155 grams of a hydrocarbon (C20H62) is burned in a bomb calorimeter that has a calorimeter...
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)
Homework Answers
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)
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155 grams of a hydrocarbon (C20H62) is burned in a bomb
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