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120    ENERGY AND THE FIRST LAW OF THERMODYNAMICS


              We are only allowed      (1)  We wish to go from the left-hand side of the reaction to the
              to make a choice of          right-hand side. We can either follow the arrow labelled
              route like this because       H r , or we pass via the elements (along the bottom line)
              enthalpy is a state          and thence back up to the ethanol.
              function.
                                       (2)  If we go along an arrow in the same direction as the arrow
                                           is pointing, then we use the value of  H as it is written.
                        (3)  If we have to go along an arrow, but in the opposite direction to the
                             direction in which it points, then we multiply the value of  H by ‘−1’.

                      In the example here, to go from the left-hand side to the right-hand side via the
                      elements, we need to go along two arrows  H f(1) and  H f(3) in the opposite directions
                      to the arrows, so we multiply the respective values of  H and multiply each by −1.
                      We then go along the arrow  H f(2) , but this time we move in the same direction as
                      the arrow, so we leave the sign of the enthalpy unaltered.
                                        And then we tie the threads together and say:
              Note there are three
              arrows, so there are            H r = (−1 ×  H f(1) ) + (−1 ×  H f(3) ) +  H f(2)
              three  H terms within
               H r .                  Inserting values into this equation:

                                                                        −1
                                                 −1
                                                                                         −1
                           H r = (−1 ×−52 kJ mol ) + (−1 ×−286 kJ mol ) + (−235 kJ mol )
                                                                        −1
                           H r = 52 kJ mol −1  + 286 kJ mol −1  + (−235 kJ mol )
                      so
                                                   H r = 103 kJ mol −1

                      We obtained this value of  H r knowing the other enthalpies in the cycle, and
                      remembering that enthalpy is a state function. Experimentally, the value of  H r =
                               −1
                      99 kJ mol , so this indirect measurement with Hess’s law provides relatively good
                      data.
                        Sometimes, these cycles are considerably harder than the example here. In such
                      cases, it is usual to write out a cycle for each reaction, and then use the results from
                      each cycle to compile another, bigger cycle.



                       How does an ‘anti-smoking pipe’ work?

                      Hess’s Law Cycles with Enthalpies of Combustion

                      Smoking causes severe damage to the heart, lungs and respiratory system. The tobacco
                      in a cigarette or cigar is a naturally occurring substance, and principally comprises
                      the elements carbon, oxygen, hydrogen and nitrogen.
                        Unfortunately, because the tobacco is contained within the bowl of a pipe or a paper
                      wrapper, complete combustion is rare, meaning that the oxidation is incomplete. One
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