Page 115 - Essentials of physical chemistry
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The First Law of Thermodynamics                                              77


                                                          (26:78 þ 26:58)

                                    a H 2   a Cl 2
                                                                         ¼ 4:00
                                     2       2                  2
                        Da ¼ a HCl              ¼ 30:68

                                                       (1:224 þ 3:722)
                             b H 2   b Cl 2                             2
                                                                     (10 ) ¼ 0:034417
                              2       2                      2
                 Db ¼ b HCl               ¼ 0:9687

                                                   ( 2:299   5:035)
                             c H 2  c Cl 2                            5             5
                                                                   (10 ) ¼ 5:423   10
                 Dc ¼ c HCl              ¼ 1:756
                              2      2                    2

                                                    (19:40 þ 31:09)
                                                                  (10 ) ¼ 33:269   10
                           d H 2    d Cl 2                            9               9
               Dd ¼ d HCl                ¼ 8:024
                            2        2                    2

                                                  ( 5:261   7:137)

                            e H 2   e Cl 2                           12             12
                                                                  (10  ) ¼ 7:286   10
                 De ¼ e HCl             ¼ 1:087
                             2       2                   2
            Next we need to compute the main term

                                   1   0       1   0
                    0
                          0
                 DH                               H (Cl 2 ) ¼ 92:31   (0 þ 0)=2 ¼ 92:31 kJ
                                       f
                   rxn  ¼ H (HCl)    H (H 2 )      f
                          f
                                   2           2
              Da ) (4:00)(1200   298:15) ¼ 3607:4 J=mol
                                    2
                                              2
                              [(1200)   (298:15) ]
              Db ) ( 0:034417)                  ¼ 23,250:51754 J=mol
                                       2
                                      3
                                                3
                                [(1200)   (298:15) ]
                              5
              Dc ) (5:423   10 )                  ¼ 30,757:8373 J=mol
                                        3
                                         4
                                                   4
                                   [(1200)   (298:15) ]
                                 9
              Dd ) ( 33:269   10 )                   ¼ 17,180:92635 J=mol
                                           4
                                                 5
                                       5
                                 [(1200)   (298:15) ]
              De ) (7:286   10  12 )               ¼ 3622:54675 J=mol
                                         5
              Sum of terms ¼ 2443.65984 J=mol ¼ 2.44365984 kJ=mol ffi 2.44 kJ=mol.
                      0
              Thus, DH (1200 K) ¼ 92:31   2:44 ¼ 94:75 kJ=mol.

                      rxn
            Other texts use more convenient shorter polynomial expansions for this reaction based on older data
            from 1934–1948 [6] but this result comes from the use of the more recent heat capacity data in the
                                    2
            CRC Handbook [8] and the R values for the polynomials used here indicate excellent numerical
            fitting to the experimental data points. As such we believe this result is more accurate than the result
                                     3
            using polynomials only up to T . This text also shows how the polynomials were determined and
                       2
            provides the R values to evaluate goodness of the fitting procedure. While that is a lot of work for a
            correction of less than 3%, we admit to a tendency of physical chemists to make extra effort to gain
            accuracy. Note the interesting alternating signs of the various correction terms that result in the net
            correction. We carried all the places on a ten place calculator to allow students to follow the
            computation, but in the end the answer was rounded to only four significant figures. The educational
            value of this exercise is that we do have a way to correct DH 0  for temperatures other than 298.158K.
                                                           rxn
            The example also teaches us that when faced with a complicated calculation it is useful to organize
            the overall process into separate steps. It is tempting to think this problem could be programmed for
            automation in Basic, f 77, or Java to make the whole process less of a chore for humans. Even so the
            whole process needs to be worked out at least once to check out any automatic program.
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