1) 1.027 g of an unknown hydrocarbon (98.1 g/mol) burns in bomb calorimeter in e

Question

1) 1.027 g of an unknown hydrocarbon (98.1 g/mol) burns in bomb calorimeter in excess oxygen. The heat capacity of the calorimeter,Cv, = 5.270 kJ/ºC and T =7.778 ºC. Find E for this reaction in kJ/mol.

2) Using the technique of the previous problem E was found to be -2,000.00 kJ/mol of an unknown liquid hydrocarbon at 298 K. In another experiment it was determined that for each mole of hydrocarbon, 5 moles of oxygen gas are consumed and 4 moles of CO2 gas and 7 moles of H2O liquid are produced. Find H per mole of this hydrocarbon (in kJ) at 298 K. Please use six significant figures, that is, to hundredths.

Explanation / Answer

change in internal energy = Cv.Temperature difference= 5.27*7.778=40.99 KJ

moles of unknown compouns= 1.027/ 98.1=0.010469

Change in internal energy/mole= 40.99/0.010469=3915.409 Kj/mol

2. Change in enthalpy= change in internal energy+PV= 3915.409+nRT

n= change in number of moles = 4+7-5-1= 5

change in enthalpy = 3915.409+5*298*8.314/1000 Kj =3928 Kj/mol

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