Assuming that the energy released by combustion of H_2 is proportional to the am
ID: 1015523 • Letter: A
Question
Assuming that the energy released by combustion of H_2 is proportional to the amount of oxygen it consumes, estimate the ratio of heat released by one mole of methane compared to one mole of hydrogen gas. Using the thermochemical information in the H_2 combustion equation above, calculate the enthalpy (heat) of combustion of hydrogen per gram, and by comparing it to that of methane (see page 235) decide which fuel is superior on a weight basis. By comparing the actual energy released by combustion per mole of gas-and hence per molar volume-decide which fuel is superior on a volume basis. For H_2 gas compressed to 700 arm at 27 degree C, calculate its density in kilograms per cubic meter, assuming (incorrectly) that it behaves as an ideal gas. Given that its actual density at that pressure is only 37 kg m^-3, calculate the percentage error in the calculated density due to the nonideal behavior of the gas under this extreme pressure. (Recall that for an ideal gas PV = nRT, and that R = 0.082 L atm mole^-1 K^-1.)Explanation / Answer
7-10. 2H2+ O2 --> 2H2O + E1
i.e. 2 mole of H2 on combustion uses 1 mol of O2, which is proportional to the corresponding release energy E1
So, 1 mole of H2 combustion uses 0.5 mol of O2
CH4 + 2O2 --> CO2 + 2H2O + E2
i.e. 1 mole of CH4 on combustion uses 2 mol of O2.
So, considering the energy released is proportional to the used O2, the relative ratio of heat released for 1 mol of CH4 to 1 mol of H2 combustion would be the ratio of O2 used i.e. 2:0.5 i.e. 4:1
7-11. Not all the information given
7-12. Considering ideal gas equation, PV = nRT = m/M * RT [m= actual mass, M=Molar mass, n= number of moles]
i.e. P= mRT/VM = (m/V) * (RT/M) = dRT/M [m/V=d=density]
i.e. d = PM/RT = [(700 atm) * (2 * 10^-3 kg mol-1)]/[(0.082 * 10^-3 m3 atm mol-1 K-1) * (273 +27)K]
= 56.91 kg/m3
Actual density = 37 kg/m3
So, error = (56.91-37)/37 * 100% = 53.81%