/*! 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 61 What is the connection between H... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 61

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

Short Answer

Expert verified
The connection between Hess's Law and the fact that enthalpy (H) is a state function lies in their dependence on the initial and final states of a system. As enthalpy is a state function, its change depends only on the initial and final states, regardless of the specific path taken. Hess's Law, which states that the overall enthalpy change for a reaction is equal to the sum of the enthalpy changes for any intermediate reactions connecting the initial and final states, is a direct consequence of this property. Together, they allow us to calculate the enthalpy change of a reaction by summing the enthalpy changes of any set of intermediate reactions connecting the initial and final states.

Step by step solution

01

A state function is a property of a system that depends only on its current state, regardless of the path taken to reach that state. Enthalpy (H) is a thermodynamic property of a system and is a state function. This means that the change in enthalpy between two states (initial and final) is the same, irrespective of the specific path taken to reach the final state. #Step 2: Defining Hess's Law#

Hess's Law states that the enthalpy change for a chemical reaction is equal to the sum of the enthalpy changes for all intermediate reactions involved in the process. In other words, the overall enthalpy change of a reaction is the sum of the enthalpy changes of the individual steps, regardless of the specific pathway. #Step 3: Connecting enthalpy as a state function with Hess's Law#
02

Since enthalpy is a state function, the total change in enthalpy depends only on the initial and final states and not on the pathway taken to reach the final state. This means that if we can find a series of intermediate reactions that lead from the initial state to the final state, the sum of the enthalpy changes for each of these intermediate steps will be equal to the overall enthalpy change for the complete reaction. This is precisely what Hess's Law states: the overall enthalpy change for a reaction is equal to the sum of the enthalpy changes for any intermediate reactions that can be used to connect the initial and final states of the system. #Step 4: Conclusion#

Hess's Law is a direct consequence of the fact that enthalpy is a state function. Because enthalpy changes depend only on the initial and final states of a system and not on the specific path taken to reach those states, we can use Hess's Law to calculate the enthalpy change of a reaction by summing the enthalpy changes of any set of intermediate reactions that connect the initial and final states.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

(a) What is meant by the term state function? (b) Give an example of a quantity that is a state function and one that is not. (c) Is the volume of a system a state function? Why or why not?

(a) What is meant by the term standard conditions with reference to enthalpy changes? (b) What is meant by the term enthalpy of formation? (c) What is meant by the term standard enthalpy of formation?

The complete combustion of ethanol, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)\), to form \(\mathrm{H}_{2} \mathrm{O}(g)\) and \(\mathrm{CO}_{2}(g)\) at constant pressure releases \(1235 \mathrm{~kJ}\) of heat per mole of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\). (a) Write a balanced thermochemical equation for this reaction. (b) Draw an enthalpy diagram for the reaction.

Consider the following reaction: $$ 2 \mathrm{CH}_{3} \mathrm{OH}(g) \longrightarrow 2 \mathrm{CH}_{4}(g)+\mathrm{O}_{2}(g) \quad \Delta H=+252.8 \mathrm{~kJ} $$ (a) Is this reaction exothermic or endothermic? (b) Calculate the amount of heat transferred when \(24.0 \mathrm{~g}\) of \(\mathrm{CH}_{3} \mathrm{OH}(g)\) is decomposed by this reaction at constant pressure. (c) For a given sample of \(\mathrm{CH}_{3} \mathrm{OH}\), the enthalpy change during the reaction is \(82.1 \mathrm{~kJ}\). How many grams of methane gas are produced? (d) How many kilojoules of heat are released when \(38.5 \mathrm{~g}\) of \(\mathrm{CH}_{4}(g)\) reacts completely with \(\mathrm{O}_{2}(g)\) to form \(\mathrm{CH}_{3} \mathrm{OH}(\mathrm{g})\) at constant pressure? 5.45 When solutions containing silver ions and chloride ions are mixed, silver chloride precipitates: $$ \mathrm{Ag}^{+}(a q)+\mathrm{Cl}^{-}(a q) \longrightarrow \operatorname{AgCl}(s) \quad \Delta H=-65.5 \mathrm{~kJ} $$ (a) Calculate \(\Delta H\) for the production of \(0.450 \mathrm{~mol}\) of \(\mathrm{AgCl}\) by this reaction. (b) Calculate \(\Delta H\) for the production of \(9.00 \mathrm{~g}\) of \(\mathrm{AgCl}\). (c) Calculate \(\Delta H\) when \(9.25 \times 10^{-4} \mathrm{~mol}\) of \(\mathrm{AgCl}\) dissolves in water.

Calcium carbide \(\left(\mathrm{CaC}_{2}\right)\) reacts with water to form acetylene \(\left(\mathrm{C}_{2} \mathrm{H}_{2}\right)\) and \(\mathrm{Ca}(\mathrm{OH})_{2}\). From the following enthalpy of reaction data and data in Appendix C, calculate \(\Delta H_{f}^{\circ}\) for \(\mathrm{CaC}_{2}(s)\) : $$ \begin{array}{r} \mathrm{CaC}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(s)+\mathrm{C}_{2} \mathrm{H}_{2}(g) \\ \Delta H^{\circ}=-127.2 \mathrm{~kJ} \end{array} $$
See all solutions

Recommended explanations on Chemistry Textbooks

Chemistry Branches

Read Explanation

Making Measurements

Read Explanation

Physical Chemistry

Read Explanation

Kinetics

Read Explanation

Nuclear Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.