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, 8 moles of oxygen gas are consumed and 6 moles of CO2 gas and 7 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

CxHyOz + 8 O2 --> 6 CO2 + 7 H2O

Using atom balance we get,

x = 6

y = 2 * 7 = 14

z = 6 * 2 + 7 - 8 * 2 = 3

Compound is C6H14O3

Heat of formation of CO2 =  -393.5 kJ/mol

Heat of formation of H2O = ?285.8 kJ/mol

Heat of formation of O2 = 0 kJ/mol

Heat of reaction of compound, ?Hr = x kJ/mol

?Hr = 6 * -393.5 kJ/mol + 7 * ?285.8 kJ/mol – Heat of formation of Compound = - 2000 kJ/mol

At 298 K, heat of formation of Compound = -2361.6 kJ/mol

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