From b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b Mon Sep 17 00:00:00 2001
From: priyanka
Date: Wed, 24 Jun 2015 15:03:17 +0530
Subject: initial commit / add all books

---
 572/CH14/EX14.4/c14_4.sce | 25 +++++++++++++++++++++++++
 1 file changed, 25 insertions(+)
 create mode 100755 572/CH14/EX14.4/c14_4.sce

(limited to '572/CH14/EX14.4')

diff --git a/572/CH14/EX14.4/c14_4.sce b/572/CH14/EX14.4/c14_4.sce
new file mode 100755
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--- /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)
\ No newline at end of file
-- 
cgit