The carbon gasification (Boudouard) reaction C + CO_2 = 2CO is important in many

Question

The carbon gasification (Boudouard) reaction C + CO_2 = 2CO is important in many pyromctallurgal extractive and refining processes as well as gasification of coal, wood or municipal waste. Make two charts of equilibrium gas composition as a function of temperature over the range 400degreeC to 1400degreeC; in the I^st use Pco / P as the y-axis and in the 2^nd log (Pco / Pco_2) Assume P = Pco + Pco_2 = 1 atm. And a_c =1. By all appearances, CO gas is stable at room temperature. Is this in agreement with thermodynamics? If not explain why. Consider the gasifier shown in the notes operating at 1000degreeC; Pco + Pco_2 ~25 atm. At lhis temperature the gasification reaction is very fast. What docs this imply for the actual exit gas composition relative to the equilibrium? Assuming the exit gas is at equilibrium would Pco / P be greater than or less than that calculated in part (i)? Justify your answer. A reactor at 1000degreeC; 1 atm. contains the equilibrium CO-CO_2 mixture calculated in part (i). Steel containing carbon at a_c- = 0.02 is placed in the reactor. Would the steel tend to decarburizc (lose carbon) or recarburize (gain carbon)? I have been asked to buy shares in Point Grey Aluminum, a new company planning to extract Al from Al_2O_3 at 1400degreeC by the reaction:

Explanation / Answer

boudouard reaction is redox reaction of chemical equillibrium mixture of carbon monoxide and carbon dioxide

2CO is reversible with CO2 + C

boudouard reaction to form carbon dioxide and carbon is exothermic at all temperatures . The standard enthalpy of

boudouard reaction becomes less negative with increasing temperature .

while the formation enthalpy of CO2 is higher than that of CO , the formation entropy is much lower . Consequently , the standard free energy of formation of CO2 from its component element is almost constant and independent of temperature , while the free energy of formation of CO decreases with temperature .

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