/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 47 Consider the combustion of liqui... [FREE SOLUTION] | 91Ó°ÊÓ

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Consider the combustion of liquid methanol, \(\mathrm{CH}_{3} \mathrm{OH}(l)\) : $$ \begin{array}{r} \mathrm{CH}_{3} \mathrm{OH}(l)+\frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) \\ \Delta H=-726.5 \mathrm{~kJ} \end{array} $$ (a) What is the enthalpy change for the reverse reaction? (b) Balance the forward reaction with whole-number coefficients. What is \(\Delta H\) for the reaction represented by this equation? (c) Which is more likely to be thermodynamically favored, the forward reaction or the reverse reaction? (d) If the reaction were written to produce \(\mathrm{H}_{2} \mathrm{O}(g)\) instead of \(\mathrm{H}_{2} \mathrm{O}(l)\), would you expect the magnitude of \(\Delta H\) to increase, decrease, or stay the same? Explain.

Short Answer

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(a) The enthalpy change for the reverse reaction is 726.5 kJ. (b) The balanced forward reaction is \(2 \mathrm{CH}_{3}\mathrm{OH}(l) + 3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g) + 4 \mathrm{H}_{2}\mathrm{O}(l)\) with an enthalpy change of -1453 kJ. (c) The forward reaction, being exothermic, is more likely to be thermodynamically favored. (d) If the reaction produced \(\mathrm{H}_{2}\mathrm{O}(g)\) instead of \(\mathrm{H}_{2}\mathrm{O}(l)\), the magnitude of the enthalpy change would decrease (i.e., the value would become less negative) due to the energy required for the phase change of water.

Step by step solution

01

*a) Enthalpy change for the reverse reaction*

For the reverse reaction, the enthalpy change will be the negative of the enthalpy change for the forward reaction. Therefore, the enthalpy change for the reverse reaction will be: \(\Delta H_{reverse} = -(-726.5 \: \mathrm{kJ}) = 726.5 \: \mathrm{kJ}\)
02

*b-1) Balance the forward reaction with whole-number coefficients*

We need to multiply the oxygen coefficient by 2 to make it a whole number. As a result, we need to multiply the other coefficients by 2 as well to keep the equation balanced. Forward Reaction: \(2 \mathrm{CH}_{3} \mathrm{OH}(l) + 3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g) + 4 \mathrm{H}_{2} \mathrm{O}(l)\)
03

*b-2) Calculate the enthalpy change*

Since we multiplied all the coefficients by 2 in the forward reaction, we also need to multiply the enthalpy change of the original reaction by 2. Therefore, the enthalpy change for the balanced forward reaction is: \(\Delta H_{balanced} = 2(-726.5 \: \mathrm{kJ}) = -1453 \: \mathrm{kJ}\)
04

*c) Thermodynamically favored reaction*

The forward reaction has a negative enthalpy change, which means that it releases heat to the surroundings. Thus, the forward reaction is exothermic. The reverse reaction, on the other hand, has a positive enthalpy change, meaning it absorbs heat from the surroundings, and it is endothermic. Generally, exothermic reactions (those with negative enthalpy changes) are more thermodynamically favored. Therefore, the forward reaction is more likely to be thermodynamically favored.
05

*d) Effect of producing \(\mathrm{H}_{2}\mathrm{O}(g)\) instead of \(\mathrm{H}_{2}\mathrm{O}(l)\) on enthalpy change*

If the forward reaction were to produce water vapor instead of liquid water, some energy would be required to change the phase of water from liquid to gas, known as the enthalpy of vaporization. In this case, less energy would be released in the reaction. Therefore, we would expect the magnitude of the enthalpy change to decrease (i.e., the value will become less negative).

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Key Concepts

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

Chemical Thermodynamics
Chemical thermodynamics is a branch of thermodynamics that focuses on energy changes, particularly the exchanges of heat and their relationship to chemical reactions. At the core of this field is the concept of enthalpy (H), which is essentially the total heat content of a system. The enthalpy change (Delta H) indicates whether a reaction is absorbing heat (endothermic) or releasing heat (exothermic) into its surroundings. In a chemical equation, the Delta H value provides crucial insights into the stability and spontaneity of a reaction, offering a window into understanding which direction a chemical process is thermodynamically favored. For instance, the negative Delta H value in the combustion of methanol implies the reaction is exothermic, tending to be spontaneous under constant pressure.
Exothermic and Endothermic Reactions
In the realm of chemical reactions, exothermic and endothermic processes serve as two fundamental categories by which we can classify the energy changes in a system. An exothermic reaction is one that releases energy, usually in the form of heat, to its surroundings, and is characterized by a negative enthalpy change (Delta H < 0). On the flip side, an endothermic reaction absorbs energy from the surroundings, resulting in a positive enthalpy change (Delta H > 0). When considering the combustion of methanol, the negative Delta H value (-726.5 kJ) clearly indicates that the reaction is exothermic; it liberates heat and is typically more favored in a thermodynamic sense compared to endothermic reactions like its reverse counterpart.
Enthalpy of Combustion
The enthalpy of combustion refers to the heat released when a substance undergoes complete combustion with oxygen under standard conditions. It is a specific case of enthalpy change and is always exothermic in nature, yielding a negative Delta H by convention. This value is often measured in a calorimeter and is a critical parameter for understanding energy production in engines and heaters. For instance, with the given Delta H of -726.5 kJ for the combustion of methanol, one can deduce how much energy per mole of methanol is released upon combustion. This has practical implications ranging from its use as a biofuel to calculations concerning energy output in thermal devices.
Stoichiometry
Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. It is based on the conservation of mass and the balancing of chemical equations, ensuring that the number of atoms for each element is conserved. For example, to balance the combustion reaction of methanol, coefficients are adjusted to maintain this balance. The enthalpy change (Delta H) associated with a reaction is also influenced by stoichiometry. When the coefficients in a chemical equation are multiplied to obtain whole numbers, the Delta H must be multiplied correspondingly, which maintains the relationship between the quantity of reactants consumed and the heat exchanged. The exact stoichiometry is crucial for applications like calculating fuel consumption and ensuring that chemical reactions occur efficiently in industrial processes.

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Most popular questions from this chapter

What is the connection between Hess's law and the fact that \(H\) is a state function?

Write balanced equations that describe the formation of the following compounds from elements in their standard states, and then look up the standard enthalpy of formation for each substance in Appendix \(\mathrm{C}\) : (a) \(\mathrm{H}_{2} \mathrm{O}_{2}(g)\), (b) \(\mathrm{CaCO}_{3}(s)\), (c) \(\mathrm{POCl}_{3}(l)\), (d) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)\).

Assume that the following reaction occurs at constant pressure: $$ 2 \mathrm{Al}(s)+3 \mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{AlCl}_{3}(s) $$ (a) If you are given \(\Delta H\) for the reaction, what additional information do you need to determine \(\Delta E\) for the process? (b) Which quantity is larger for this reaction? (c) Explain your answer to part (b).

In what two ways can an object possess energy? How do these two ways differ from one another?

Suppose an Olympic diver who weighs \(52.0 \mathrm{~kg}\) executes a straight dive from a 10 -m platform. At the apex of the dive, the diver is \(10.8 \mathrm{~m}\) above the surface of the water. (a) What is the potential energy of the diver at the apex of the dive, relative to the surface of the water? (b) Assuming that all the potential energy of the diver is converted into kinetic energy at the surface of the water, at what speed, in \(\mathrm{m} / \mathrm{s}\), will the diver enter the water? (c) Does the diver do work on entering the water? Explain. 5.93 The air bags that provide protection in automobiles in the event of an accident expand because of a rapid chemical reaction. From the viewpoint of the chemical reactants as the system, what do you expect for the signs of \(q\) and \(w\) in this process?

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