Question

Using the technique of the previous problem AE 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, 7 moles of oxygen gas are consumed and 5 moles of CO2 gas and 7 moles of H2O liquid are produced. Find AH 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(1) +2.502(g) →2CO2(g)+ 3H2O(1)

Answer 1

Here we have to use the equation $\small \Delta H$ = $\small \Delta E$ + $\small \Delta nRT$

for the given question(top question)

1 mole of hydrocarbon, 7 moles of oxygen gas are consumed and 5 moles of carbon dioxide and 7 moles water are produced

$\small Hydrocarbon + 7O_2 \rightarrow 5CO_2+7H_2O$

so the difference in number of moles of product and reactant($\small \Delta n$) =$\small \sum$number of moles of product - $\small \sum$ number of moles of reactant = (7+5) - (7+1) = 4

R = 8.314 $\small Jmol^{-1}K^{-1}$

T = 298 K

given $\small \Delta E$ = -2000 kJ/mol = $\small -2000*10^3$ J/mol

plugging everything into the firs equation:

$\small \Delta H$ = ($\small -2000*10^3$) + (4*8.314*298) = $\small -2000*10^3$ + 9910.28 = -1990090 J/mol = -1990.09 kJ/mol

$\small \Delta H$ per mole for this hydrocarbon is -1990.09 kJ/mol