Question

Calcium Oxide (CaO) is widely used in the production of cement, steel, medicines and many other familiar materials. It is usually produced by heating and decomposing limestone (CaCO3), a cheap and abundant material, in a calcination process:

CaCO3(s) -> CaO(s) + CO2(g)

CaCO3 at 298 K is fed to a continuous reactor. The calcination is complete, and the products leave at 1000 K. Taking 800 gmol of limestone as a basis and elemental species [Ca(s), C(s), O2(g)] at 298 K as reference for enthalpy calculations.

Ccalculate the required heat transfer to the reactor in MJ.

Answer 1

for the reaction CaCO3----> CaO+ CO2, enthalpy change/mole= 178.49 Kj/mole

moles of lime stone = 800 gmole, enthalpy change =178.49*800 =142792 Kj

moles of CaO and CO2 formed = 800 gmole

Specific heat of CaO at 1000 K= 51.47 J/kmole, at 298 K= 0.316 J/mole.K

Average speciifc heat = (51.47+0.316)/2=25.893 J/mol.K

heat required for CaO to change the temperature from 289K to 1000K = 800*25.893*(1000-298)=1.45*10⁷ joules

for CO2, CP/R= 5.457+1.045*10^-3T

enthalpy change = 800( moles of CO2)* CpdT between 298 and 1000K

hence enthalpy change = 800*8.314* {5.457*(1000-298) +1.045*10^-3*(1000²-298²)/2}=2.846*10⁷ joules

heat transferrted to reactor = enthalpy of reactans+ enthalpy of products + standard heat of reaction

since reference temperature is 298K,   which is also the intlet temperature of reactants, enthalpy of reactants = 0

hence heat transferred = 0+1.45*10⁷ +2.846*10⁷ ( enthalpy of products)+142792*1000 j =1.86*10⁸ joules

1.86*10⁸/10⁶MJ =186 Mj