Example 33.3 Chemical problem Chemical reaction:1 mol CO and 0.5 mol O2 at p=1 atm,=25 C,final product 1 mol CO2.measured rejected heat 283 190 J,no work output. Assuming the molar basis entropy of CO,O2,and CO2 are 197.67 J/(mol-K), 205.167 J/(mol-K),and 213.82 J/(mol-K)at (p.t),respectively. Question:are these data correct/reliable?(surrounding to=25 C) Solution:Take CO,O2,COz and the surrounding as system C0+02=C02 a-wo+Sj片as Isolated system =1mol×213.82J/mol.K)-[1mol×197.67J/mol·K) 283190J +0.5mol×205.16J/mol·K)]+ =863.4J/K>0 (273+25)K So,not violate the 2nd law.Increase of entropy principle also applies to chemical reaction. 上游充通大学 April 23,2019 9 SHANGHAI JLAO TONG UNIVERSITYApril 23, 2019 9 Chemical reaction: 1 mol CO and 0.5 mol O2 at p = 1 atm, t= 25 ℃, final product 1 mol CO2 . measured rejected heat 283 190 J, no work output. Assuming the molar basis entropy of CO, O2 , and CO2 are 197.67 J/(mol·K), 205.167 J/(mol·K) , and 213.82 J/(mol·K) at (p, t), respectively. Question: are these data correct/reliable?(surrounding t0= 25 ℃) Solution： 2 2 1 CO+ O =CO 2 Take CO, O2 , CO2 and the surrounding as system 2 2 iso CO CO O 0 1 2 S S S S S             1 mol 213.82 J/(mol K) [1 mol 197.67 J/(mol K) 283 190 J 0.5 mol 205.16 J/(mol K)] 863.4 J/K 0 (273 25) K              So, not violate the 2nd law. Increase of entropy principle also applies to chemical reaction. Example 33.3 Chemical problem Isolated system