Question

The heat of combustion of hydrogen by the reaction 2H+O=H2O is quoted as 34.18 kilogram calories per gram of hydrogen. (a) Find how many Btu per pound this is with the conversions 1 Btu=0.252 kcal, 1 lb=454 grams. (b) Find how many joules per gram this is noting 1 cal=4.184 J. (c) Calculate the heat of combustion in eV per H2 molecule. Note: Recall the number of particles per gram of molecular weight, Avogadro’s number, NA=6.02x10^23.

Answer 1

a)
E = 34.18 Kcal
   = 34.18/0.252 Btu
    = 135.63 btu

m = 1 g
     =1/454 lb
     = 2.203*10^-3 lb

Btu per pound = 135.63 / (2.203*10^-3)
                              = 61576 Btu/pound

b)
E = 34.18 Kcal
   = 34.18*4.184 KJ
    = 143.009 KJ
    = 143009 J

m = 1 g

Joules/g = 143009 J / 1g
                  =143009 J/g

c)
E = 143009 J/g
143009 J = 143009/(1.6*10^-19) eV
                  = 8.938*10^23 eV

1 g of H2 = 0.5 mol of H2 since molar mass of H2 is 2 g
number of H2 = 0.5*6.022*10^23 = 3.011*10^23 molecules

ev per H2 = 8.938*10^23 eV / 3.011*10^23 molecules
                    = 2.97 eV/molecules