/*! 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 73 Using values from Appendix \(\ma... [FREE SOLUTION] | 91Ó°ÊÓ

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Using values from Appendix \(\mathrm{C},\) calculate the standard enthalpy change for each of the following reactions: (a) \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{SO}_{3}(g)\) (b) \(\mathrm{Mg}(\mathrm{OH})_{2}(s) \longrightarrow \mathrm{MgO}(s)+\mathrm{H}_{2} \mathrm{O}(l)\) (c) \(\mathrm{N}_{2} \mathrm{O}_{4}(g)+4 \mathrm{H}_{2}(g) \longrightarrow \mathrm{N}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g)\) (d) \(\mathrm{SiCl}_{4}(l)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{SiO}_{2}(s)+4 \mathrm{HCl}(g)\)

Short Answer

Expert verified
The standard enthalpy changes (∆H°) for the given reactions are: (a) ∆H°(reaction) = -198 kJ/mol (b) ∆H°(reaction) = -37 kJ/mol (c) ∆H°(reaction) = -1284 kJ/mol (d) ∆H°(reaction) = -291 kJ/mol

Step by step solution

01

(Reaction a) Calculate ∆H° for 2SO2(g) + O2(g) → 2SO3(g)

First, look up the standard enthalpy values for the reactants and products in Appendix C. Then apply the formula mentioned above: ∆H°(reaction) = [2 × ∆H°(SO3)] - [2 × ∆H°(SO2) + ∆H°(O2)] Substitute the values from Appendix C into the equation and solve for ∆H°(reaction).
02

(Reaction b) Calculate ∆H° for Mg(OH)2(s) → MgO(s) + H2O(l)

Look up the standard enthalpy values for the reactants and products in Appendix C. Then apply the formula: ∆H°(reaction) = [∆H°(MgO) + ∆H°(H2O)] - [∆H°(Mg(OH)2)] Substitute the values from Appendix C into the equation and solve for ∆H°(reaction).
03

(Reaction c) Calculate ∆H° for N2O4(g) + 4H2(g) → N2(g) + 4H2O(g)

Look up the standard enthalpy values for the reactants and products in Appendix C. Then apply the formula: ∆H°(reaction) = [∆H°(N2) + 4 × ∆H°(H2O)] - [∆H°(N2O4) + 4 × ∆H°(H2)] Substitute the values from Appendix C into the equation and solve for ∆H°(reaction).
04

(Reaction d) Calculate ∆H° for SiCl4(l) + 2H2O(l) → SiO2(s) + 4HCl(g)

Look up the standard enthalpy values for the reactants and products in Appendix C. Then apply the formula: ∆H°(reaction) = [∆H°(SiO2) + 4 × ∆H°(HCl)] - [∆H°(SiCl4) + 2 × ∆H°(H2O)] Substitute the values from Appendix C into the equation and solve for ∆H°(reaction).

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Key Concepts

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

Chemical Reactions
Chemical reactions are processes where substances are transformed into new substances. In these processes, the bonds between atoms are rearranged, resulting in the formation of new compounds. Each reaction involves reactants, which change into products. Consider a chemical equation, which highlights the transformation from reactants to products.

Understanding chemical reactions is crucial because they are happening around us all the time. For example, burning wood, cooking food, and even the rusting of iron involve chemical reactions. These processes are essential in various fields such as biology, chemistry, and environmental science.

In the context of chemistry, reactions show how substances interact at the molecular level. The balanced equations help predict the amounts of products and reactants involved, allowing scientists to conduct energy calculations like enthalpy, which further explain the nature of reactions.
Enthalpy Calculation
Enthalpy is a measure of the heat content of a system. In chemistry, calculating the enthalpy change (\( ΔH° \) ) during a reaction shows whether a reaction is endothermic or exothermic. An endothermic reaction absorbs heat, while an exothermic reaction releases heat into the surroundings.

The calculation involves taking the sum of the enthalpies of the products and subtracting the sum of the enthalpies of the reactants. Use the formula:\[ΔH°(reaction) = ΣΔH°(products) - ΣΔH°(reactants)\]This formula requires values from standardized tables, such as Appendix C, providing the standard enthalpy values for common substances. By knowing these values, students can analyze and describe reactions more effectively, allowing them to predict energy changes in various chemical processes.

Understanding these calculations helps in grasping how energy flows in reactions, which is crucial for both academic and practical applications in fields like engineering, environmental science, and more.
Thermodynamics
Thermodynamics is the branch of physics and chemistry that deals with the laws of energy transfer within chemical reactions. This field provides crucial insights into how reactions occur and the conditions under which they proceed. The key concept in thermodynamics is the conservation of energy, meaning energy cannot be created or destroyed, only transformed.

In thermodynamic studies, the concept of enthalpy is vital as it quantifies the heat energy involved in chemical transformations. It allows scientists to understand the potential energy change during a reaction and helps to predict whether a reaction will be spontaneous based on enthalpy change and other factors like entropy and temperature.

Thermodynamics involves understanding complex equations that can predict system behavior under various conditions. These predictions allow chemists and engineers to design processes that are efficient and safe. Its principles are applied in everything from power generation to understanding biological processes, demonstrating its vast importance in both science and everyday life.

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Most popular questions from this chapter

A 1.800 -g sample of phenol \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}\right)\) was burned in a bomb calorimeter whose total heat capacity is \(11.66 \mathrm{~kJ} /{ }^{\circ} \mathrm{C}\). The temperature of the calorimeter plus contents increased from \(21.36^{\circ} \mathrm{C}\) to \(26.37{ }^{\circ} \mathrm{C}\). (a) Write a balanced chemical equation for the bomb calorimeter reaction. (b) What is the heat of combustion per gram of phenol? Per mole of phenol?

The automobile fuel called E85 consists of \(85 \%\) ethanol and \(15 \%\) gasoline. \(\mathrm{E} 85\) can be used in so-called "flex-fuel" vehicles (FFVs), which can use gasoline, ethanol, or a mix as fuels. Assume that gasoline consists of a mixture of octanes (different isomers of \(\mathrm{C}_{8} \mathrm{H}_{18}\) ), that the average heat of combustion of \(\mathrm{C}_{8} \mathrm{H}_{18}(l)\) is \(5400 \mathrm{~kJ} / \mathrm{mol}\), and that gasoline has an average density of \(0.70 \mathrm{~g} / \mathrm{mL}\). The density of ethanol is \(0.79 \mathrm{~g} / \mathrm{mL}\). (a) By using the information given as well as data in Appendix C, compare the energy produced by combustion of \(1.0 \mathrm{~L}\) of gasoline and of \(1.0 \mathrm{~L}\) of ethanol. (b) Assume that the density and heat of combustion of \(\mathrm{E} 85\) can be obtained by using \(85 \%\) of the values for ethanol and \(15 \%\) of the values for gasoline. How much energy could be released by the combustion of \(1.0 \mathrm{~L}\) of E85? (c) How many gallons of E85 would be needed to provide the same energy as 10 gal of gasoline? (d) If gasoline costs \(\$ 3.10\) per gallon in the United States, what is the break-even price per gallon of \(\mathrm{E} 85\) if the same amount of energy is to be delivered?

Consider the combustion of liquid methanol, \(\mathrm{CH}_{3} \mathrm{OH}(l):\) $$ \begin{aligned} \mathrm{CH}_{3} \mathrm{OH}(l)+\frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) & \\ \Delta H=&-726.5 \mathrm{~kJ} \end{aligned} $$ (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.

An aluminum can of a soft drink is placed in a freezer. Later, you find that the can is split open and its contents frozen.Work was done on the can in splitting it open. Where did the energy for this work come from?

Consider the conversion of compound \(A\) into compound \(B\) : \(\mathrm{A} \longrightarrow \mathrm{B}\). For both compounds \(\mathrm{A}\) and \(\mathrm{B}, \Delta H_{f}^{\mathrm{o}}>0 .\) (a) Sketch an enthalpy diagram for the reaction that is analogous to Figure \(5.23 .\) (b) Suppose the overall reaction is exothermic. What can you conclude? [Section 5.7]

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