Chemistry
Hess' Law
Hess' Law states that the total enthalpy change for a chemical reaction is the same, regardless of the number of steps taken to achieve the reaction. This principle allows for the calculation of the overall enthalpy change of a reaction by summing the enthalpy changes of individual steps. It is a fundamental concept in thermodynamics and is used to determine the energy changes in chemical reactions.
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12 Key excerpts on "Hess' Law"
eBook - PDF- Young, William Vining, Roberta Day, Beatrice Botch(Authors)
- 2017(Publication Date)
- Cengage Learning EMEA(Publisher)
WCN 02-300 Unit 10 Thermochemistry 295 10.5 Hess’s Law 10.5a Hess’s Law Enthalpy is a state function , which means its value for a system depends only on the current state of the system and not on its history. Therefore, the change in enthalpy for a process depends only on the initial and final states for a system and not on the path between the two. For example, the altitude of a given location is like a state function: the difference in altitude between the top and bottom of a mountain has a fixed value, but the length of the hike from bottom to top does not, because there is more than one possible route, each with a different length. The state function nature of enthalpy is expressed in Hess’s law , which states that if a chemical reaction can be expressed as the sum of two or more chemical reactions, the enthalpy change for the net reaction is equal to the sum of the enthalpy changes for the individual steps. Consider the enthalpy changes for three reactions involved in the formation of formic acid (Interactive Figure 10.5.1). Reaction 1: C(s) 1 ½ O 2 (g) S CO(g) D H 1 5 2 110.5 kJ Reaction 2: CO(g) 1 H 2 O( ℓ ) S HCO 2 H(g) D H 2 5 1 33.7 kJ Reaction 3: HCO 2 H(g) S HCO 2 H( ℓ ) D H 3 5 2 62.9 kJ The net reaction is the production of formic acid from carbon, oxygen, and water. Net reaction: C(s) 1 ½ O 2 (g) 1 H 2 O( ℓ ) S HCO 2 H( ℓ ) To apply Hess’s law, sum reactions 1, 2, and 3 and cancel any species common to the products and reactants in the reaction. 1 1 2 1 3: C(s) 1 ½ O 2 (g) 1 CO(g) 1 H 2 O( ℓ ) 1 HCO 2 H(g) S CO(g) 1 HCO 2 H(g) 1 HCO 2 H( ℓ ) Cancel: C(s) 1 ½ O 2 (g) 1 CO(g) 1 H 2 O( ℓ ) 1 HCO 2 H(g) S CO(g) 1 HCO 2 H(g) 1 HCO 2 H( ℓ ) Net reaction: C(s) 1 ½ O 2 (g) 1 H 2 O( ℓ ) S HCO 2 H( ℓ ) Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-300
eBook - PDF- R. Prasad(Author)
- 2016(Publication Date)
- Cambridge University Press(Publisher)
Hess used this principle to give his law for chemical reactions. According to Hess’s law the change in standard enthalpy in a given chemical reaction is equal to the sum of the changes in standard enthalpies of all other chemical reactions that add up to the given reaction. Let DH A B C D + Æ + ( ) f represent the change in standard enthalpy of the reaction (A + B Æ C+D) and DH A B C D 1 1 1 1 + Æ + ( ) f , DH A B C D 2 2 2 2 + Æ + ( ) f , DH A B C D 3 3 3 3 + Æ + ( ) f , change in standard enthalpies, respectively, for reactions, A B C D 1 1 1 1 + Æ + ( ) ; A B C D 2 2 2 2 + Æ + ( ) and ( A B C D 3 3 3 3 + Æ + ), then DH A B C D + Æ + ( ) f = D D D H H H A B C D A B C D A B C D 1 1 1 1 2 2 2 2 3 3 3 3 + Æ + ( ) + Æ + ( ) + Æ + ( ) + + f f f , provided the reactions, ( ) ( ) ( ) A B C D A B C D A B C D 1 1 1 1 2 2 2 2 3 3 3 3 + Æ + + + Æ + + + Æ + = A B C D + Æ + This law is frequently used to compute the change in standard enthalpy of reactions for which the changes in standard entropies of the component reactions are known. Solved Examples 6. Using the data provided, calculate the change in the standard enthalpy for the reaction H O H O 2 2 liq gas gas ( ) Æ ( ) + ( ) 1 2 2 . Data: H O H O 2 gas gas gas ( ) Æ ( ) + ( ) 2 2 1 2 DH rea f = 41 kJ (1) H O H O 2 2 liq gas ( ) Æ ( ) DH rea f = 240 kJ (2) 376 Classical and Quantum Thermal Physics Solution: It may be observed that adding the two reactions given in the data one gets the reaction for which the change in standard enthalpy is required. H O H O H O H O H O H O 2 2 2 2 2 2 2 1 2 ( ) ( ) ( ) ( ) ( ) ( ) ( gas gas gas liq gas liq ga Æ + Æ Æ s gas ) ( ) + 1 2 2 O Now from Hess’s law DH desired reaction Δ = D D H H given reaction given reaction - Δ - Δ + 1 2 = 41 240 281 kJ kJ kJ + = The change in the standard enthalpy of the desired reaction = 281 kJ 7. Change in standard enthalpies for reactions marked (a) and (b) are respectively, DH a Δ = –162.0 kJ and DH b f = – 1012.0 kJ.
eBook - PDFChemistry
The Molecular Nature of Matter
- Neil D. Jespersen, Alison Hyslop(Authors)
- 2014(Publication Date)
- Wiley(Publisher)
The chief use of Hess’s law is to calculate the enthalpy change for a reaction for which such data cannot be determined experimentally or are otherwise unavailable. Often this requires that we manipulate equations, so let’s recapitulate the few rules that govern these operations. Rules for Manipulating Thermochemical Equations 1. When an equation is reversed—written in the opposite direction—the sign of ΔH ° must also be reversed. 4 2. Formulas canceled from both sides of an equation must be for the substance in identi- cal physical states. 3. If all the coefficients of an equation are multiplied or divided by the same factor, the value of ΔH ° must likewise be multiplied or divided by that factor. Hess’s Law Manipulating thermochemical equations Carbon monoxide is often used in metallurgy to remove oxygen from metal oxides and thereby give the free metal. The thermochemical equation for the reaction of CO with iron(III) oxide, Fe 2 O 3 , is Fe 2 O 3 (s) + 3CO( g) h 2Fe(s) + 3CO 2 ( g) ΔH ° = -26.7 kJ Example 6.8 Using Hess’s Law 4 To illustrate, the reverse of the equation, C(s) + O 2 ( g) h CO 2 ( g) ΔH ° = -394 kJ is the following equation. CO 2 ( g) h C(s) + O 2 ( g) ΔH ° = +394 kJ 282 Chapter 6 | Energy and Chemical Change Use this equation and the equation for the combustion of CO, CO( g) + 1 2 O 2 ( g) h CO 2 ( g) ΔH ° = -283.0 kJ to calculate the value of ΔH ° for the following reaction: 2Fe(s) + 3 2 O 2 ( g) h Fe 2 O 3 (s) Analysis: We are given two thermochemical equations and are asked to find ΔH ° for the third equation. We will need to manipulate the first two equations in order for them to give the third equation when they are added together. Assembling the Tools: The tools will be Hess’s law and the rules for manipulating thermochemical equations. Solution: In this problem, we cannot simply add the two given equations, because this will not produce the equation we want.
eBook - PDFGeneral Chemistry I as a Second Language
Mastering the Fundamental Skills
- David R. Klein(Author)
- 2015(Publication Date)
- Wiley(Publisher)
You will rarely encounter this, but you should keep it in mind. In a case where you do encounter this, the trans- formation of A (g) to A (l) would be considered as a separate reaction that would need to be added to give the sum total reaction. Now, you should be ready to practice more problems in your textbook. Good luck. . . . 5.5 HESS’S LAW 161 5.6 ENTHALPY OF FORMATION In the previous section, you saw situations where you were given the H° values for two or more reactions, and you had to manipulate those reactions to find H° for another reaction. But there is another common way to calculate H° for a re- action. To see how it works, let’s revisit the Lego analogy we saw earlier in this book. Let’s say that you have a Lego structure that is already built, and you want to take it apart and build a different structure using the exact same materials: 162 CHAPTER 5 ENERGY AND ENTHALPY 2 3 Notice that the final structure has all of the same pieces as the initial structure. In other words, this is a balanced reaction. Now we will combine this Lego analogy with Hess’s Law to understand the second important way to calculate enthalpy changes for reactions. Recall that Hess’s Law tells us that we can take any path to get to the products, and the total enthalpy change should be the same as if we took any other path. So, let’s say we want to calculate the enthalpy change for a reaction, for example: H 2 SO 4 SO 3 H 2 O H° ? If we know H° for breaking apart H 2 SO 4 into its elements (hydrogen, sulfur, and oxygen), and if we know H° for recombining those elements to form SO 3 and H 2 O, then we have all the information we need in order to calculate H° for the reaction above. Once again, we are using Hess’s Law, which tells us that any path that gets us from reactants to products will have the same H° as any other path that gets us to the end result.
eBook - PDF- Rose Marie O. Mendoza(Author)
- 2020(Publication Date)
- Arcler Press(Publisher)
Principles of Thermochemistry CHAPTER 2 CONTENTS 2.1. Introduction ...................................................................................... 30 2.2. Hess’s Law ........................................................................................ 36 2.3. Calorimetry ....................................................................................... 38 2.4. Laws of Thermochemistry .................................................................. 42 2.5. Systems And Processes ...................................................................... 46 2.6. State Function Definition .................................................................. 49 2.7. Enthalpy ........................................................................................... 50 2.8. Energy, Temperature Changes And Changes of State .......................... 52 2.9. Heat (q) ............................................................................................ 52 2.10. Conclusion ..................................................................................... 56 References ............................................................................................... 58 Thermochemistry 30 This section discusses the principles of thermochemistry: the various important concepts, laws, and principles related to thermochemistry. The chapter also provides discussion on heat transfer and temperature changes. It also provides techniques in determining specific heat capacity and its role in a chemical reaction. The chapter also throws some light on Hess’s law. The meaning of calorimetry and how do calorimeters work is also tackled as well as the laws of thermochemistry, the meaning of state function and some other important also discussed.
eBook - ePub- A.P.H. Peters(Author)
- 2010(Publication Date)
- CRC Press(Publisher)
When enthalpy was introduced, it was agreed that it was a function of state, such that changes of it did not depend on route. In the present example, it follows that ∆ H for Route I and that for Route II must be identical. That is to say that the overall change in H is equal to the sum of the values of ∆ H for each step. Thus, Δ H (I)=Δ H (IIa)+Δ H (IIb). Now Equation I represents formation of CO 2 and Equation IIa represents formation of CO. Values are: Δ f, H 0 (CO 2 (g))= − 395.5 kJ mo1 − 1. Δ f, H 0 (CO(g))= − 110.5 kJ mo1 − 1. Thus, (− 395.5)=(− 110.5)+Δ H (IIb). Δ H (IIb)= − 283.0 kJ mo1 − 1. That is, the combustion of CO to give gaseous CO 2 is accompanied by evolution of 283 kJ mol -1. This, the principle of constant heat summation, often known as Hess’s law, is thus seen to lead directly from the fact that H is a function of state. This idea is immensely powerful, because it enables Δ H 298 0 values to be determined for any reaction, as long as the enthalpies of formation are known for each reactant and product. Consider an equation of the general form: A+B... → P+Q... for which we wish to know ∆ H 0. Imagine this to take place in two stages, first the “unformation” of A, B, etc., into their elements, then the formation of P, Q, etc., from those same elements. That is, A+B... → Δ H 2 0 (elements in standard states) → Δ H 2 0 P+Q.... Now Δ H 1 0 = − [ Δ f H 0 (A)+Δ f H 0 (B)+... ] (unformation) and Δ H 2 0 = [ Δ f H 0 (P)+Δ f H 0 (Q)+... ] (formation). Thus, for the overall. reaction: Δ H 0 = Δ H 1 0 +Δ H 2 0 = [ Δ f H 0 (P)+Δ f H 0 (Q) ] − [ Δ f H 0 (A)+Δ f H 0 (B) ]. This may be written in the abbreviated form: Δ H 0 = ∑ [ Δ f H 0 (products) ] − ∑ [ Δ f H 0 (reactants) ]. (2.10) This general equation is seen to fit the prototype Equation 1.1, and will be used often. Example 2.6 Fluorine has been considered for use as an oxidant in rocket propellants. Its high reactivity ensures high energy release, and consequently high thrust
eBook - PDF- David Ball(Author)
- 2014(Publication Date)
- Cengage Learning EMEA(Publisher)
In Example 2.18 above, the state function of interest was enthalpy, and by applying Hess’s law and the definition of formation reactions, we were able to develop a procedure for determining the changes in enthalpy and internal energy for a chemical process. What is the relationship between D H and D U for a chemical reaction? If one knows the D f U and D f H values for the products and reactants, one can simply com-pare them using the products-minus-reactants scheme of equations 2.55 and 2.56. There is another way to relate these two state functions. Recall the original defini-tion of H from equation 2.16: H 5 U 1 pV We also derived an expression for dH as dH 5 dU 1 d 1 pV 2 dH 5 dU 1 p dV 1 V dp where the second equation above was obtained by applying the chain rule. There are sev-eral ways we can go with this. If the chemical process occurs under conditions of con-stant volume, then the p dV term is zero and dU 5 dq V (because work 5 0). Therefore, dH V 5 dq V 1 V dp (2.57) The integrated form of this equation is D H V 5 q V 1 V D p (2.58) Because dU 5 dq under constant volume conditions, this gives us one way to calculate how dH differs from dU. Under conditions of constant pressure, the equation becomes D H p 5 D U 1 p D V (2.59) and we have a second way of evaluating D H and D U for a chemical process. If the chemical process occurs isothermally, then by assuming the gases involved are acting ideally, d 1 pV 2 5 d 1 nRT 2 5 dn # RT where dn refers to the change in the number of moles of gas that accompanies the chemical reaction. Because both R and T are constant, the chain rule of calculus In expressions like these, it is important to keep track of all of the negative signs. Evaluating: D rxn H ° 5 2 2799 kJ By noting that the coefficients from the balanced chemical reaction are the num-ber of moles of products and reactants, we lose the moles in the denominator of the D rxn H ° .
- Peter V. Hobbs(Author)
- 2000(Publication Date)
- Cambridge University Press(Publisher)
2 Chemical thermodynamics Heat can be released or absorbed during a chemical reaction. This pro-vides a powerful method for studying chemical equilibrium by means of chemical thermodynamics. Thermodynamics is based on a few funda-mental postulates, called the first, second, and third laws of thermo-dynamics. We will discuss these laws first, and then apply them to chemical equilibria. 2.1 The first law of thermodynamics; enthalpy In addition to the macroscopic kinetic and potential energy that a body or system as a whole may possess, it also contains internal energy due to the kinetic and potential energy of its molecules or atoms. Increases in internal kinetic energy in the form of molecular motions are manifested as increases in the temperature of the system, while changes in the poten-tial energy of the molecules are caused by changes in their relative configurations. Let us suppose that a system of unit mass takes in a certain quantity of heat energy q (measured in joules). As a result, the system may do a certain amount of external work w (also measured in joules). The excess energy supplied to the system, over and above the external work done by the system, is q - w. Therefore, if there is no change in the macro-scopic kinetic and potential energy of the system, it follows from the principle of conservation of energy that the internal energy of the system must increase by q -w. That is, q - w = u 2 -Mi (2.1) where u x and u 2 are the internal energies of a unit mass of the system before and after the change. In differential form Eq. (2.1) becomes 17 18 Chemical thermodynamics dq-dw = du (2.2) where dq is the differential increment of heat added to a unit mass of the system, dw the differential increment of work done by a unit mass of the system, and du the differential increment in internal energy of a unit mass of the system. Equations (2.1) and (2.2) are statements of the first law of thermodynamics.
eBook - ePubChemical Thermodynamics
Theory and Applications
- W.J. Rankin(Author)
- 2019(Publication Date)
- CRC Press(Publisher)
4 The first lawIn this house, we obey the laws of thermodynamics!Homer Simpson (‘The PTA Disbands’ episode of The Simpsons TV series)Scope
This chapter explains the first law of thermodynamics and its implications for chemical thermodynamics.Learning objectives
- Understand quantitatively the meaning of the first law of thermodynamics.
- Understand the concepts of the internal energy, enthalpy and heat capacity of substances.
- Be able to calculate the enthalpy of a substance from its heat capacity.
- Be able to calculate the enthalpy change of a system when substances react.
- Understand how heat capacity and enthalpy are determined experimentally.
4.1 Introduction
Historically, the discipline of thermodynamics developed in order to better understand the relationships between heat and mechanical energy, particularly in steam engines. It was only later that the principles so developed were applied to chemical systems, in particular to the relationships between heat, temperature, pressure and the composition of systems. This is the domain of chemical thermodynamics. There are two main aspects to chemical thermodynamics. The first is concerned with the thermal effects associated with changes in temperature, pressure and chemical composition. These relations are derived from the first law of thermodynamics and are the subject of this chapter. The second aspect is concerned with the tendency for a change in a system (for example, a chemical reaction or physical transformation) to take place, the extent to which it will take place, the equilibrium composition of a system and the effects of temperature and pressure on the equilibrium position. These relations are derived from the second law of thermodynamics and are discussed in Chapter 7 onwards.4.2 The first law
Development of the understanding that led to the first law of thermodynamics occurred over many years. An early important development was the realisation that heat and mechanical energy are related. In 1798, at the arsenal in Munich, Benjamin Thompson, later Count Rumford,*
eBook - ePubElectrochemical Energy Storage
Physics and Chemistry of Batteries
- Reinhart Job(Author)
- 2020(Publication Date)
- De Gruyter(Publisher)
U with respect to the volume at constant pressure is not zero, and it is called the internal pressure of the system (or of the gas). This internal pressure is a measure of the cohesive forces within a system.Similar considerations can be done with regard to the enthalpy. In case of an ideal gas, the partial derivate of the enthalpy with respect to the pressure at constant temperature is zero. Only if interaction forces are existing between the gas particles, a change of the gas pressure can cause a change of the enthalpy.2.2.3 Second law of thermodynamics
As briefly described in the last section, the first law of thermodynamics deals with the change of the internal energy upon a transition between two states of a system. In this context, theoretically, the two types of energy named heat and work can be completely converted into each other. According to the first law of thermodynamics, such a conversion is not dependent on the direction of the chemical or physical process, that is, the conversion of heat into work and vice versa could occur in any order – there are no restrictions on base of the first thermodynamic law. However, in nature, real chemical or physical processes occur only in a certain direction.For example, if during a thunderstorm a roof tile is falling to the ground, that is, potential energy is converted to kinetic energy that in turn is finally converted to heat, after the tile came to a stop on the ground. According to the first law of thermodynamics, the reversal process could be also possible: A tile lying on the ground could absorb heat from the surroundings that subsequently could be transformed to potential energy, that is, the tile could spontaneously jump from the ground to the roof. Of course, such a process does not occur in nature. Perpetual motion machines of the second kind – that are machines which spontaneously convert thermal energy into mechanical work – are impossible. Hence, the thermodynamic laws have to be extended; and the second law of thermodynamics comes into play. This law describes the elotrophy or directionality that cannot be explained by the first thermodynamic law.
eBook - PDFChemistry
Structure and Dynamics
- James N. Spencer, George M. Bodner, Lyman H. Rickard(Authors)
- 2011(Publication Date)
- Wiley(Publisher)
7.8 THE ENTHALPY OF A SYSTEM 281 that occurs in the steel reaction vessel (¢E sys ) is measured by determining the change in the temperature of the water in the insulated container that surrounds the reaction vessel (¢E surr ). Because the reaction is run in a sealed container at constant volume, no work can be done. As a result, the change in the energy of the system (¢E sys ) must have the same magnitude but the opposite sign of the change in the energy of the surroundings (¢E surr ). By monitoring the change in temperature of a water bath that surrounds the reac- tion container, we can calculate the change in the energy of the surroundings, ¢E surr . This is equal in magnitude and opposite in sign to the change in the energy of the system that occurs during the chemical reaction. As noted in Section 7.4, thermodynamicists use the symbol q to represent the heat transferred from the system to its surroundings or vice versa. Because no work is done when the reaction is run in a calorimeter at constant volume, the heat given off or absorbed in a chemical reaction is equal to the change in the energy of the system. Chemists, however, seldom work in sealed containers at constant volume. They usually carry out reactions in containers such as beakers or flasks that are open to the atmosphere. These reactions occur under conditions of constant pres- sure. When the volume of a system can change, not all of the heat supplied to or taken from the system goes into producing a temperature change; some of the heat is used to do work. When describing a system at constant pressure, chemists therefore look for changes in the enthalpy of the system, ¢H sys . Enthalpy, like energy, is a state function. The change in enthalpy that occurs during a chemical reaction at constant pressure is exactly equal to the heat given off or absorbed by the reaction. A constant-pressure calorimeter can be used to collect data to calculate the value of ¢H for a reaction.
eBook - PDF- Arther Adamson(Author)
- 2012(Publication Date)
- Academic Press(Publisher)
4 -6 ENTHALPY. AN ALTERNATIVE FORM OF THE FIRST LAW 115 4-6 Enthalpy. An Alternative Form of the First Law The first law takes on a particularly simple form for an isochoric process, in that dE is simply C v dT. Actual experiments are more often done under constant-pressure conditions, and the following alternative statement of the first law is very useful. We define a new molar quantity Η called the enthalpy: Η = E+ PV. (4-25) Since the volume and pressure of a system are determined solely by its state, as is E, Η must also be a state function. The differential form of Eq. (4-25) is dH = dE+ PdV + VdP (4-26) and, using Eq. (4-24), we have dH = 8q + VdP. (4-27) We then have the following special cases, corresponding to the standard processes. (1) Adiabatic Since q = 0, it follows that dH = VdP. (4-28) (2) Isochoric dH = C v dT+ V dP. (4-29) (3) Isobaric Since dP is zero, 3H dH = C P dT [or (-^r) p = C P ] . (4-30) Since Η is a state function, its value depends on those of the state variables, Ρ, V, and T, any one of which can be eliminated by means of the equation of state of the system. In the case of E, it is natural to pick V and Τ as the independent variables; in the case of H, the natural choices are Ρ and T. Since dH is an exact differential, we have Λ -(ΐ) Γ +(ττ), Λ ' (4 31 > and 3P _ (dH/dT) P ^T ) H (dH/dP) 7 (4-32) The introduction of the enthalpy function allows the following derivation of an important relationship for the difference between C P and C v : C r -C y -(-Βτ or, from the definition of H, Eq. (4-25), 116 CHAPTER 4: CHEMICAL THERMODYNAMICS. THE FIRST LAW OF THERMODYNAMICS This equation is discussed further in the Commentary and Notes section. 4-7 Applications of the First Law to Ideal Gases A. Internal Energy Change for an Ideal Gas The application of the first law to the case of an ideal gas not only provides simple illustrations of its use but turns out to be of considerable importance generally.
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