To get the best deal on Tutoring, call 1-855-666-7440 (Toll Free)
Top

Thermochemistry

It is well known that nearly all chemical reactions are accompanied by energy changes. These changes appear ordinarily in the form of evolution or absorption of heat.

The energy given out during a chemical change appears in the form of heat. While that which is absorbed may be in the form of thermal electrical or photo energy. the amount of energy evolved or absorbed during a chemical change always remains same for the same quantities of reacting substances. All thermochemical reactions are governed by two laws, which are given below.

 

Thermochemistry Definition

Back to Top
Thermochemistry is defined as "the branch of chemistry which deals with the study of energy changes accompanying chemical reactions."
Thermochemistry is based on the first law of thermodynamics. The energy changes in chemical reactions are generally due to the breaking up of existing bonds between the atoms and the formation of new bonds. Thus, thermochemistry provides important information regarding bond energies.

Thermochemical Equations

Back to Top
An equation which indicates the evolution or absorption of heat in the reaction or process is called a thermochemical equation. For example, the following equation

C + O2 CO2 + 393.5kJ

reveals that when carbon burns in oxygen to form carbon dioxide, 393.5kJ of heat are set free per one mole of carbon dioxide produced. Similarly the equation

C + 2S CS2 - 92.0kJ

reveals that one mole of carbon combines with 2 moles of sulfur to form one mole of carbon disulphide with the absorption of 92.0kJ of heat. Thus the equations written above are the thermochemical equations.

Laplace Law

Back to Top

A.L.Lavoisier and P.S.Laplace gave this law in 1780 which states that "the enthalpy of a reaction is exactly equal but opposite in sign for the reverse reaction."

For example, if DH is the enthalpy change in going from A to B then the enthalpy change for the process B to A would be -DH. Thus, the enthalpy of formation of a compound is numerically equal but opposite in sign to the enthalpy of decomposition of the compound.

S(s) + O2(g) $\rightarrow$ SO2(g) $\Delta$H = -296.9 kJ

SO2(g) $\rightarrow$ S(s) + O2(g) $\Delta$H = +296.9 kJ

Whenever a thermochemical equation is reversed the sign of DH also gets reversed.

G.H.Hess proposed a law regarding the heat or enthalpies of reaction in 1840 called the Hess's law. This law states that "the heat change in a particular reaction is the same whether it takes place in one step or several steps."

For example, a reactant 'A' changes to a product 'B' in one step and the heat change during this process is DH. If the reaction is carried out in two steps where 'A' first changes to 'C' an intermediate stage and then 'C' changes to 'B' in the following step then let the heat change during the formation of 'A' to 'C' be DH1 and that from 'C' to 'B' be DH2. From Hess's law the heat change for the reaction is given as

DH = DH1 + DH2

Hess's Law

This means that the amount of heat evolved or absorbed in a chemical reaction depends only upon the energy of the initial reactants and the final products. The heat change is independent of the path or the manner in which the change has taken place.

The formation of carbon dioxide from carbon and oxygen can be illustrated as follows. Carbon can be converted into carbon dioxide in two ways. Firstly solid carbon combines with sufficient amount of oxygen to form CO2. The same reaction when carried out in the presence of lesser amount of oxygen, gives carbon monoxide which then gets converted to CO2 in step two, in the presence of oxygen.


DH = DH1 + DH2

Thus, one can conclude that thermochemical equations can be added, subtracted or multiplied like algebraic equations to obtain the desired equation.

→ Read More

Application of Hess's Law

Back to Top

Hess's law has been useful in determining the heat changes of reactions, which cannot be measured directly with a calorimeter. Some of its applications are

1. Determination of Heat of Formation

Compounds whose heats of formation cannot be measured directly using colorimetric methods because they cannot be synthesized from their elements easily e.g. methane, carbon monoxide, benzene etc are determined using Hess's Law.

For example, the heat of formation of carbon monoxide can be calculated from the heat of combustion data for carbon and carbon monoxide as shown above.

2. Determination of Heat of Transition

The heats of transition of allotropic modification of compounds such as diamond to graphite, rhombic sulphur to monoclinic sulfur, yellow phosphorous to red phosphorous etc. can be determined using Hess's Law.

For example, the heat of transition of diamond to graphite can be calculated from the heat of combustion data for diamond and graphite, which is -395.4 kJ and -393.5 kJ respectively.

The thermochemical equations showing the combustion reaction of diamond and graphite are

thermochemical equations showing the combustion reaction of diamond
thermochemical equations showing the combustion reaction of graphite
The conversion that is required is

This can be obtained by subtracting the second equation from the first one.
the heat of transition of diamond to graphite


3. Determination of Heat of Formation

The heat of hydration of substances is calculated using Hess's law.

For example the heat of hydration of copper sulphate can be calculated from the heat of solution of anhydrous and hydrated salts of copper. The heat of solution of CuSO4 and CuSO4.5H2O are -66.5 and -11.7 kJ mol-1. The corresponding thermochemical equations are:
 heats of solution of CuSO4
 heats of solution of CuSO4.5H2O

The process of hydration can be expressed as

process of hydration of copper sulphate

According to Hesss law, DH1 = DH + DH2

DH = DH1 - DH2

= - 66.5 11.7 = - 78.2 kJ/mol

4. Determination of Heat of Various Reactions


Hess's law is useful in calculating the enthalpies of many reactions where direct measurement is difficult or impossible.

Thermochemistry Problems

Back to Top
Below you could see problems

Solved Examples

Question 1: Calculate the standard heat of formation of carbon disulphide (l). Given that the standard heats of combustion of carbon (s) sulfur (s) and carbon disulphide (l) are 393.3, -293.72 and -1108.76kJ mol-1respectively.
Solution:
 
The given data can be written in thermochemical equation form as







The required equation is



Multiplying equation (ii) by 2 and adding to equation (i) we get,



Subtracting equation (iii) from the above equation we have




 

Question 2: Calculate lattice energy for the change,



Given that DHsubl. of Li = 160.67 kJ mol-1, DHDissociation of

Cl2 = 244.34 kJ mol-1, DHionisation of Li(g) = 520.07 kJ mol-1,
DHE.A of Cl(g) = - 365.26 kJ mol-1, DHof of LiCl(s) = - 401.66 kJ mol-1.


Solution:
 
Considering the different changes that occur in the formation of solid lithium chloride based on the data given the lattice energy of the above can be constituted as





or



= - 839.31 kJ mol-1

 

More topics in Thermochemistry
Internal Energy Bond Enthalpy
Bond Dissociation Energy Enthalpy of Combustion
NCERT Solutions
NCERT Solutions NCERT Solutions CLASS 6 NCERT Solutions CLASS 7 NCERT Solutions CLASS 8 NCERT Solutions CLASS 9 NCERT Solutions CLASS 10 NCERT Solutions CLASS 11 NCERT Solutions CLASS 12
Related Topics
Chemistry Help Chemistry Tutor
*AP and SAT are registered trademarks of the College Board.