JEE Main & Advanced Chemistry Thermodynamics / रासायनिक उष्मागतिकी Laws Of Thermochemistry

(1) Levoisier and Laplace law : According to this law enthalpy of decomposition of a compound is numerically equal to the enthalpy of formation of that compound with opposite sign, For example,

\[C(s)+{{O}_{2}}\to C{{O}_{2}}(g);\Delta H=-94.3\,kcal\]

\[C{{O}_{2}}(g)\to C(s)+{{O}_{2}}(g);\,\Delta H=+94.3kcal\]

(2) Hess's law (the law of constant heat summation) : This law was presented by Hess in 1840. According to this law “If a chemical reaction can be made to take place in a number of ways in one or in several steps, the total enthalpy change (total heat change) is always the same, i.e. the total enthalpy change is independent of intermediate steps involved in the change.” The enthalpy change of a chemical reaction depends on the initial and final stages only. Let a substance A be changed in three steps to D with enthalpy change from A to B, \[\Delta {{H}_{1}}\] calorie, from B to \[C,\,\Delta {{H}_{2}}\] calorie and from C to \[D,\,\Delta {{H}_{3}}\] calorie. Total enthalpy change from A to D will be equal to the sum of enthalpies involved in various steps. Total enthalpy change \[\Delta {{H}_{\text{steps}}}=\Delta {{H}_{1}}+\Delta {{H}_{2}}+\Delta {{H}_{3}}\]

Now if D is directly converted into A, let the enthalpy change be \[\Delta {{H}_{\text{direct}}}.\] According to Hess's law \[\Delta {{H}_{\text{steps}}}+\Delta {{H}_{\text{direct}}}=0,\] i.e. \[\Delta {{H}_{\text{steps}}}\] must be equal to \[\Delta {{H}_{\text{direct}}}\] numerically but with opposite sign. In case it is not so, say \[\Delta {{H}_{\text{steps}}}\](which is negative) is more that \[\Delta {{H}_{\text{direct}}}\](which is positive), then in one cycle, some energy will be created which is not possible on the basis of first law of thermodynamics. Thus, \[\Delta {{H}_{\text{steps}}}\] must be equal to \[\Delta {{H}_{\text{direct}}}\] numerically.

(i) Experimental verification of Hess's law

(a) Formation of carbon dioxide from carbon

First method : carbon is directly converted into \[C{{O}_{2}}(g).\]

\[C(s)+{{O}_{2}}(g)=C{{O}_{2}}(g);\,\,\Delta H=-94.0\,kcal\]

Second method : Carbon is first converted into \[CO(g)\] and then \[CO(g)\] into \[C{{O}_{2}}(g)\], i.e. conversion has been carried in two steps,

\[C(s)+\frac{1}{2}{{O}_{2}}=CO(g)\] ; \[\Delta H=-26.0\,kcal\]      

\[CO(g)+\frac{1}{2}{{O}_{2}}=C{{O}_{2}}(g);\] \[\Delta H=-\,68.0\,kcal\]

Total enthalpy change \[C(s)\] to \[C{{O}_{2}}(g);\] \[\Delta H=-94.0\,kcal\]

(b) Formation of ammonium chloride from ammonia and hydrochloric acid:

First method               

\[\frac{\begin{align} & N{{H}_{3}}(g)+HCl=N{{H}_{4}}Cl(g);\Delta H=-\,42.2\,kcal\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \\ & N{{H}_{4}}Cl\left( g \right)+aq=N{{H}_{4}}Cl\left( aq \right);DH=+4.0kcal \\ \end{align}}{~N{{H}_{3}}(g)+HCl(g)+aq=N{{H}_{4}}Cl(aq);\Delta H=-38.2\,kcal}\]

Second method  

\[\frac{\begin{align} & N{{H}_{3}}(g)+aq=N{{H}_{3}}(aq);\Delta H=-8.4\,kcal\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \\ & HCl(g)+aq=HCl(aq);\Delta H=-17.3\,kcal \\  & N{{H}_{3}}(aq)+HCl(aq)\,=N{{H}_{4}}Cl(aq);\Delta H=-12.3\,kcal \\ \end{align}}{N{{H}_{3}}(g)+HCl(g)+aq=N{{H}_{4}}Cl(aq);\Delta H=-38.0\,kcal}\]

(ii) Applications of Hess's law

(a) For the determination of enthalpies of formation of those compounds which cannot be prepared directly from the elements easily using enthalpies of combustion of compounds.

(b) For the determination of enthalpies of extremely slow reactions.

(c) For the determination of enthalpies of transformation of one allotropic form into another.

(d) For the determination of bond energies.

\[\Delta {{H}_{\text{reaction }}}=\Sigma \]Bond energies of reactants – \[\Sigma \]Bond energies of products.

(e) For the determination of resonance energy.

(f) For the determination of lattice energy.