University Of Mauritius

BEng (Hons.) Chemical and Environmental Engineering - E400

Chemistry &Basic

Chemical EngineeringCHE 1001Y (3)Assignment 3:Explain the Hesss Law while calculating heat of reaction.Tutor: Dr D. SurroopName: Fanilo Antra Tia Razafindralambo

Index number: 1312636TABLE OF CONTENTS32.INTRODUCTION

33.APPLICATION OF HESSS LAW

3Evaluating Enthalpy changes

3Calculating Heat of Reaction

3Finding Lattice Energy

34.CONCLUSIONS

35.REFERENCES

1. INTRODUCTIONGermain Henri Hess(1802 - 1850) is important primarily for his thermochemical studies.Hess' Lawstates thatthe heat evolved or absorbed in a chemical process is the same whether the process takes place in one or in several steps. This is also known asthe law of constant heat summation. According to this law, it is permissible to write stoichiometric equations, together with the enthalpy changes, and to treat them as mathematical equations, thereby obtaining thermochemically valid result.

2. APPLICATION OF HESSS LAWFew reactions occur on a simple manner, following a simple chemical equation, with the result that the enthalpy changes corresponding to a simple chemical equation cannot be measured directly. For many of these the enthalpy changes can be calculated from the values for other reactions by making use of Hesss Law. To illustrate Hess's law, the simplified thermal equations and the energy level diagrams are shown below:

For example, suppose that a substance A reacts with B according to the equation

1. A + B ( X DH1= -10kJ mol-1Suppose that X reacts with an additional molecule of A to give another product Y:

2. A + X ( Y DH2= -20 kJ mol-1Then according to Hesss Law, it is permissible to add these two equations and obtain:

3. 2A + B ( Y DH3= DH1 + DH2 = -30 kJ mol-1DHn= Enthalpy change for reaction n.

The law follows at once from the Principle of conservation of Energy and from the fact that enthalpy is a state function. Thus, if reactions 1 and 2 occur, there is a net evolution of 30 kJ, when 1 mole of Y is produced. In principle we could reconvert Y into 2A + B by the reverse of reaction 3. If the heat required to do this were different from 30 kJ, we should have obtained the starting materials with a net gain or loss of heat, and this would violate the principle of conservation of Energy.

Evaluating Enthalpy changes

The majority of reactions and processes that we encounter will be conducted at constant pressure, Enthalpy describes the energy changes at constant pressure, represented with the symbol H. Although we are interested we are interested in enthalpy changes during chemical processes, the knowledge of how much heat a reaction needs-or how much it evolves-can be much more significant (at a larger scale).As a straightforward example, let us consider the oxidation of carbon monoxide (at room conditions). We could conduct the reaction directly and measure DH.

(1) CO(g) + O2(g) ( CO2(g) DH1= -283.0 kJ mol-1Alternatively, the reaction of graphite with oxygen could be carried out under two different stoichiometries:

(2) C(graph.) + O2(g) ( CO(g) DH2= -110.5 kJ mol-1(3) C(graph.) + O2(g) ( CO2(g) DH3= -393.5 kJ mol-1Hesss Law allows us to treat the chemical equations and the enthalpy changes in the same way as algebraic equations. Subtracting (2) from (3) gives the equivalent of (1) and leads to the same result for DH.[C(graph.) + O2(g)] - [C(graph.) + O2(g)] ( [CO2(g)] [CO(g)]

CO(g) + O2(g) ( CO2(g)

DH = (-393.5) - (-110.5)= -283.0 kJ mol-1Or DH3 = DH1 + DH2Another method of displaying these enthalpy changes and performing the calculations is to use graphical methods of the energy levels. This method has a number of important applications in chemistry. The most obvious consequences of Hesss law (and the First Law of Thermodynamics) is that, for a reaction, DHforward= -DHreverse. A more important use is that, the calculation of enthalpy changes for reactions which would be experimentally difficult or even impossible to carry out in practice,Calculating Heat of ReactionThe enthalpy of reaction is sometime called heat of reaction, especially for combustion reactions. It is the heat evolved when a combustion reaction occurs.

As an example, let us consider the enthalpy change when methane is formed from both its elements, both reactants and product being in their standard states. This is called the Standard enthalpy of Formation and written DHf0.C(s) + 2H2(g) ( CH4(g)

The standard state of each element is the most stable form at 1 atm and the temperature specified (usually 298 K).

The direct reaction cannot conveniently be carried out, it is relatively easy to measure the enthalpy of combustion of methane in an apparatus called a flame calorimeter.

As DH = (q) (heat), when methane is burnt with oxygen the heat produced gives the enthalpy of combustion directly (Heat of reaction).

CH4(g) + O2(g) ( CO2(g) + 2H2O(l); DH = - 890.4 kJ mol-1C(s) + O2(g) ( CO2(g); DH= - 393.5 kJ mol-1H2(g) + O2(g) ( 2H2O(l); DH= -571.6 kJ mol-1Combining the enthalpy of combustion of methane with that of carbon and hydrogen enables us to apply Hesss law to the problem. We can see that D Hf0 = DH1 DH2, in the figure below.

DH1 is the enthalpy of combustion of one molecule of carbon and two moles of hydrogen. Therefore

DHf0(CH4) = [- 393.5 571.6 + 890.4] kJ mol-1=- 74.7 kJ mol-1Note: If reverse chemical reaction is considered, the enthalpy change reverses its sign.Finding Lattice Energy

Many substances are composed of ions. These exist in a three-dimensional array called Lattice. The Lattice Energy is a measure of the energetic stability of the crystal with respect to the ions from which it is made when they are present in gaseous state.The procedure of finding lattice energy using Hesss law is called Born-Haber cycle. This method makes use of energy level diagrams for better understanding.

We choose Potassium fluoride, KF, as an example illustrated below

Energy/ kJ mol-1-600

-500

-400

-300

-200

-100

-0

- -100

- -200

- -300

- -400

- -500

- -600

This method simplifies significantly the calculation of lattice energy. The principle is easy, positive enthalpy changes move upwards while negative value downwards. Knowing each value of enthalpy from tables, we can find the value of lattice energy with minimal effort.Or simply by using - (lattice energy) = - (heat of formation)

+heats of atomization

+ionization energies

+electron affinities3. CONCLUSIONS

We can, therefore, conclude that Hesss law is an important consideration in thermochemistry, whether in organic or non-organic. It can be used to find (a) The heat of reaction (b) The Enthalpy of reaction (c) Lattice energy and even more.

4. REFERENCES

1) Graham Hill, J. H., 1995. Chemistry in Context. 4th ed. Surrey: Thomas Nelson & Sons Ltd.2) Keith Laidler, J. M., 1995. Physical Chemistry. 2nd ed. New Jersey: Houghton Mifflin Company.

3) Matthews, P., 1997. Advanced Chemistry. Cambridge: Cambridge University Press.

4) Price, G., 1998. Thermodynamics of Chemical Processes. New York: Oxford University Press.

5) Smith, B., 2004. Basic Chemical Thermodynamics. 5th ed. London: Imperial College Press.

+ 2O2

DH

CH4(g)

DHf0

CO2(g) + 2H2O(l)

+ 2O2

DH1

C(s) + 2H2(g)

K+(g) + F-(g)

Atomisation of F

Electron affinity of fluorine

K+(g) + F2 (g)

Ionisation energy of potassium

K+(g) + F-(g)

K+(g) + F-(g)

K(g) + F2 (g)

Atomisation of K

K(s) + F2 (g)

K(s) + F2 (g)

Lattice energy

Lattice energy

Heat of formation

Heat of formation

KF(s)

KF(s)

(b)

(a)