Standard reaction entropy

Just as the standard reaction enthalpy is

\Delta H_r^{\; o}=\sum_Pv_P\left (\Delta H_f^{\; o}\right )_P-\sum_Rv_R\left (\Delta H_f^{\; o}\right )_R

the standard reaction entropy of a system is

\Delta S_r^{\; o}=\sum_Pv_P\left (S_m^{\; o}\right )_P-\sum_Rv_R\left (S_m^{\; o}\right )_R\; \; \; \; \; \; \; \; 1

where vP and vR are the stoichiometric coefficients of the products and reactants respectively.

The difference between ΔHfo and Smo is that the former is the relative energy between a substance and the constituents of the substance in their reference states, while the latter is the absolute energy of the substance. This is why ΔHfo[C(graphite)] = 0 but Smo[C(graphite)] ≠ 0. Despite the difference, the quantities of ΔHro and ΔSro are comparable, as the zero-enthalpies of reference states in \sum_Pv_P\left ( \Delta H_f^{\; o} \right )_p and \sum_Rv_R\left ( \Delta H_f^{\; o} \right )_R cancel out when we compute ΔHro (see articles on standard enthalpy change of formation and Hess Law for details).



Calculate the standard reaction entropy of C(s)+\frac{1}{2}O_2(g)\rightarrow CO(g) given S_m^{\; o}[C(s)]=5.7\,JK^{-1}mol^{-1}, S_m^{\; o}[O_2(g)]=205.0\,JK^{-1}mol^{-1} and S_m^{\; o}[CO(g)]=197.6\,JK^{-1}mol^{-1}.


Using eq1,

\Delta S_r^{\; o}=197.6-[5.7+\frac{1}{2}(205.0)]=+89.4\, JK^{-1}mol^{-1}


The positive change in standard reaction entropy of the incomplete combustion of carbon illustrates the dependency of the change in entropy of a system on the change in the number of moles of gas in a reaction. In the above example, the double in number of moles of gas outweighs the removal of a mole of carbon solid in the system, as the increase in the number of moles of gas leads to the dispersal of energy over a greater number of translational energy states of the system. This implies that ΔSro can be negative when we have a reaction that results in a decrease the number of moles of gas.


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