Physical Chemistry , 1st ed.

If the standard enthalpy of formation of glucose is 1277 kJ/mol, what is the

rxnH° for this process? You will need to get fH° values from Appendix 2.

Solution

The fHvalues for CO 2 (g) and H 2 O () are 393.51 and 285.83 kJ/mol,

respectively. Therefore, we use equation 2.55 and find

rxnH°6(393.51) + 6(285.83)  (1277) kJ for the process

fH° (products) fH° (reactants)

In expressions like these, it is important to keep track of all of the negative

signs. Evaluating:

rxnH°2799 kJ

By noting that the coefficients from the balanced chemical reaction are the

number of moles of products and reactants, we lose the moles in the denomi-

nator of the rxnH°. Another way to consider it is to say that 2799 kJ of energy

are given off when 1 mole of glucose reacts with 6 moles of oxygen to make 6

moles of carbon dioxide and 6 moles of water. This eliminates the question

“moles of what?” that would be raised if a kJ/mol unit were used for rxnH°.

The products-minus-reactants tactic is a very useful one in thermodynamics. It
is also a useful idea to carry along with respect to other state functions: the change
in any state function is the final value minus the initial value. In Example 2.17
above, the state function of interest was enthalpy, and by applying Hess’s law and
the definition of formation reactions, we were able to develop a procedure for de-
termining the changes in enthalpy and internal energy for a chemical process.
What is the relationship between Hand Ufor a chemical reaction? If one
knows the fUand fHvalues for the products and reactants, one can simply
compare them using the products-minus-reactants scheme of equations 2.55
and 2.56. There is another way to relate these two state functions. Recall the
original definition ofHfrom equation 2.16:

HU+ pV

We also derived an expression for dHas

dHdU+ d(pV)

dHdU+ p dV+ V dp

where the second equation above was obtained by applying the chain rule. There
are several ways we can go with this. If the chemical process occurs under con-
ditions of constant volume, then the p dVterm is zero and dUdqV(because
work 0). Therefore,
dHVdqV+ V dp (2.57)

The integrated form of this equation is

HVqV+ Vp (2.58)

Since dUdqunder constant volume conditions, this gives us one way to cal-
culate how dHdiffers from dU. Under conditions of constant pressure, the
equation becomes
HpU+ pV (2.59)

and we have a second way of evaluating Hand Ufor a chemical process.

2.10 Chemical Changes 57

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