Combustion of natural gas (primarily methane) occurs in most household heaters.

Question

Combustion of natural gas (primarily methane) occurs in most household heaters. The heat given off in this reaction is used to raise the temperature of the air in the house.

Assuming that all the energy given off in the reaction goes to heating up only the air in the house, determine the mass of methane required to heat the air in a house by 10.0 C. Assume each of the following: house dimensions are 35.0 m ×35.0 m × 3.0 m ; specific heat capacity of air is 30 J/Kmol; 1.00 mol of air occupies 22.4L for all temperatures concerned

Explanation / Answer

Solution :-

Balanced reaction equation

CH4(g)+ 2O2(g) ---- > CO2(g) +2 H2O(g)

Using the Standard enthalpy of formations lets calculate the enthalpy change of the combustion

Delta Hrxn = Sum of delta H product – sum of delta H reactant

                  = [(CO2*1)+(H2O*2)]-[CH4*1]

                 = [(-393.5 *1)+(-285.8*2)]-[-74.87 *1]

                = -890 kJ

Therefore 1 mole CH4 gives 890 kJ heat

Now lets calculate the volume of the house

Volume = h*l*w

               = 35 m*35 m *3 m

               = 3675 m3

3675 m3 * 1000 L / 1 m3 = 3675000 L

Now lets calculate the moles of air

367500 L * 1 mol / 22.4 L = 167045.5 mol

Now lets calculate the heat needed to raise the temperature of the air by 10 degree C

q= specific heat of air * moles of air

   = 30.0 J /mol * 167045.5 mol

   = 5011364 J

5011364 J * 1 kJ / 1000 J = 5011 kJ

Now lets calculate the moles of methane

5011 kJ * 1 mol / 890 kJ = 5.63 mol CH4

Now lets calculate the mass of the CH4

Mass of CH4 = 5.63 mol * 16.04 g per mol = 90.3 g

Therefore the mass of methane need to be combusted is 90.3 g

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