1 files changed, 63 insertions, 0 deletions
diff --git a/572/CH14/EX14.1/c14_1.sce b/572/CH14/EX14.1/c14_1.sce
new file mode 100755
index 000000000..c101c0225
--- /dev/null
+++ b/572/CH14/EX14.1/c14_1.sce
@@ -0,0 +1,63 @@
+//(14.1)  Evaluate the equilibrium constant, expressed as log10K, for the reaction at (a) 298 K and (b) 2000 K. Compare with the value obtained from Table A-27.
+  
+  
+//solution
+
+//The reaction is CO + .5O2  ---->   CO2
+
+//part(a)
+T = 298                                                                         //in kelvin
+Rbar = 8.314                                                                    //universal gas constant in SI units
+//from table A-25
+
+hfbarCO2 = -393520                                                              //in kj/kmol
+hfbarCO = -110530                                                               //in kj/kmol
+hfbarO2 = 0                                                                     //in kj/kmol
+deltahbarCO2 = 0                                                                //in kj/kmol
+deltahbarCO = 0                                                                 //in kj/kmol
+deltahbarO2 = 0                                                                 //in kj/kmol
+sbarCO2 = 213.69                                                                //in kj/kmol.K
+sbarCO = 197.54                                                                 //in kj/kmol.K
+sbarO2 = 205.03                                                                 //in kj/kmol.K
+
+deltaG = [hfbarCO2-hfbarCO-.5*hfbarO2] + [deltahbarCO2-deltahbarCO-.5*deltahbarO2] - T*(sbarCO2-sbarCO-.5*sbarO2)
+lnK = -deltaG/(Rbar*T)
+logK = (1/log(10))*lnK
+//from table A-27
+logKtable = 45.066
+printf('part(a) the value of equilibrium constant expressed as log10K is:  %f',logK)
+printf('\nthe value of equilibrium constant expressed as log10K from table A-27 is:  %f',logKtable)
+
+//part(b)
+T = 2000                                                                        //in kelvin
+//from table A-23
+hfbarCO2 = -393520                                                              //in kj/kmol
+hfbarCO = -110530                                                               //in kj/kmol
+hfbarO2 = 0                                                                     //in kj/kmol
+deltahbarCO2 = 100804-9364                                                      //in kj/kmol
+deltahbarCO = 65408 - 8669                                                      //in kj/kmol
+deltahbarO2 = 67881 - 8682                                                      //in kj/kmol
+sbarCO2 = 309.210                                                               //in kj/kmol.K
+sbarCO = 258.6                                                                  //in kj/kmol.K
+sbarO2 = 268.655                                                                //in kj/kmol.K
+
+deltaG = [hfbarCO2-hfbarCO-.5*hfbarO2] + [deltahbarCO2-deltahbarCO-.5*deltahbarO2] - T*(sbarCO2-sbarCO-.5*sbarO2)
+lnK = -deltaG/(Rbar*T)
+logK = (1/log(10))*lnK
+//from table A-27
+logKtable = 2.884
+printf('\n\npart(b) the value of equilibrium constant expressed as log10K is:  %f',logK)
+printf('\nthe value of equilibrium constant expressed as log10K from table A-27 is:  %f',logKtable)
+
+
+
+
+
+
+
+
+
+
+
+
+