You don\'t need to do the Excel work but please show me how to do each parts (sa

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You don't need to do the Excel work but please show me how to do each parts (sample calculation will help me a lot).

You don't need to do the Excel work but please show me how to do each parts (sample calculation will help me a lot).

Thanks!

Question 1 The carbon gasification (Boudouard) reaction, C+ CO2-2CO, is important to many extractive and refining processes as well as gasification of coal and biomass (wood or municipal waste) a) Make two charts of equilibrium gas composition as a function of temperature over the range 400°C to 1400°C; in the 1st use pco/P as the y-axis and in the 2nd log (pco/pco2); Assume P-poo + pc02-1 atm. and ac-1 b) By all appearances, CO gas is stable at room temperature. Is this in agreement with thermodynamics? If not explain why CO seems stable c) Consider the gasifier shown in the notes operating at 1000°C; pco + pco2 ~ 0.25 atm. At this temperature the gasification reaction is very fast. What does 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 d) A reactor at 1000°C; 1 atm. contains the equilibrium CO-CO2 mixture calculated in part (i) Steel containing carbon at ac 0.02 is placed in the reactor. Would the steel tend to decarburize (lose carbon) or recarburize (gain carbon)? e) While surfing the net you see an offer to buy shares in Point Grey Aluminum, a new company planning to extract Al from Al203 at 1400°C by the reaction log K =-7.554 The process will operate at P pcoPco2 1 atnm Should you hand over your $$? Why or why not? Reaction (J/mole) 112,877 394,762 AS° (J/mole-°K) 86.514 C0.5O,-CO 0.836 Summary of Thermodynamic Data

Explanation / Answer

a. The entahlpy change and entropy change for the required reaction is obtained by:

(reaction 1) x2 - (reaction 2)

For eg.

Accordingly calculate H = 2(-112877) - (-394762) = 369008 J/mol

S = 2(86.514) - 0.836 = 144.651 J/mol-K

At T= 400 C

We have G= H - TS = 369008 - (400 + 273)*144.651 = 271657.9 J/mol

Now, we have G = -RT ln(Kp)

From this, we get Kp = 8.2152 x 10-22

Also, Kp = [P(CO)]2/[P(CO2)]

Since the Kp values is so small, we assume hardly any CO is formed. Major part of the 1 atm is CO2.

Thus, we get it simplified to Kp = (0.5*P(CO))2 (0.5 is added for the reaction stoichiometry. ratio of CO2:CO is 0.5)

P(CO) = 2*(Kp)1/2 = 2.866 x 10-11 atm

Hence make the caluclations

b. Due to highly positive G, the reaction is non-spontaneous, and thermodynamically, there should be minimal CO in such conditions. However, in CO molecule, both atoms have a completed octet, hence it is so stable. Plus, the trople bond ( including one coordinate bond) makes it very stable.

c. The reaction is faster implies higher rate, and rate constant. This implies a greater conversion. Hence, PCO/P would be higher than that calculated.

d. The equilibrium conversion and hence CO pressure would yet reduce due to the activity of carbon. The reaction would be shifted backwards. Hence, the steel would tend to gain carbon

e. Log(K) = -7.554 indicates that K is in order of 10-8. This indicates that the reaction will hardly proceed and there will be no extraction of Al. It is not advisable to invest here.

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