Standard Heats of Reaction |

Chemical scientists have agreed upon a standard reference set of conditions
for temperature and pressure. These conditions have of course been chosen
so that experiments can be done easily. Thus the standard reference temperature
is 25 ^{o}C, which is just slightly above normal room temperature.
This is easily controlled in a water bath. The reference pressure is 1 atmosphere
or 101.325 kPa. |

When the enthalpy change of a reaction is determined with all reactants
and products at 1 atm and some given temperature, and when the scale of the
reaction is in the moles specified by the coefficients of the equation,
then H is the standard enthalpy change or the standard heat of reaction.
To show that a pressure of 1 atm is used, the symbol H is given a superscript,
^{o}, to make the symbol H^{o}. Values of H^{o }usually
correspond to an initial and final temperature of 25^{o}C, unless
otherwise specified. |

The units of H ^{o }are normally kilojoules. For example, a
reaction between gaseous nitrogen and hydrogen produces gaseous ammonia according
to the equation |

N _{2}(g) + H_{2}(g) ----> 2 NH_{3}(g) |

When 1.00 mol of N _{2 }react with 3.00 mol of H_{2 }to
form 2.00 mol of NH_{3 }at 25^{o }and 1 atm, the reaction
releases 92.38 kJ. Hence for the reaction as written H^{o }= -92.38
kJ. |

Thermochemical Equations |

Often it is useful to make the enthalpy change of a reaction part
if its equation. When we do this we have to be very careful about the coefficients,
and we must indicate the physical states of all the reactants and products.
The reaction between gaseous nitrogen and hydrogen to form gaseous ammonia,
for example, releases 92.38 kJ is 2.00 mol of NH _{3 }forms. |

But if we were to make twice as much, or 4.00 mol, of NH
_{3 }from
2.00 mol of N_{2 }and 6.00 mol of H_{2}, then twice as much
heat (184.8 kJ) would be released. On the other hand, if only 0.50 mol of
N_{2 }and 1.50 mol of H_{2 }were to react to form only 1.00
mol of NH_{3}, then only half as much heat (46.19 kJ) would be released. |

An equation that includes its value of H
^{o }is called a thermochemical
equation. The following three thermochemical equations for the formation
of ammonia, for example, give the quantitative data describe in the preceding
paragraph and correctly specify the physical states of all substances. |

N _{2}(g) + 3 H_{2}(g) ----> 2 NH_{3}(g) H^{o
}= -92.38 kJ |

2 N _{2}(g) + 6 H_{2}(g) ----> 4 NH_{3}(g) H^{o
}= -184.8 kJ |

½ N _{2}(g) + 1½ H_{2}(g) ----> NH_{3}(g)
H^{o }= -46.19 kJ |

When you read a thermochemical equation, always interpret the coefficient
as moles. This is why we must use fractional coefficients in such an
equation, where normally we try to avoid them (because we cannot have fractions
of molecules). In thermochemical equations, however, fractions are allowed,
because we can have fractions of moles. |

Thermochemical Equations for Experimentally Difficult
Reactions |

Once we have the thermochemical equation for a particular reaction,
we automatically have all the information we need for the reverse reaction.
The thermochemical equation for the combustion of carbon to give carbon dioxide
is: |

C(s) + O _{2}(g) -------> CO_{2}(g) H^{o }= -393.5
kJ |

The reverse reaction would be experimentally impossible to preform,
for reasons that will become clear later. But from the above equation we can
write the reverse reaction. |

CO _{2 }(g) ------> C(s) + O_{2 }(g) H^{o }= +393.5
kJ |

If we have the H ^{o }for a given reaction, the H^{o }for
the reverse reaction has the same numerical value, but its algebraic sign
is reversed. |