Question

Coke can be converted into CO in following reaction CO2(g) + C(s) -> 2CO(g). A coke that contains 84% carbon by mass and the balance noncombustible ash is fed to a reactor with a stoichiometric amount of CO2. The coke is fed at 77 oF, and the CO2 enters at 400 oF. Heat is transferred to the reactor in the amount of 5200 btu/lbm coke fed. The gaseous products and the solid reactor effluent (the ash and unburned carbon) leave the reactor at 1830 oF. The heat capacity of the solid is 0.24 btu/(lbm.oF) Calculate the percentage conversion of the carbon in the coke. (Provide your answer in decimal format, NOT percentage format!)

Answer 1

First write down the basis for the calculation:

Basis: 100 lb of coke feed (84 lb carbon, 16 lb of ash)_Mass basis.

So convert into mole basis. Molecular weight of carbon is 12, so 84 / 12 = 7 lb moles of carbon.

From the reaction, 1 mole of CO2 is fed for each moles of Carbon.

Therefore,

Let us first findout the Heat of reaction;

(Del.H)_r (77 °F) = (Del.H) _CO2 - (2(Del.H))_CO

It is at 25 ° C (77 ° F), so the standard heat of formation of CO2 is -393.5 kJ/mol and of CO is -282.99 kJ/mol.

= [-393.5 -(2*(-282.99))](kJ/mol)*0.9486(Btu/kJ)*453.6(mols/lbmole) = 74210 Btu / lb mole.

Let us assume x - fractional conversion of C and CO2:

n1 = 7x (lb moles Carbon reacted) * 2 lb moles CO formed / 1 lb mole C reacted = 14x lb moles of CO.

n2 = 7 * (1 - x) lb moles of CO2

n3 = 7 * (1 - x) lb moles of C

Let us get the Enthalpies from the references,

CO2 @ 400 ° F, H_CO2 = 3130 Btu / lb mole

CO2 @ 1830 ° F, H_CO2 = 20880 Btu / lb mole

CO @ 1830 ° F, H_CO = 13280 Btu / lb mole

Solid @ 1830 ° F = H_S = m. Cp . Del. T

Since, Cp = 0.24 Btu / lb ° F

H_S = 0.24 * (1830 -77) = 420 Btu / lb

Mass of solids (emerging),

= 7 (1 - x) lb moles of C * (12 lb / 1 lb mole) + 16 lb = (100 - 84x) lb

Energy balance:

Q = Del.H = (Del.H)_S +(Σn_iH_i)_out −(Σn_iH_i)_in

520000 Btu = (7x * 74210) + [(7*(1-x)*20880) + (14x*13280) + ((100-84x)*420)] - (7*3130)

Conversion, x = 0.67516 (i.e. 67.52 %)