Question
Sat February 02, 2013 By: Kaustubh Deshpande
 

determine the enthalpy of a reaction C3h8(g)+H2(g)-C2h6(g)+Ch4(g) at 25*c .using given enthalpies of combustion for:- H2(g)= -285.5kj/mol Ch4(g)= -890kj/mol C2h6(g)=-1560kj/mol c(s)=-393.5 kj/mol enthalpy of formation of C3h8 gas is -103.8kj/mol

Expert Reply
Sat May 04, 2013
The required equation can be obtained by the following manipulationsdH = 3 dH4 -dH5 +5dH1-dH2-dH3dH=3(-393.5)-(-103)+5(-285.8)+890 +1560      =-1180.5 +103.8 -1429.0 + 890 +1560      = -2609.5 + 2553.8       =-55.7 kJ/mol
Ask the Expert