Saved Bookmarks
| 1. |
Derive a relation DeltaH = DeltaU + DeltanRT . Derive a relationq_(p) = q_(v) + DeltanRT |
|
Answer» Solution :Consider a reaction in which `n_(1)` moles of gaseous REACTANT in initial state change to `n_(2)` moles of gaseous product in the final state. Let `H_(1),U_(1),P_(1),V_(1) and H_(2),U_(2),P_(2),V_(2)` represententhalpies, internal energies, pressures and volumes in the initial and final statesrespectively then, `underset(H_(1),U_(1),P_(1),V_(1))(n_(1) A_((g))) overset(T) to underset(H_(2),U_(2),P_(2),V_(2))(n_(2)B_((omega)))` The heat of reaction is given by ENTHALPY change `DeltaH` as, `DeltaH = H_(2) - H` By definition, H = U + PV `therefore H_(1) = U_(1) +P_(1)V_(1) and H_(2) = U_(2)P_(2)V_(1)` `therefore DeltaH = (U_(2) +P_(2)V_(2)) - (U_(1) +P_(1)V_(1))` ` =(U_(2) - U_(1)) +(P_(2)V_(2) - P_(1)V_(1))` Now `DeltaU = U_(2) - U_(1)` SincePV = nRT . For initial state `P_(1)V_(1) = n_(1)RT` For final state ,`P_(2)V_(2) = n_(2)RT` `therefore P_(2)V_(2) =n_(2)RT-n_(2)RT` `=(n_(2) - n_(1)) RT` `= DeltanRT` Where `Deltan = [{:("Number of MOLE"),("of gaseous products"):}]-[{:("Number of moles of"),("gaseous reactants"):}]` `therefore DeltaH = DeltaU + DeltanRT` If `q_(p)` and `q_(v)` are the the heats INVOLVED in the reaction at constantpressure and constant volume respectively, then since `q_(p)= DeltaH` and`q_(v) = DeltaU` `thereforeq_(p) = q_(v) = DeltanRT` |
|