25. 357 points Ch. Ex. 92-cp92 Octane Vs. The The gas is then barsed as a fsel A

Question

25. 357 points Ch. Ex. 92-cp92 Octane Vs. The The gas is then barsed as a fsel A primary advastage of hydrogen as a fael is that it is opolluting. A major dinadvastage in hat it is a gas and therefore is harder to store than Eiquids or solids. Calcalate the gas at 21°C and 1.00 atm required to produce an amoust of eaergy equivaleat to that prodaced by the combustion of a gallon of octane (GH13). The density of octane is 2.66 kgial, and its standard eathalpy of formation is-249.9 klimol Assume that the products of the combestion f octane are Co,(e)and H0u) voleme of ydroge O Type here to search

Explanation / Answer

first we have to calculate the heat of combustion of octane

density= 2.66kg/gal

delHformation=-249.9KJ/mol

combustion reaction of octane

C8H18 + 12.5O2 -----> 8CO2 + 9H2O

delHc = delHf(prod) - delHf(react)

we need delHf at 21oC i.e, 294K

at 298K, delHf of CO2, H2O is  -393.5 kJ mol-1 and –285.8 kJ mol-1 respectively.

to find at 294K, dT =298-294= 4

delHf for CO2 = delHf (at 298) + Cp(dT)

= -393.5+ 3.5R*4

= -277.1KJ/mol

similarly for H2O delHf = -285.8 + 3.5R*4

= -169.4KJ/mol

now, at 294K, delHc = delHf(prod) - delHf(react)

= {8*(-277.1) + 9*(-169.4)} - {133.54 + 12.5*(0)}

delHc= 3607.9KJ/mol

volume of hydrogen calculation

P= 1atm, T=294K, delH= 3607.9KJ/mol

delH = delHprod- delHreact

= delH (H2O) - (delHH2 + delHO2)

= x*(-169.4) - x*(-158.68) + 0

3607.9= -169.4x + 158.68x

= -10.72x

x= 3607.9/10.72

x= 336 moles

336 moles of hydrogen

using ideal gas equation,

PV=nRT

V= nRT/P

=336* 0.0821* 294/1

= 8110L of hydrogen is needed

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