Question

155 grams of a hydrocarbon (C20H62) is burned in a bomb calorimeter that has a calorimeter constant of 3250.0 J/oC The calorimeter undergoes a 1.95 oC temperature increase as the hydrocarbon is burned. Determine the hydrocarbon's heat of combustion, ΔHcomb in kJ/mol. (Closest answer)

Answer 1

Mass of sample used for combustion experiment in bomb calorimeter = 155 g ( C₂₀H₆₂= molar mass 302 g/mol)

calorimeter constant = 3250.0 J/oC

Rise in temperature , dT =  1.95 ^oC

Heat energy released in the above experiment( absorbed by calorimeter) = calorimeter constant x rise in temperature

heat gained by bomb calorimeter equals caolrimeter constant multiplied by rise in temperature . Heat of combustion is exothrermic reaction . It bears - ve sign

= 3250.0 J/oC x 1.95 ^oC

= 6337.5 joules

= 6.3375 kj

This heat energy released by combustion of 155 g of the hydrocarbon given  C₂₀H₆₂ )

Heat of combustion of 1 mole of sample = 302 g/mol x 6.3375 kj / 155 g

= 12.348 kj /mol

   = -12.348 kj /mol ( exothermic)