9.2. The standard heat of combustion of liquid n-octane to form CO2 and liquid w

Question

9.2. The standard heat of combustion of liquid n-octane to form CO2 and liquid water at 25°C and 1 atm is AH 5471 kJ/mol (a) Briefly explain what that means. Your explanation may take the form "When (specify quantities of reactant species and their physical states) react to form (quantities of product species and their physical states), the change in enthalpy is (b) Is the reaction exothermic or endothermic at 25C? Would you have to heat or cool the reactor to keep the temperature constant? What would the temperature do if the reactor ran adiabatically?

Explanation / Answer

9.2 Combustion of liquid octane

(a) Combustion means bruning a compound in air to CO2 and H2O. When 2 moles of liqud octane is burnt in oxygen we get 16 moles of CO2 and 18 moles of water vapor formed. The enthalpy change for the reaction is -5471 kJ/mol of octane.

(b) the sign of enthalpy is -ve, that means the reaction is exothermic in nature. The temperature is maintained by cooling the reactor.

(c) Power KW = kJ/s

with 25 mol/s octane feed

Power = -5471 x 25 = -136775 KW

Here the reactor pressure was assumed to be 1 atm

(d) the heat of combustion for octane vapor is 57 kJ/mol greater than the combustion of liquid octane. This means that we have to supply some energy to convert liquid octane to first convert into vapor octane before combustion starting with octane vapor.

(e) The liquid octane would have some part stay in the vapor phase at 25 oC even when the temperature has not reached the boiling pont of liquid n-octane.

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