Question

A typical frostless refrigerator uses 655 kWh of energy per year in the form of electricity. Suppose that all of this electricity is generated at a power plant that burns coal containing 3.5% sulfur by mass and that all of the sulfur is emitted as SO2when the coal is burned.

Part A If all of the SO2 goes on to react with rainwater to form H2SO4, what mass of H2SO4 is produced by the annual operation of the refrigerator? (Hint: Assume that the remaining percentage of the coal is carbon and begin by calculating ΔH∘rxn for the combustion of carbon.)

Answer 1

Part A:

The fuel contains 3.5 % suflur and rest coal

The reactions are,

C + O_{2 -}--> CO₂ ; ∆H = -393.31 kJ/mole = 393.31*1000 j/mol

S(s) + O₂(g) --> SO₂(g) ∆H = -296.7 kJ/mole = -296.7 * 1000 joules

Since, the fuel contains 96.5% Coal and 3.5 % Sulfur, taking basis as 100 kg of Coal

1 mole of Carbon gives 393.31*1000 j

12gms of Carbon gives 393.31*1000 j

96.5 gms gives 393*1000*96.5/12 joules = 3160375 joules

1mole of S gives 296.7*1000 joules

32gms   S gives 296.7*1000 joules

3 gms of S give 3*296.7*1000/32 joules = 27815.63 joules

Total heat produced per 100 gms of fuel = 3160375 + 27815.63

= 3188190.63 joules

655 KWh correspond to 239400849.422 joules

239400849.422 joules correspond to =100 * 2394008849.422 / 3188190.63

= 75089.89 gms of Coal

= 75.089 Kg of Coal

Amount of sulphur in coal = 3 * 75.089 / 100

= 2.252 kg = 2252 gm

= 2252 / 32 mole

= 70.37 moles

S+O₂ ---> SO₂

SO₂+1/O₂+ H₂O -----> H₂SO₄

Hence, 1 mole of S gives 1 mole of H₂SO₄

70. 37 moles of Sulfur gives 70.37 moles of H2SO4 = 70.37 * 98

=6896.26 gm