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Enthalpy Of Formation, Thermochemical Equations And Hess's Law

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Enthalpy Of Formation, Thermochemical Equations And Hess's Law - Lesson Summary

The enthalpy change associated with the formation of one mole of a compound from its constituent elements, all substances being in their standard states, that is, at two hundred and ninety eight Kelvin and one bar pressure is called the standard molar enthalpy of formation, it is represented ΔfH°.

               

      Substance     Δf H° / (KJmol-1)       C 3H 8(g)       -103.85       n-C 4H 10(g)       -126.15       HgS(s) red       -58.2       H 2(g)        0       H 2O(g)        -241.82       H2O(l)        -285.83       HF(g)        -271.1       HCl(g)        -92.31       HBr(g)        -36.40         HI(g)        +26.48       KCl(s)        -436.75       KBr(s)        -393.8       MgO(s)        -601.70       Mg(OH) 2(s)        -924.54       NaF(s)        -573.65       NaCl(s)        -411.15       NaBr(s)       - 361.06

Ex: When carbon and oxygen in their elemental states react to form 1 mole of carbon dioxide, 395 kJ of heat is produced. Hence the standard enthalpy formation of gaseous carbon dioxide is -395 kJ per mole.

By convention, the standard enthalpy of formation of all elements is assumed to be zero.

The amount of heat required to decompose calcium carbonate to lime and carbon dioxide, given that all the substances are in their standard state.

CaCo3(s)    ∆ →   CaO(s) + CO2(g)

ΔrH = ΔHproducts - ΔHreactants
            
(CaO + CO2)     (CaCO3)
      = (- 635.09 + - 393.51) - (- 1206.92)
      = (- 1028.6) - (- 1206.92)

 ΔrH = + 178.3 kj mol-1

A balanced chemical equation which indicates the amount of heat evolved or absorbed in a chemical reaction, is known as a thermo chemical equation.

For exothermic reactions ΔH is negative and for endothermic reactions it is positive.

The physical states of the reactants and products must be mentioned in the equation. This is because the physical states of the reactants and products cause appreciable difference in the value of ΔH.

Ex: The heat of formation of one mole of water in the liquid state is -85.8 kJ, while it is -241.8 kJ for water in the gaseous state.

The coefficients in the balanced thermo chemical equation refer to the number of moles of the substances involved in the reaction and the numerical value of ΔHr corresponds to these coefficients.

In the year 1840, a Russian Chemist, G.H. Hess performed certain experiments and stated that the energy change that occurs in a chemical reaction is independent of substance changes from one state to another.

Hess' law states that, if a chemical change can be made to take place in two or more different ways whether in one step or more than one step, the amount of total heat change is the same, no matter by which method the change is brought about.

When sulphur in its elemental state burns to form sulphur trioxide in one step, the heat evolved during the chemical equation is -395.4 kJ per mole.

Break the reaction into two steps. In the first step, sulphur reacts with oxygen to form sulphur dioxide. The heat evolved during the chemical equation is -297.5 kJ per mole.

In the second step, sulphur dioxide reacts with oxygen to form sulphur trioxide. The heat evolved during the chemical equation is -97.9 kJ per mole.

On addition of the heat evolved from both steps obtained -395.4 kJ mole.

   S(Rhombic(s)) + 3 2 O2(g) → SO3(g); ΔrH° = -395.4 KJ mol-1
 Step 1:
      S(Rhombic(s)) +O2(g) → SO2(g); ΔrH1 = -297.5 KJ mol-1
Step 2:
      SO2(g) + 1 2 O2(g) → SO3(g); ΔrH2 = -97.9 KJ mol-1

ΔrH = ΔrH1 + ΔrH2
       
= (-297.5) + (- 97.9)
ΔrH = -395.4 kj mol-1  

The amount of heat evolved through the two steps is the same as the amount of heat evolved in one step, thus proves Hess's law.

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