1 files changed, 34 insertions, 0 deletions
diff --git a/3012/CH14/EX14.2/Ex14_2.sce b/3012/CH14/EX14.2/Ex14_2.sce
new file mode 100755
index 000000000..23b2160e2
--- /dev/null
+++ b/3012/CH14/EX14.2/Ex14_2.sce
@@ -0,0 +1,34 @@
+
+// Given:-
+// Applying conservation of mass, the overall balanced chemical reaction equation is
+// CO + .5O2    ------    zCO + (z/2)O2 + (1-z)CO2
+
+// At 2500 K, Table A-27 gives
+log10K = -1.44
+// Part(a)
+p = 1.0                                                                         // in atm
+// Calculations
+K = (10.0)**(log10K)                                                            // equilibrium constant
+// Solving equation K = (z/(1-z))*(2/(2 + z))^.5 *(p/1)^.5 gives
+z = 0.129
+yCO = 2.0*z/(2.0 + z)
+yO2 = z/(2.0 + z)
+yCO2 = 2.0*(1.0 - z)/(2.0 + z)
+
+// Results
+printf( ' Part(a) mole fraction of CO is: %.3f ',yCO)
+printf( ' Mole fraction of O2 is:  %.3f',yO2)
+printf( ' Mole fraction of CO2 is:  %.3f',yCO2)
+
+// Part(b)
+p = 10.0                                                                          // in atm
+// Solving equation K = (z/(1-z))*(2/(2 + z))^.5 *(p/1)^.5 gives
+z = 0.062
+yCO = 2.0*z/(2.0 + z)
+yO2 = z/(2.0 + z)
+yCO2 = 2.0*(1.0 - z)/(2.0 + z)
+
+// Results
+printf( ' Part(b) mole fraction of CO is:  %.3f',yCO)
+printf( ' Mole fraction of O2 is:  %.3f',yO2)
+printf( ' Mole fraction of CO2 is: %.3f ',yCO2)