Question

What is the work associated with decomposition of trinitrotoluene (TNT) upon detonation according to the following reaction at 300.15 K? C7H5N3O6(s)⟶3/2 N2(g)+5/2 H2(g)+3O2(g)+7C(s) Express your answer using four significant figures. (Answer must be in kJ/mol)

Answer 1

1 mole C7H5N3O6 gives 3/2 moles N2 and 5/2 moles H2 and 3 moles O2 gas

Total gas moles produced = N2 moles + H2 moles + O2 moles

                                     = 1.5 + 2.5 + 3 = 7

work done = - P x dV    where dV is change in volume  

we use ideal gas equation PV = nRT           ( since V is nothing but volume occupied by all gases)

work done = - nRT    where R = gas constant = 0.08206 liter atm/molK

              = - 7 mol x 0.08206 liter atm/molK x 300.15 K

                  = - 172.412 atm liter /mol

we have 1 atmliter = 101.325 J

hence work done = - 172.412 atm liter /mol x ( 101.325 J/atmliter)

                = - 17470 J/mol

                 = - 17.47 KJ/mol

( -ve sign indicates expansion work )