Page 296 - Advanced Thermodynamics for Engineers, Second Edition
P. 296

12.11 DISSOCIATION PROBLEMS          285




               nitric oxide is positive, i.e. the reaction which forms it is endothermic, and hence combination of
               nitrogen and oxygen tends to reduce the Gibbs energy of the mixture. Note that other compounds tend
               to dissociate at high temperature because their energies of formation are negative, i.e. the reactions
               which form them are exothermic.


               Example
               Methane is burned with a stoichiometric quantity of air and achieves a pressure of 99.82 bar and a
               temperature of 2957 K after combustion, if dissociation of the carbon dioxide and water vapour are
               taken into account. The mole fractions of the constituents are 7.495% CO 2 , 1.890% CO, 18.183%
               H 2 O, 0.6425% H 2 , 1.267% O 2 and 70.58% N 2 . Calculate the amount of nitric oxide formed at this
               temperature, neglecting the effect of the NO formation on the dissociation of carbon monoxide and
               water. Estimate the effect of the NO formation on the temperature of the products. The energy of the
               reactants, U R (T R ) ¼ 179377 kJ.
               Solution:
                  Chemical equation without dissociation

                                    CH 4 þ 2ðO 2 þ 3:76N 2 Þ/CO 2 þ 2H 2 O þ 7:52N 2      (12.152)
                  Chemical equation with dissociation

                   CH 4 þ 2ðO 2 þ 3:76N 2 Þ

                   /n P ð0:07495CO 2 þ 0:0189CO þ 0:18183H 2 O þ 0:006425H 2 þ 0:01267O 2 þ 0:7058N 2 Þ
                        |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}
                                                           a 1  a 2
                                                             þ
                                                   n P ¼10:52þ
                                                           2    2
                                                                                          (12.153)
                  It is possible to evaluate the total amount of substance in the products as follows.
                                        1   a 1             a 1
                                                               ; giving n P ¼ 10:655:
                               0:07495 ¼      ; and 0:0189 ¼
                                          n P               n P
               and hence

                    CH 4 þ 2ðO 2 þ 3:76N 2 Þ
                                                                                          (12.154)
                    / 0:7986CO 2 þ 0:2014CO þ 1:9315H 2 O þ 0:06845H 2 þ 0:13495O 2 þ 7:52N 2
                       |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}
                                                   n P ¼10:655
                  Assume that combination of the nitrogen and oxygen occurs to form nitric oxide.

                                                      a 3    a 3
                                              a 3 NO 5          N 2                       (12.155)
                                                       2  O 2 þ  2
                  The temperature of the products is T P ¼ 2956 K, giving
                                                             p rNO
                                                                   :                      (12.156)
                                             K p r3  z 0:11279 ¼  1=2 1=2
                                                            p  p
                                                             rO 2 rN 2
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