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

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, 4 moles of oxygen gas are consumed and 9 moles of CO2 gas and 6 moles of H2O liquid are produced. Find ?H per mole of this hydrocarbon (in kJ) at 298 K.

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

?E = -20000 KJ

?H =?E +?n(gas) * R*T

or?H = -2000 + [(9-4-1)*8.314*298/1000]

or?H = -1990.0897 KJ/mol

here we didn't reduce include H2O mole because it is in liquid form.

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