A 5.000 g sample of a hydrocarbon (100.0 g/mol) is completely combusted in a bomb calorimeter with a heat capacity (excluding water) of 1500. J/°C.

Question

A 5.000 g sample of a hydrocarbon (100.0 g/mol) is completely combusted in a bomb calorimeter with a heat capacity (excluding water) of 1500. J/°C. The calorimeter contains 2.500 kg of water (s = 4.184 J/°C•g). The combustion causes a temperature increase of 7.550°C. Calculate the enthalpy of combustion in kJ per mole of the hydrocarbon combusted.  

Please show steps.

Also, is this exothermic or endothermic? Should my answer have a + or - sign in front of i?

Answer 1

I believe the previous answer neglected to include the heat capacity of the calorimeter. Thus, one should have q = mC∆T + C_cal∆T = (2500g)(4.184 J/g/deg)(7.55 deg) + (1500 J/deg)(7.55 deg) =90,298 J

This is generated from 0.05 mol of hydrocarbon, so ∆H_combustion = 90,298 J/0.05 mole x 1kJ/1000 J = 1806 kJ/mole. And yes, the sign should be - since it is an exothermic reaction.

Answer 2

If the water bath was raised by 7.55˚C and there were 2,500 g of water then the heat energy acquired by the water is:

q=mcΔT = (2,500g) * (4.184J/˚C*g) * (7.55˚C) = 78,973 J

This energy was provided by the combustion (always exothermic) of 5g of a 100g/mol hydrocarbon (0.05 moles).

In other words there were 78,973 J released per 0.05 moles: (78,973 J)/ (0.05 mol) = 1579.46 kJ/mol

note: since the heat is leaving (being subtracted from) the system (the hydrocarbon) the sign should be negative (-)