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Diffstat (limited to '572/CH14/EX14.4')
| -rwxr-xr-x | 572/CH14/EX14.4/c14_4.sce | 25 |
1 files changed, 25 insertions, 0 deletions
diff --git a/572/CH14/EX14.4/c14_4.sce b/572/CH14/EX14.4/c14_4.sce new file mode 100755 index 000000000..74401fc65 --- /dev/null +++ b/572/CH14/EX14.4/c14_4.sce @@ -0,0 +1,25 @@ +//(14.4) One kilomole of carbon monoxide reacts with the theoretical amount of air to form an equilibrium mixture of CO2, CO, O2, and N2 at 2500 K and 1 atm. Determine the equilibrium composition in terms of mole fractions, and compare with the result of Example 14.2.
+
+
+
+//solution
+
+//For a complete reaction of CO with the theoretical amount of air
+//CO + .5 O2 + 1.88N2 -----> CO2 + 1.88N2
+//Accordingly, the reaction of CO with the theoretical amount of air to form CO2, CO, O2, and N2 is
+//CO + .5O2 + 1.88N2 ---> zCO + z/2 O2 + (1-z)CO2 + 1.88N2
+
+K = .0363 //equilibrium constant the solution to Example 14.2
+p =1 //in atm
+pref = 1 //in atm
+
+//solving K = (z*z^.5/(1-z))*((p/pref)*2/(5.76+z))^.5 gives
+z = .175
+yCO = 2*z/(5.76 + z)
+yO2 = z/(5.76 + z)
+yCO2 = 2*(1-z)/(5.76 + z)
+yN2 = 3.76/(5.76 + z)
+printf('the mole fraction of CO is: %f',yCO)
+printf('\nthe mole fraction of O2 is: %f',yO2)
+printf('\nthe mole fraction of CO2 is: %f',yCO2)
+printf('\nthe mole fraction of N2 is: %f',yN2)
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