Using the technique of the previous problem E was found to be -2,000.00 kJ/mol o

Question

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, 9 moles of oxygen gas are consumed and 9 moles of CO2 gas and 8 moles of H2O liquid are produced. Find H per mole of this hydrocarbon (in kJ) at 298 K.

Hint given in feedback.

Not used to answer question: This question is a bit awkward and unrealistic for molar amounts, but allows for random numbers. An example of a possible reaction is:
C2H4(OH)2(l) +2.5O2(g) 2CO2(g)+ 3H2O(l)

Explanation / Answer

Given that;

each mole of hydrocarbon, 9 moles of oxygen gas are consumed and 9 moles of CO2 gas and 8 moles of H2O liquid are produced means it is an example of combustion reaction.

All CO2 comes from hydrocarbon thus there are 9 mole C in hydrocarbon.

One water molecule has 2H, thus there are 16 H mole in hydrocarbon.

According to the problem, 9 mole O2 is consumed but product side total O atom = 18+8=6 atom, and here 9 mole O2 or 18 O atoms are consumed means 26- 18 O = 8 O atom present in hydrocarbon.

The molecular formula is C9H16O8 ; Which has molar mass 252.219 g/ mole

C9H16O8 + 9O2 = 9CO2   +8H2O

Reactant side mole = 10 mole

Product side mole = 17 mole

n = 17- 10 = 7

H = E+PV = E + nRT for gases.

Then;

H = E + nRT

= -2,000.00 kJ/mol + 7 * 8.3145/ 1000 *298

= -2,000.00 kJ/mol

= - 1982.66 KJ/ mol

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