Question

Calcium oxide (Lime) is usually produced by heating and decomposing limestone (CaCO3), a cheap and abundant mineral, in a process: CaCO3 (s) CaO (s)+CO2 (g) a) Limestone at 25°C is fed to a continuous calcination reactor. The calcination is complete, and the products leave at 900 °C. Taking 1 metric ton (1000 kg) of limestone as a basis and elemental species [Ca(s), C(s), O2(g)] at 25°C as references for enthalpy calculations, prepare and fill in an inlet-outlet enthalpy table and prove that the required heat transfer to the reactor is 2.7x106 kJ b) In a common variation of this process, hot combustion gases containing oxygen and carbon monoxide (among other components) are fed into the calcination reactor along with the limestone. The carbon monoxide is oxidized in the reaction: CO (g) 2 O2(g) - CO2 (g) Suppose The combustion gas fed to a calcination reactor contains 75 mole% N2, 2.0%。2 9.0% CO, and 14% CO2 · .The gas enters the reactor at 900°C in a feed ratio of 20 kmol gas/kmol limestone .The calcination is complete All of the oxygen in the gas feed is consumed in the CO oxidation reaction; The reactor effluents are at 900°C. . Again taking a basis of 1 metric ton of limestone calcined, prepare and fill in an inlet- outlet enthalpy table for this process [don't recalculate enthalpies already calculated in part (a)] and calculate the required heat transfer to the reactor.

Answer 1

Solution (a)

CaCO₃(s) === CaO(s) + CO₂(g)

Mass of Lime stone = 1MT = 1000 kg = 10 k mol = 10⁴ mol

Δ_fH^o of lime stone = - 1207 kJ /mol

        Of quick lime (CaO) = - 151.9 kJ / mol

        Of CO₂   = - 393.3 kJ / mol

1 mol of lime stone on thermal decomposition gives 1 mole of quick lime and 1 mol of carbon dioxide.

The ΔH^o of the reaction = ∑ Δ_fH^o_products - ∑ Δ_fH _reactants

= - 1028.8 – ( - 1207) = + 178.2 kJ / mol

When 10 k mol of lime stone undergoes thermal decomposition, the heat required is 1.78 x 10⁶ kJ

( 178.2 x 10000 = 1.78 x 10⁶ kJ)

Note : Heat transfer to the reactor is given is 2.7 x 10⁶ kJ and heat transfer as per calculation is 1.78 x10⁶ and the small variation may be due to standard heat of formation data used in the calculation.

Solution (b)

Mass of lime stone = 10 k mol

200 k mol gas fed into the reactor has the composition : 150 k mol of N₂, 4 k mol of O₂ , 18 k mol of CO and 28 k mol of CO₂

Nitrogen is a spectator and its effect may assumed to be nil ; 28 k mol of CO₂ also have no effect and 10 k mol of CO also remains unaffected.

CO (g) + ½ O₂(g) == CO₂(g) ΔH = - 282.8 kJ / mol

It is given that 4 k mol of Oxygen gas is completely utilised in the oxidation of CO ; as per the stoichiometry, 4 k mol of O₂ oxidises 8 k mol of CO to 8 k mol of CO₂. In this combustion, 2.26 x 10⁶ kJ heat is liberated.

The heat transfer to the reactor is calculated as follows;

CaCO₃(s) === CaO(s) + CO₂(g)

(i)10 k mol of lime stone decomposes by absorbing heat 1.78 x 10⁶ kJ

(ii) 20 k mol of combustion gas fed into the reactor, liberates 2.26 x 10⁶ kJ heat ( Oxygen and CO react as explained earlier)

Net Heat transfer = - 2.26 x 10⁶ + 1.78 x 10⁶ = - 0.18 x 10⁶ kJ   i.e.,    1.8 x 10⁵ kJ heat is liberated.       

Assuming 2.7 x 10⁶ kJ , 0.44 x 10⁶ or 4.4 x 10⁵ kJ heat transfer to the reactor takes place