Hess’s law states that if we add two or more thermochemical equations to yield a final equation for the reaction, and then the sum of the enthalpy changes for the individual reactions 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.
S(S) + 3O2 (g) → SO2 (g) ∆ H=-297kJ
2SO3 (g) →2SO2 (g) + 3O2 (g ∆ H=+198kJ
Firstly, chemical equations are required to have substances contained 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
(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:
In summary, the sum of two equations gives: 2S(S) + 3O2 (g) 2SO3 (g) ∆ H=-792kJ
Standard Enthalpy of Formation
With the use of 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 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.
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)