Question

Calaculate Delta H -delta E for combustion C6H6 in KJ/mol at 27degreeC?? What is Delta H-DeltaE=?

Answer 1

delta H= delta E + delta nRT

now let's see the reaction,

C6H6 + 15/2O2 --------------> 6CO2 + 3H2O

delta n= 6-15/2= -1.5

temp.= 27+ 273= 300

delta H- delta E =

= delta nRT

= -1.5 × 8.314 × 300

= -3741.3 joule