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Question 14.4.1: Estimating the Energy Released in a Vapor-Phase Explosion Es...

Estimating the Energy Released in a Vapor-Phase Explosion

Estimate the energy released in a vapor-phase explosion of 1 kg of ethylene with a stoichiometric amount of air.

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The reaction stoichiometry is

 

C _{2} H _{4}+3 O _{2}+3 \times \frac{0.79}{0.21} N _{2}=2 CO _{2}+2 H _{2} O +3 \times \frac{0.79}{0.21} N _{2}

 

where we have recognized that with 1 mole of oxygen in air there is 0.79/0.21 moles of nitrogen. To proceed further we note that N^{ V , i}=1+3+3 \times(0.79 / 0.21)=15.29 moles of gas present initially, and N^{ V , f}=2+2+3 \times(0.79 / 0.21)=15.29 moles of gas present finally, so that
there is no contribution to the total energy release as a result of a change in the number of moles present in the gas phase. Next,

 

\begin{aligned}G\left(\underline{N}^{2}\right)=& \Delta_{ f } \underline{G}_{ C _{2} H _{4}}^{\circ}+3 \times \Delta_{ f } G_{ O _{2}}^{\circ}+11.29 \times \Delta_{ f } \underline{G}_{ N _{2}}^{\circ} \\&+R T\left[1 \times \ln \left(\frac{1}{1+3+11.29}\right)+3 \times \ln \left(\frac{3}{1+3+11.29}\right)\right.\\&\left.\quad+11.29 \times \ln \left(\frac{11.29}{1+3+11.29}\right)\right] \\=& 68.5+3 \times 0+11.29 \times 0+8.314 \times 10^{-3} \times 298.15 \times(-2.7-4.9-3.4\\=& 68.5-27.3=41.2 kJ / mol\end{aligned}

 

and

 

\begin{aligned}G\left(\underline{N}^{f}\right)=& 2 \times \Delta_{ f } \underline{G}_{ CO _{2}}^{\circ}+2 \times \Delta_{ f } \underline{G}_{ H _{2} O }^{\circ}+11.29 \times \Delta_{ f } \underline{G}_{ N _{2}}^{\circ} \\&+R T\left[2 \times \ln \left(\frac{2}{2+2+11.29}\right)+2 \times \ln \left(\frac{2}{2+2+11.29}\right)\right.\\&\left.+11.29 \times \ln \left(\frac{11.29}{2+2+11.29}\right)\right] \\=& 2 \times(-394.4)+2 \times(-228.6)+11.29 \times 0 \\&+8.314 \times 10^{-3} \times 298.15 \times(-4.1-4.1-3.4) \\=&-488.8-457.2-28.8=-1274.8 kJ / mol\end{aligned}

 

Therefore,

 

W=-1274.8-41.2=-1316.0 \frac{ kJ }{ mol }

 

and the total energy released by the explosion of 1 kg of ethylene is

 

W=-1316.0 \frac{ kJ }{ mol } \times \frac{1 mol }{28.054 g } \times \frac{1000 g }{1 kg }=-46910 \frac{ kJ }{ kg }

 

Note that the energy released on the explosion of ethylene, 46 910 kJ/kg, is more than 10 times that for TNT, which is 4570 kJ/kg. It is of interest to note that in this example of the total energy release of −1316 kJ/mol, the amount −1314.5 [= 2×(−394.4)+2×(−228.6)−68.5] kJ/mol results from the standard-state Gibbs energy change on reaction, which is the dominant contribution, and only −28.8 − (−27.3) = −1.5 kJ/mol is from the entropy-of-mixing term.

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