Question: #4. (20%) Analyze the NOx equilibrium by first constructing the corresponding van't Hoff graph and comparing it with the information available in Chapter 16 and

van't Hoff graph and comparing it with the information available in Chapter

16 and then summarizing the NO/NO2 (and N2O/NO?) dependence on P and

T. Explain how this confirms, of course, the validity of the Le

#4. (20%) Analyze the NOx equilibrium by first constructing the corresponding van't Hoff graph and comparing it with the information available in Chapter 16 and then summarizing the NO/NO2 (and N2O/NO?) dependence on P and T. Explain how this confirms, of course, the validity of the Le Chatelier principle. -Do some of your graphs look like this? Comments? (See also Q3 comments above...) 1.0E-01 0.0004 0.0005 0.0006 0.0007 0.0008 0.0009 0.0010 1.0E-02 1.0E-03 -y = 1.044E-04e6.998E+03x R2 = 1.000E+00 y = 4.580E+00e-1.087E+04x R2 = 1.000E+00 = 1.0E-04 1.0E-05 ---0.5N2+0.502=NO -NO+0.502-NO2 1.08-06 y = 1.576E-04e-9.975E+03x R2 = 9.999E-01 -N2+0.502-N20 = 1.0E-07 1.0E-08 10000 --T=2000 K -T=1300 K 3000 Comments: -NO concentrations vary from 9468 ppm at 2500 K and 100 atm to 184 at 1300 K and 0.01 atm. The concentrations of NO2 and N2O are always much lower than that Effect of P (atm) on equilibrium NO/NO, ratio (YN2=0.76, YO2=0.04) 0.01 10 300 2500 Effect of T (K) on equilibrium NO/NO, ratio (yN2=0.76, yO2=0.04) 2000 --P=1 atm P=10 atm 3800 --P=100 atm 3000 8 500 a 500 700 300 1100 1300 1500 1700 2100 2100 2500 2700 3 of 4 P100 atm -Pa 10 atm Palatm Effect of T (K) on equilibrium N20/NO ratio (YN2=0.76, yO2=0.04) 10 10 TO -T=1300 K 2000 K LE 200 Effect of P (atm) on equilibrium N,0/NO ratio (YN2=0.76, YO2=0.04) 30 #4. (20%) Analyze the NOx equilibrium by first constructing the corresponding van't Hoff graph and comparing it with the information available in Chapter 16 and then summarizing the NO/NO2 (and N2O/NO?) dependence on P and T. Explain how this confirms, of course, the validity of the Le Chatelier principle. -Do some of your graphs look like this? Comments? (See also Q3 comments above...) 1.0E-01 0.0004 0.0005 0.0006 0.0007 0.0008 0.0009 0.0010 1.0E-02 1.0E-03 -y = 1.044E-04e6.998E+03x R2 = 1.000E+00 y = 4.580E+00e-1.087E+04x R2 = 1.000E+00 = 1.0E-04 1.0E-05 ---0.5N2+0.502=NO -NO+0.502-NO2 1.08-06 y = 1.576E-04e-9.975E+03x R2 = 9.999E-01 -N2+0.502-N20 = 1.0E-07 1.0E-08 10000 --T=2000 K -T=1300 K 3000 Comments: -NO concentrations vary from 9468 ppm at 2500 K and 100 atm to 184 at 1300 K and 0.01 atm. The concentrations of NO2 and N2O are always much lower than that Effect of P (atm) on equilibrium NO/NO, ratio (YN2=0.76, YO2=0.04) 0.01 10 300 2500 Effect of T (K) on equilibrium NO/NO, ratio (yN2=0.76, yO2=0.04) 2000 --P=1 atm P=10 atm 3800 --P=100 atm 3000 8 500 a 500 700 300 1100 1300 1500 1700 2100 2100 2500 2700 3 of 4 P100 atm -Pa 10 atm Palatm Effect of T (K) on equilibrium N20/NO ratio (YN2=0.76, yO2=0.04) 10 10 TO -T=1300 K 2000 K LE 200 Effect of P (atm) on equilibrium N,0/NO ratio (YN2=0.76, YO2=0.04) 30

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