From the following information, determine \({\bf{\Delta S}}_{{\bf{298}}}^{\bf{^\circ }}\)for the following:

\(\begin{array}{*{20}{c}}{{\bf{N(g) + O(g)}} \to {\bf{NO(g)}}}&{{\bf{\Delta S}}_{{\bf{298}}}^{\bf{^\circ }}{\bf{ = ?}}}\\{{\bf{\;}}{{\bf{N}}_{\bf{2}}}{\bf{(g) + }}{{\bf{O}}_{\bf{2}}}{\bf{(g)}} \to {\bf{2NO(g)}}}&{{\bf{\Delta S}}_{{\bf{298}}}^{\bf{^\circ }}{\bf{ = 24}}{\bf{.8\;J/K}}}\\{{{\bf{N}}_{\bf{2}}}{\bf{(g)}} \to {\bf{2\;N(g)}}}&{{\bf{\Delta S}}_{{\bf{298}}}^{\bf{^\circ }}{\bf{ = 115}}{\bf{.0\;J/K}}}\\{{{\bf{O}}_{\bf{2}}}{\bf{(g)}} \to {\bf{2O(g)}}}&{{\bf{\Delta S}}_{{\bf{298}}}^{\bf{^\circ }}{\bf{ = 117}}{\bf{.0\;J/K}}}\end{array}\)

• According to Hess' law, also known as Hess' law of constant heat summation, "at constant temperature, heat energy changes (enthalpy – Hrec) accompanying a chemical reaction will remain constant, regardless of how the reactants react to form product."
• Hess' law is based on enthalpy's state function character and the first law of thermodynamics. A system's (molecule's) energy (enthalpy) is a state function. As a result, the enthalpy of reactant and product molecules is constant and does not change depending on the origin and path of formation.The standard entropy change (S°) of any process can be calculated using the standard entropies of its reactant and product species, as shown below:

ΔS° = ∑vS° ₂₉₈(products) - ∑vS° ₂₉₈(reactants)

Here to calculate the reaction entropy change, we can use the Hesse law. The Hesse law states that the reaction entropy for a reaction equals the sum of all the reaction entropies of elementary reaction to which we can subdivide the chemical reaction. This means that if we need to calculate the \(\Delta S\)for the reaction \(A \to B\) and we know the \(\Delta S\)for the reaction \(A \to C\)and \(C \to B\)we can calculate the \(\Delta S\)for the \(A \to B\) reaction by simply adding the \(\Delta S\)for the two reactions that "make up" the reaction \(A \to B\) This means that we can rearrange the reactions in order to create a reaction scheme in which one of the chemical reactions is the sum of all the others. One more thing: if we change the direction of the chemical reaction (turn the products to reactants and vice versa), we can calculate the entropy of this new reaction by multiplying the old\(\Delta S{\rm{ }}by{\rm{ }} - 1\).

The reaction

\(N(g) + O(g) \to NO(g)\)

will be our "sum" reaction. Now we have to rearrange all the other chemical reactions so that their sum would equal this reaction which we have written above. Firstly, we have to have \(N(g)\)and \(O(g)\)among the reactants, and they are the products of the last two reactions in the problem. So we have to change the direction of these reactions, getting:

\(2N(g) \to {N_2}(g)\Delta {S^{298}} = - 115.0\frac{J}{K}\)

\(2O(g) \to {O_2}(g)\Delta {S^{298}} = - 117.0\frac{J}{K}\)

We also have to divide the reactions by two because in our "sum" reaction we do not have \(2N\)and \(2O\), but only \(N\)and\(O\).

\(N(g) \to \frac{1}{2}{N_2}(g)\Delta {S^{298}} = - 57.5\frac{J}{K}\)

\(O(g) \to \frac{1}{2}{O_2}(g)\Delta {S^{298}} = - 58.5\frac{J}{K}\)

The second reaction in the problem is in the right direction, however we need to divide it by two for the same reason.

\(\frac{1}{2}{N_2}(g) + \frac{1}{2}{O_2}(g) \to NO(g)\Delta {S^{298}} = 12.4\frac{J}{K}\)

Now if we combine all the three reaction, we can see that \({N_2}\)and \({O_2}\)cancel each other out and that we get our sum reaction. Now we can calculate the deltas for our reaction using Hesse law.

\(\Delta {S^{298}} = ( - 57.5 + ( - 58.5) + 12.4)\frac{J}{K}\)

\(\Delta {S^{298}} = - 103.6\frac{J}{K}\)

Therefore the value of \({\bf{\Delta S}}_{{\bf{298}}}^{\bf{^\circ }}\) for \({\rm{N}}(g) + {\rm{O}}(g) \to {\rm{NO}}(g)\) is \( - 103.6\frac{J}{K}\).