4. Rod Bagley was co-inventor of the catalytic converter along with two other en

Question

4. Rod Bagley was co-inventor of the catalytic converter along with two other engineers at Corning. The catalytic converter converts 95% of exhaust pollutants into more benign emissions vastly improving air quality. Upon learning that Rod Bagley is an alumnus from the MSE department at the University of Utah (BS 1960, PhD 1964) you feel inspired to make the world a better place and excitedly take a job at a new environmentally friendly company designing diesel engines. Your boss is convinced that increasing the combustion temperature from 1000K to 1500K will decrease the amount of NO formed (a harmful emission that can convert to NO2 and other toxic compounds). You check the literature and find that for N2+O2 2NO the AGO for NO and that of molecular oxygen is 0.06 in the combustion gas (there is also 17% co and 8% Hao in the combustion gas but these do not enter the calculation). Calculate the equilibrium concentrations of NO at both temperatures, is your boss right? 5. Water is a precious resource on the desert planets like Tatooine located in the Outer Rim Territories. You decide to apply your skills as a Materials Scientist to manufacture water on a planet with an atmosphere of 80% H2 and 18% CO2 and 1% H20 and 1% CO using the reverse water gas shift reaction which consumes CO20e) and H20e) to generate H20og and Code). (a) Use the table of free energy values to determine whether or not H20 production is spontaneous at 298K. Justify. AGO H20 at 298K AGO H2 at 298K. AGO CO at 298K AGO CO2 at 298K -137.16 kJ/mol -3 0 kJ/mol (b) What voltage will be produced or required from the reaction in part a?

Explanation / Answer

At 1000k, deltaGO= -RT1lnK1, where T1 and lnK1 are 1000 K and equilibrium constnt at 1000K. R units are J/mole.K. Hence deltaGO also needs to be converted into J/mole.

lnK1= -detaGO/RT= - 77.7*1000/(1000*8.814)= -9.31

K1= exp(-9.31)= 9.05*10-5

For the reaction, N2+O2--à2NO

K1= [PNO]2/ [PN2] [PO2]

[PNO], [PN2] [PO2] are partial pressures of NO, N2 and O2 respectively,

= [PNO]2/ (0.69*0.06) =9.05*10-5

[PNO]2/ 0.0414 =9.05*10-5 , [PNO] = .00193

Let K2= equilibrium cosntant at 1500K

lnK2= -71.4*1000/(8.314*1500)= =5.72, K2= 0.00324

For the reaction, N2+O2--à2NO

K2= [PNO]2/ [PN2] [PO2] = [PNO]2/ (0.69*0.06) =0.00324

[PNO]2/ 0.0414 =0.00324 , [PNO] = 0.0115

The boss is incorrect since upon increasing the temperature, the concentrations of [NO] decreased as shown by the calculations of partial pressures of [NO].

5.

The water gas shift reaction is

CO2 +H2ß-> CO+H2O

deltaG= 1* deltaG of CO + 1* deltaG of H2O-{ 1* deltaG of CO2 + 1* deltaG of H2}

where 1,1,1 and 1 are moles CO, H2O, CO2 and H2 respectively.

deltaG= -137.6 + 394.39+237.14 =493.93 Kj/mol. The reaction is not spontaneous as indicated by the postive value of deltaG for the reaction.

deltaG= -nFE

493.93*1000= -96500*1* E

E= -5.11 Volts

Related Questions

Navigate

• Browse (All)
• Browse 4
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.