Estimating the Energy Released on the Explosion of a Liquid Nitromethane (a liquid) undergoes the following reaction on explosion:
CH _{3} NO _{2} \rightarrow 0.2 CO _{2}+0.8 CO +0.8 H _{2} O +0.7 H _{2}+0.5 N _{2}Estimate the energy released on the explosion of nitromethane.
Data:
\Delta_{ f } \underline{G}_{ NM }^{ o }=-13.0 kJ / mol \Delta_{ f } \underline{H}_{ NM }^{ o }=-111.7 kJ / molThere are a number of terms that contribute to the total energy released. First we note that N^{ V , f}= 0.2+0.8+0.8+0.7+0.5 = 3 moles of gas present finally, and N^{ V , i}=0 0moles of gas initially. Next,
\begin{aligned}G\left(\underline{N}^{i}\right)=& \Delta_{ f } \underline{G}_{ NM }^{\circ}=-13.0 kJ / mol \\G\left(\underline{N}^{f}\right)=& 0.2 \Delta_{ f } \underline{G}_{ CO _{2}}^{\circ}+0.8 \Delta_{ f } \underline{G}_{ CO }^{\circ}+0.8 \Delta_{ f } \underline{G}_{ H _{2} O }^{\circ}+0.7 \Delta_{ f } \underline{G}_{ H _{2}}^{\circ}+0.5 \Delta_{ f } \underline{G}_{ N _{2}}^{\circ} \\&+R T\left[0.2 \ln \frac{0.2}{3}+0.8 \ln \frac{0.8}{3}+0.8 \ln \frac{0.8}{3}+0.7 \ln \frac{0.7}{3}+0.5 \ln \frac{0.5}{3}\right] \\=&-371.5 \frac{ kJ }{ mol }+8.314 \frac{ J }{ mol K } \times 298.15 K \times(-4.571) \times \frac{1 kJ }{1000 J } \\=&-371.5-11.3=-382.8 \frac{ kJ }{ mol }\end{aligned}
Therefore,
W=-382.8-(-13.0)-3 \times \frac{8.314 \times 298.15}{1000}=-377.2 \frac{ kJ }{ mol }
Comments
It is again interesting to assess the contributions of the various terms to the total energy release on explosion. By far the largest contribution is the difference in Gibbs energies of the components, that is
\begin{aligned}\Delta G &=0.2 \Delta_{ f } \underline{G}_{ CO _{2}}^{\circ}+0.8 \Delta_{ f } \underline{G}_{ CO }^{\circ}+0.8 \Delta_{ f } \underline{G}_{ H _{2} O }^{\circ}+0.7 \Delta_{ f } G_{ H _{2}}^{\circ}+0.5 \Delta_{ f } \underline{G}_{ N _{2}}^{\circ}-\Delta_{ f } \underline{G}_{ NM }^{\circ} \\&=-371.5-(-13.0)=-358.5 kJ / mol\end{aligned}
The Gibbs energy change due to the entropy of mixing is −11.1 kJ/mol, and the energy released due to the vapor-phase mole number change is −7.4 kJ/mol. Therefore, the majority of the explosive energy release comes from the Gibbs energies of formation of the species involved. A reasonable approximation in many cases, accurate to ±10 percent, is to neglect the gas-formation and entropy-of-mixing terms and merely use W = \Delta_{ rxn } G.
A second condition for an explosion is that the heat of reaction be negative (i.e., the reaction should be exothermic). For the reaction here, we have
\begin{aligned}\Delta_{ rxn } H=& 0.2 \Delta_{ f } \underline{H}_{ CO _{2}}^{\circ}+0.8 \Delta_{ f } \underline{H}_{ CO }^{\circ}+0.8 \Delta_{ f } \underline{H}_{ H _{2} O }^{\circ}+0.7 \Delta_{ f } \underline{H}_{ H _{2}}^{\circ}+0.5 \Delta_{ f } \underline{H}_{ N _{2}}^{\circ}-\Delta_{ f } \underline{H}_{ NM }^{\circ} \\=& 0.2 \times(-393.5)+0.8 \times(-110.5)+0.8 \times(-241.8) \\&+0.7 \times(0)+0.5 \times(0)-(-111.7) \\=&-78.7-88.4-193.4+111.7=-248.8 \frac{ kJ }{ mol }\end{aligned}
which is large and negative. Therefore, nitromethane is a likely candidate for a chemical explosion.