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| author | priyanka | 2015-06-24 15:03:17 +0530 |
|---|---|---|
| committer | priyanka | 2015-06-24 15:03:17 +0530 |
| commit | b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (patch) | |
| tree | ab291cffc65280e58ac82470ba63fbcca7805165 /572/CH14/EX14.9/c14_9.sce | |
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initial commit / add all books
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| -rwxr-xr-x | 572/CH14/EX14.9/c14_9.sce | 20 |
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diff --git a/572/CH14/EX14.9/c14_9.sce b/572/CH14/EX14.9/c14_9.sce new file mode 100755 index 000000000..6e1cac474 --- /dev/null +++ b/572/CH14/EX14.9/c14_9.sce @@ -0,0 +1,20 @@ +//(14.9) As a result of heating, a system consisting initially of 1 kmol of CO2,.5 kmol of O2, and 5 kmol of N2 forms an equilibrium mixture of CO2, CO, O2, N2, and NO at 3000 K, 1 atm. Determine the composition of the equilibrium mixture.
+
+
+
+//solution
+
+//The overall reaction can be written as
+//CO2 + .5O2 + .5N2 ----> aCO + bNO + (1-a)CO2 + .5(1+a-b)O2 + .5(1-b)N2
+
+//At 3000 K, Table A-27 provides
+log10K1 = -.485 //equilibrium constant of the reaction CO2 <--> CO + .5O2
+log10K2 = -.913 //equilibrium constant of the reaction .5O2 + .5N2 <-->NO
+
+K1 = 10^log10K1
+K2 = 10^log10K2
+
+//solving equations K1 = (a/(1-a))*((1+a-b)/(4+a))^.5 and K2 = 2b/((1+a-b)*(1-b))^.5
+a = .3745
+b = .0675
+printf('The composition of the equilibrium mixture, in kmol per kmol of CO2 present initially, is then 0.3745CO, 0.0675NO, 0.6255CO2, 0.6535O2, 0.4663N2.')
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