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clear;
clc;
// Stoichiometry
// Chapter 5
// Energy Balances
// Example 5.28
// Page 261
printf("Example 5.28, Page 261 \n \n");
// solution
lv1 = 26694 // kj/kmol
Tc = 466.74
lv2 = lv1*((Tc-298.15)/(Tc-307.7))^.38/1000 // kJ/mol
Hf = -252 // kJ/mol
Hf1 = Hf-lv2 // kJ/kmol
printf("Heat of formation of liquid di ethyl ether = "+string(Hf1)+" kJ/mol.")
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