The standard heat of the combustion reaction of Liquid n hexane to form CO2 (g) and H2O (l), with all reactants and products at 77F and I atm, is H 1 791 x 106 Btu lb mole The heat of vaporization of hexane at 77F is 13,550 Btu lb mole and that of water is 18,934 Btu lb mole (a) Is the reaction exo...

a b C Exothermic The reactor will have to be cooled to ke ...

The standard heat of the combustion reaction of Liquid n-hexane to form CO2 (g) and H2O (l),

Question:

The standard heat of the combustion reaction of Liquid n-hexane to form CO2 (g) and H2O (l), with all reactants and products at 77°F and I atm, is ΔH˚г = —1.791 x 106 Btu/lb-mole. The heat of vaporization of hexane at 77°F is 13,550 Btu/lb-mole and that of water is 18,934 Btu/lb-mole.

(a) Is the reaction exothermic or endothermic at 77°F? Would you have to heat or cool the reactor to keep the temperature constant? What would the temperature do if the reactor ran adiabatically? What can you infer about the energy required to break the molecular bonds of the reactants and that released when the product bonds form?

(b) Use the given data to calculate ΔH˚г (Btu/lb-mole) for the combustion of n-hexane vapor to form CO2 (g) and H2O (g).

(c) If Q = ΔH, at what rate in Btu/s is heat absorbed or released (state which) if 120lbm/s of O2 is consumed in the combustion of hexane vapor, water vapor is the product, and the reactants and products are all at 77°F?

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