/*! 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 22 The enthalpy change for a reacti... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 22

The enthalpy change for a reaction is a state function and it is an extensive property. Explain.

Short Answer

Expert verified
The enthalpy change for a reaction, denoted by 螖H, is a state function because it depends only on the initial and final states of the system and not on the path taken to reach those states. It represents the difference in enthalpy between the products and reactants (螖H = H_products - H_reactants). Moreover, enthalpy change is an extensive property as it depends on the amount of material involved in the reaction, meaning it increases proportionally with the quantity of reactants.

Step by step solution

01

Definition of State Function

A state function is a property of a system that depends only on its current state, and is independent of the path taken to reach that state. In other words, a state function does not depend on the process or the history of the system, only on its present condition.
02

Definition of Extensive Property

An extensive property is a property of a system that depends on the amount of material present in the system. It means that if the quantity of the system increases, the value of the extensive property will also increase proportionally.
03

Enthalpy Change as a State Function

Enthalpy (H) is a thermodynamic property that represents the total energy of a system, including its internal energy (U) and the energy associated with the pressure-volume work (PV): H = U + PV. The enthalpy change for a reaction, denoted by 螖H, is the difference in enthalpy between the final state (products) and the initial state (reactants), given by 螖H = H_products - H_reactants. Since enthalpy change depends only on the initial and final states of the system in a reaction and not on the path taken to reach those states, it is a state function.
04

Enthalpy Change as an Extensive Property

Enthalpy change is an extensive property because it depends on the amount of material involved in the reaction. If the quantity of reactants involved in a reaction increases, the amount of energy transferred as heat during the reaction (which is represented by the enthalpy change) will also increase proportionally. In summary, the enthalpy change for a reaction is a state function because it depends only on the initial and final states of the system and not on the path taken to reach those states. It is also an extensive property because it depends on the amount of material involved in the reaction.

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

In a coffee-cup calorimeter, 100.0 \(\mathrm{mL}\) of 1.0 \(\mathrm{M}\) NaOH and 100.0 \(\mathrm{mL}\) of 1.0 \(\mathrm{M} \mathrm{HCl}\) are mixed. Both solutions were originally at \(24.6^{\circ} \mathrm{C}\) . After the reaction, the final temperature is \(31.3^{\circ} \mathrm{C}\) . Assuming that all the solutions have a density of 1.0 \(\mathrm{g} / \mathrm{cm}^{3}\) and a specific heat capacity of \(4.18 \mathrm{J} / \mathrm{C} \cdot \mathrm{g},\) calculate the enthalpy change for the neutralization of \(\mathrm{HCl}\) by NaOH. Assume that no heat is lost to the surroundings or to the calorimeter.

The sun supplies energy at a rate of about 1.0 kilowatt per square meter of surface area \((1 \text { watt }=1 \mathrm{Js} \text { ). The plants in an }\) agricultural field produce the equivalent of \(20 . \mathrm{kg}\) sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) per hour per hectare \(\left(1 \mathrm{ha}=10,000 \mathrm{m}^{2}\right) .\) Assuming that sucrose is produced by the reaction $$ \begin{aligned} 12 \mathrm{CO}_{2}(g)+11 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(s)+& 12 \mathrm{O}_{2}(g) \\ & \Delta H=5640 \mathrm{kJ} \end{aligned} $$ calculate the percentage of sunlight used to produce the sucrose-that is, determine the efficiency of photosynthesis.

The best solar panels currently available are about 19\(\%\) efficient in converting sunlight to electricity. A typical home will use about \(40 .\) kWh of electricity per day \((1 \mathrm{kWh}=1 \text { kilowatt }\) hour; \(1 \mathrm{kW}=1000 \mathrm{J} / \mathrm{s}\) ). Assuming 8.0 hours of useful sunlight per day, calculate the minimum solar panel surface area necessary to provide all of a typical home's electricity. (See Exercise 132 for the energy rate supplied by the sun.)

Are the following processes exothermic or endothermic? a. the combustion of gasoline in a car engine b. water condensing on a cold pipe c. \(\mathrm{CO}_{2}(s) \longrightarrow \mathrm{CO}_{2}(g)\) d. \(\mathrm{F}_{2}(g) \longrightarrow 2 \mathrm{F}(g)\)

Given the following data $$\mathrm{Ca}(s)+2 \mathrm{C}(\text { graphite }) \longrightarrow \mathrm{CaC}_{2}(s)$$ \(\Delta H=-62.8 \mathrm{kJ}\) $$ \mathrm{Ca}(s)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{CaO}(s) $$ \(\Delta H=-635.5 \mathrm{kJ}\) $$ \mathrm{CaO}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(a q) $$ \(\Delta H=-653.1 \mathrm{kJ}\) $$ \mathrm{C}_{2} \mathrm{H}_{2}(g)+\frac{5}{2} \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) $$ \(\Delta H=-1300 . \mathrm{kJ}\) $$ \mathrm{C}(\text {graphite})+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(\mathrm{g}) $$ \(\Delta H=-393.5 \mathrm{kJ}\) calculate \(\Delta H\) for the reaction $$ \mathrm{CaC}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(a q)+\mathrm{C}_{2} \mathrm{H}_{2}(g) $$
See all solutions

Recommended explanations on Chemistry Textbooks

Making Measurements

Read Explanation

Nuclear Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Organic Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Chemical Reactions

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.