Question

Consider the combustion of 1-hexanol, C₆H₁₄O:

C₆H₁₄O(l) + 9O_2(g) ------> 6O_2(g) + 7H₂O(g)

How much heat is released, in units of kJ, when 1-hexanol is burned at constant pressure, when 1.00*10³ L (1.00m³) of CO₂ is produced at 1.00 bar and 25^oC ?

Answer 1

C6H14O(l) + 9O2(g) -----> 6CO2(g) + 7H2O(g)

Delta H(rxn) = 6*Delta H(CO2) + 7*Delta H(H2O) - Delta H(C6H14O) - 9*Delta H(O2)

Delta H(rxn) = 6*(-393.5) + 7*(-241.82) - (-377.5) -9*(0)

Delta H(rxn) = -3676.24 KJ

Please note that in reaction H2O is in gas form. If it is in liquid form then Delta H(H2O) = -285.8 KJ/mol

Then Delta H(rxn) = -3984.37 KJ

But here I am considering H2O as gas.

1 bar = 0.987 atm

Using ideal gas equation for CO2

PV = nRT

n = PV/RT = 0.987*1000 / 0.0821*298

n = 40.342 moles of CO2

From reaction;

6 mole CO2 produces -3676.24 KJ heat

So, 40.342 moles will produce = -3676.24*40.342/6 = -24717.83 KJ .....Answer

Now, using Delta H = -3984.37 KJ

Heat produced = -3984.37*40.342/6 = -964425.41 KJ

Let me know if any doubts.