Hess’s Law of constant heat summation

Hess’s law states that if we add two or more thermochemical equations to yield a final equation for the reaction, then the sum of the enthalpy changes for the individual reactions is greater than the enthalpy change for the final reaction. For example, Hess’s law can be applied in calculating the enthalpy change for the production of SO3.

2S(S) + 3O2 (g)   2SO3 (g)     H=?


Consider the following equations

  1. S(S) + 3O2 (g)           SO2 (g)              H=-297kJ

  2.  2SO3 (g)           2SO2 (g) +  3O2 (g                               H=+198kJ

  3. Firstly, chemical equations are required to have substances that are found in the desired equation and have the enthalpy values that are known to us. From equation (a), two moles of sulfur reacted to yield two moles of SO3 (g). This implies that equation (i) must be multiplied by two together with the corresponding enthalpy since energy will be released twice the original amount. Therefore equation (i) becomes  2S(S) + 3O2 (g)        2SO2 (g)           H=-594kJ                             

  4. (iii)Equation (ii) on the other hand is reversed because we are interested with determining the H in which SO3is the product but not a reactant. 2SO2 (g) + O2 (g)   2SO3 (g)        H=-198kJ         

Lastly, we find the sum of equation (iii) and (iv) so that we obtain the desired equation. Remember to cancel any like terms on RHS and LHS shown below:

            2S(S) + 3O2 (g)           2SO2 (g)                                                                     H=-594kJ

            2SO2 (g) + O2 (g)           2SO3 (g)                                                               H=-198kJ

           2SO2 (g) + 3O2 (g) +             2SO2 (g) + 2SO3 (g)                                   H=-792Kj

In summary, the sum of two equations gives: 2S(S) + 3O2 (g)        2SO3 (g)        H=-792kJ



Standard Enthalpy of Formation

By using Hess’s law, we can compute the unknown value of H. The standard state of a substance refers to the normal physical state of the substance that is measured at a specific atmospheric pressure .  Usually, the H value for reactions is called the standard heat formation of the compound. 


The standard heat of formation

The standard heat of formation refers to the change in the heat that accompanies the formation of one mole of the compound in its standard state through the constituent elements.

For example,

2S(S) + 32 O2 (g)          2SO3 (g)     Hf0=-396kJ is the typical standard heat of formation of one mole of SO3.

The standard heat of formation is the difference between the standard heat of formation of all the reactants and products.


Heat of solution

The molar heat of solution refers to the heat change caused by the dissolution of one mole of substances. It is usually denoted by Hsoln.  A good example is dissolving sodium hydroxide pellets in normal water. When one mole of NaOH is dissolved in water, the solution becomes so hot that it steams. The heat from this process is released as sodium ions and hydroxide ions in a separate and dissociate manner. Simultaneously, the temperature of the solution increases, releasing approximately 445-kilo joules of heat as the molar heat of solution.


NaOH (aq)    Na+ (aq) + OH-(aq)                                                   H=-445kJmol-




Calculate the enthalpy change for the process CCl4 (g)        C (g) + 4Cl (g) and calculate the bond enthalpy of  C-Cl in CCl4 (g)

. vapH0 (CCl4) =30.5 kJmol-1

fH0 (CCl4) = -135.5 kJmol-1

aH0 (C) = 715.0 kJmol-1

aH0 (Cl) = 242 kJmol-1




  1. CCl4 (l)             CCl4 (g)        vapH0 (CCl4) =30.5 kJmol-1

  2. C(S)            C(g)   aH0 (C) = 715.0 kJmol-1

  3. Cl2(g)            2Cl(g)         aH0 (Cl )= 242 kJmol-1

  4. C(g)    + 4Cl (g)               CCl4 (g)        fH0 (CCl4) = -135.5 kJmol-1

Enthalpy change for the given process can be calculated using algebraic calculation as follows:

Equation (ii) + 2 Χ  equation (iii) – equation (i) – equation (iv)

= (715.0 kJmol-1)+2(242 kJmol-1)-( 30.5 kJmol-1)-( -135.5 kJmol-1)

  H = 1304 kJmol-1

Bond enthalpy of C-Cl bond in CCl4 (g) becomes 13044  kJmol-1

  = 326 kJmol-1


Enthalpy of combustion

Enthalpy of combustion refers to the amount of heat released or absorbed when one mole of a substance undergoes combustion in the presence of oxygen. Consider the combustion of ethane.

C2H4(g)           2CO2(g)  + 2H2O(l)              H= -1390kJmol-1


Enthalpy of neutralization

Enthalpy of neutralization refers to the amount of heat energy released or absorbed when one equivalent of an acid is completely neutralized by one equivalent of the base.





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