and beyond.

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Most popular questions from this chapter

A piece of iron weighing \(20.0 \mathrm{~g}\) at a temperature of \(95.0^{\circ} \mathrm{C}\) was placed in \(100.0 \mathrm{~g}\) of water at \(25.0^{\circ} \mathrm{C}\). Assuming that no heat is lost to the surroundings, what is the resulting temperature of the iron and water? Specific heats: iron \(=.108 \mathrm{cal} / \mathrm{g}-\mathrm{C}^{\circ}\); water \(=1.0 \mathrm{cal} / \mathrm{g}-\mathrm{C}^{\circ}\).

In a pound-for-pound comparison, which would you prefer as a rocket fuel, hydrogen \(\left(\mathrm{H}_{2}\right)\) or dimethythydrazine \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NNH}_{2} ? \mathrm{H}_{2}(\mathrm{~g})+(1 / 2) \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\ell)\) \(\Delta \mathrm{H}=-68.3 \mathrm{~K} \mathrm{cal} / \mathrm{mole}\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NNH}_{2}+4 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{N}_{2}(\mathrm{~g})+\) \(4 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})+2 \mathrm{CO}_{2}(\mathrm{~g}), \Delta \mathrm{H}=-404.9 \mathrm{~K}\) cal/mole.

The following are physical properties of methy 1 alcohol, \(\mathrm{CH}_{3} \mathrm{OH}\) : freezing point \(-98^{\circ} \mathrm{C}\); boiling point \(65^{\circ} \mathrm{C}\); specific heat of liquid \(0.570 \mathrm{cal} / \mathrm{g}\) - degree; heat of fusion \(22.0\) \(\mathrm{cal} / \mathrm{g}\); and heat of vaporization \(263 \mathrm{cal} / \mathrm{g}\). Calculate the number of kilocalories required to convert one mole of methyl alcohol solid at \(-98^{\circ} \mathrm{C}\) to vapor at \(65^{\circ} \mathrm{C}\).

At constant volume, the heat capacity of gas differs from the heat capacity at constant pressure. Why?

In a hydrogen bomb, deuterium nuclei fuse to produce a helium nucleus. In the process, \(5 \mathrm{MeV}\) of energy is released per deuterium. If, \(1 / 7000\) of natural \(\mathrm{H}\) is \(\mathrm{D}\), how many liters of gasoline \(\left(\mathrm{C}_{8} \mathrm{H}_{18}\right)\) are needed to be burned to produce the same amount of energy as derived from 1 liter of water. Assume the following: the heat of combustion of gasoline is \(5100 \mathrm{KJ} /\) mole the density of gasoline is \(.703 \mathrm{~g} / \mathrm{cm}^{3}\); \(\mathrm{leV}=1.60 \times 10^{-19} \mathrm{~J}\)

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