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Ch.5 - Thermochemistry
Brown - Chemistry: The Central Science 14th Edition
Brown14th EditionChemistry: The Central ScienceISBN: 9780134414232Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoBrown 14th EditionCh.5 - ThermochemistryProblema 47a
Capitolo 5, Problema 47a

Consider the combustion of liquid methanol, CH3OH(l): CH3OH(l) + 3/2 O2(g) → CO2(g) + 2 H2O(l) ΔH = -726.5 kJ (a) What is the enthalpy change for the reverse reaction?

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Identify the given reaction: CH_3OH(l) + \(\frac{3}{2}\) O_2(g) \(\rightarrow\) CO_2(g) + 2 H_2O(l) with \(\Delta\) H = -726.5 \(\text{ kJ}\).
Understand that the reverse reaction is: CO_2(g) + 2 H_2O(l) \(\rightarrow\) CH_3OH(l) + \(\frac{3}{2}\) O_2(g).
Recall that the enthalpy change for a reverse reaction is the negative of the enthalpy change for the forward reaction.
Apply this concept: \(\Delta\) H_{reverse} = - (\(\Delta\) H_{forward}).
Substitute the given \(\Delta\) H_{forward} value into the equation to find \(\Delta\) H_{reverse}.

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Enthalpy Change

Enthalpy change (ΔH) is a measure of the heat content of a system at constant pressure. It indicates the amount of energy absorbed or released during a chemical reaction. A negative ΔH value, like -726.5 kJ in the combustion of methanol, signifies that the reaction is exothermic, releasing energy to the surroundings.
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Enthalpy of Formation

Reverse Reaction

The reverse reaction is the process in which the products of a chemical reaction are converted back into the original reactants. For the combustion of methanol, the reverse reaction would involve converting carbon dioxide and water back into methanol and oxygen. The enthalpy change for the reverse reaction is equal in magnitude but opposite in sign to that of the forward reaction.
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Reversible Changes in Matter

Thermodynamic Principles

Thermodynamic principles govern the relationships between heat, work, and energy in chemical reactions. According to the first law of thermodynamics, energy cannot be created or destroyed, only transformed. This principle helps in understanding that the enthalpy change for a reaction and its reverse are related, allowing us to calculate the enthalpy change for the reverse reaction directly from the forward reaction's ΔH.
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First Law of Thermodynamics
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Domanda del libro di testo

At one time, a common means of forming small quantities of oxygen gas in the laboratory was to heat KClO3: 2 KClO3(s) → 2 KCl(s) + 3 O2(g) ΔH = -89.4 kJ (c) The decomposition of KClO3 proceeds spontaneously when it is heated. Do you think that the reverse reaction, the formation of KClO3 from KCl and O2, is likely to be feasible under ordinary conditions? Explain your answer.

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Domanda del libro di testo

Consider the combustion of liquid methanol, CH3OH(l): CH3OH(l) + 3/2 O2(g) → CO2(g) + 2 H2O(l) ΔH = -726.5 kJ (b) Balance the forward reaction with whole-number coefficients. What is ΔH for the reaction represented by this equation?

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At one time, a common means of forming small quantities of oxygen gas in the laboratory was to heat KClO3: 2 KClO3(s) → 2 KCl(s) + 3 O2(g) ΔH = -89.4 kJ For this reaction, calculate H for the formation of (a) 1.36 mol of O2

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At one time, a common means of forming small quantities of oxygen gas in the laboratory was to heat KClO3: 2 KClO3(s) → 2 KCl(s) + 3 O2(g) ΔH = -89.4 kJ For this reaction, calculate H for the formation of (b) 10.4 g of KCl.

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Consider the decomposition of liquid benzene, C6H6(l), to gaseous acetylene, C2H2(g): C6H6(l) → 3 C2H2(g) ΔH = +630 kJ (a) What is the enthalpy change for the reverse reaction?

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Consider the combustion of liquid methanol, CH3OH(l): CH3OH(l) + 3/2 O2(g) → CO2(g) + 2 H2O(l) ΔH = -726.5 kJ (d) If the reaction were written to produce H2O(g) instead of H2O(l), would you expect the magnitude of ΔH to increase, decrease, or stay the same? Explain.

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