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

Question

25. 357 points Ch. Ex. 92-cp92 Octane Vs. The The gas is then barsed as a fael advaatage of hydrogen as a fuel is that it is sonpollating, A major disadvastage is that voleme of ydroge it is a gas and therefore is harder to store than Eiquids or solds. Calcalate the gas at 21°C and 1.00 atm required to produce an amoust of eaergy equivaleat to hat produced by the combustion of a gallon of octane (GH1). The density of octane is 2.66 kgial, and its standard eathalpy of formation is-249.9 kJ/mol Assume that the products of the combustion of octane are Co,(e) and H0u). O Type here to search

Explanation / Answer

1.

Basis : 1 gallon of Octane, density = 2.66 kg/gal, mass of octane = density* volume= 2.66*1= 2.66 kg

Moles octane = mass/molar mass =2.66Kg*1000gm/kg/114=23.33 moles

Enthalpy of combustion of octane = 249.9 Kj/mole, energy produced from octane = moles* molar heat of combustion = 249.9*23.33 Kj=5823 Kj

From the reaction 2H2(g)+O2 -----à2H2O(l)

heat of formation of H2O(l)= -285.8 Kj/mole,

standard heat of formation when 2 moles of H2O is produced = -285.8*2= -571.6 Kj

571.6 Kj is produced from 2 moles of H2

5823 Kj is produced from 2*5823/571.6 moles of H2=500 moles of H2

From gas law, PV =nRT, n= 500, R=0.0821 L.atm/mole.K, T= 21deg.c =21+273= 294K, P= 1atm

V= nRT/P =500*0.0821*294/1 =12069 L

2.

Heat added to water is in the form of latent heat of vaporization.

Moles of liquid nitrogen = mass of Liquid nitrogen/molar mass =40/28=1.43

Heat given to liquid nitrogen = moles of liquid nitrogen* molar heat of vaporization

Heat lost by water is gained by liquid nitrogen, Heat lost by water= mass of water* specific heat of water* change in temperature = 200*4.184*(50.5-41) joules =7950 joules

Hence 1.43* latent heat of liquid nitrogen= 7950

Latent heat of liquid nitrogen = 7950/1.43 =5559 J/mole= 5.559 Kj/mole

3.

Energy intake = 4.7*1000 Kj, latent heat of water = 44 Kj/mole

Water has to take up all this heat by getting evaporated. Hence

Moles of water to be evaporated = 4.7*1000/44 moles =106.8 moles of water

Molar mass of water= 18 g/mole, mass of water required = 106.8*18 gm =1922.4 gm

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