Page 253 - Advanced thermodynamics for engineers
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240    CHAPTER 11 CHEMISTRY OF COMBUSTION




                Hence, from Fig. 11.1,
                                     X        X              X          X
                           ðDH f Þ  ¼   DH a      DHðX   YÞ      DH res    DH latent      (11.4)
                                CO 2
                In this case there is a resonance energy but the latent energy (i.e. latent heat) is zero.
                Then

                                       h      i      h     i                  h   i
                         ðDH f Þ  ¼ DH a C graphite þ DH a O ¼ O   2ðC ¼ OÞ  DH res CO 2
                              CO 2
                                 ¼ 717:2 þ 435:4   2   698:1   137:9                      (11.5)
                                 ¼ 381:5MJ=kmol
                The tabulated value is  393.4 MJ/kmol.
              Water (H 2 O)
                This is formed from the hydrogen and oxygen molecules in the reaction
                                                  1
                                           H 2 ðgÞþ O 2 ðgÞ/H 2 OðgÞ                      (11.6)
                                                  2
                Thus

                                     X        X               X         X
                           ðDH f Þ H 2 O  ¼  DH a    DHðX   YÞ   DH res    DH latent      (11.7)
                In this case DH res ¼ 0 but DH latent depends upon the phase of the water. First, consider the water
             is in vapour phase, when DH latent ¼ 0. Then
                                X        X
                     ðDH f Þ H 2 O  ¼  DH a    DHðX   YÞ
                                            1
                              ¼ DH a ½H   HŠþ DH a ½O ¼ OŠ ½H   OHŠ ½O   HŠ               (11.8)
                                            2
                                       1
                              ¼ 435:4 þ ð498:2Þ  497:5   428:7
                                       2
                              ¼ 241:7MJ=kmol
                This compares well with the tabulated values of  241.6 MJ/kmol.
                If the water was in liquid phase then Eqn (11.6) becomes
                                                  1
                                           H 2 ðgÞþ O 2 ðgÞ/H 2 Oð[Þ                      (11.9)
                                                  2
                                     ðDH f Þ   ¼ðDH f Þ    þðm w Þ  h
                                          H 2 OðlÞ    H 2 OðgÞ    H 2 O fg
                                               ¼ 241:7   18   2441:8=1000                (11.10)

                                               ¼ 285:65 MJ=kmol
                The tabulated value is  285.6 MJ/kmol.
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