Question

Question 5 Calcium oxide (Lime) is usually produced by heating and decomposing limestone (CaCO3), 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 2.7*106 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)-»CO2 (g) Suppose The combustion gas fed to a calcination reactor contains 75 mole% N2, 2.0% O2, 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 oflimestone 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

Species	moles in	$\Delta$H in	moles out	$\Delta$H out
CaCO3 (s)	1	-1206.9 KJ/mol	0	0
CaO (s)	0	0	1	Ha
CO2 (g)	0	0	1	Hb

To calculate Ha & Hb:-

Heat of formation of CaO (s) at 25 deg c is -635.09 KJ/mol, but it is mentioned that product temperature is 900 deg C

Therefore, Ha = Hf of CaO at 25 deg C + cp of CaO * $\Delta$T = -635.09 KJ/mol + (53.08 J/mol k ) (900-25)/1000

Ha = -635.09+46.445 = -588.645 KJ/mol

similarly, heat of formation of CO₂ at 25 deg C is -393.509 KJ/mol

Hb= Hf of CO2 at 25 deg C + cp of CO2 * $\Delta$T = -393.509 + (50.87 /1000)(900-25)

Hb = -348.998 KJ/mol

Now, required heat transfer is calculated as per Hess Law of constant heat summation;

$\Delta$H = heat of product - Heat of reactant

$\Delta$H = 1000 [(-588.645)+(-348.998)-(- -1206.9)]

$\Delta$H= 1000[269.257]

$\Delta$H = 2.7 x 10⁵ KJ