Page 130 - Physical chemistry understanding our chemical world

INTRODUCTION TO THERMODYNAMICS: INTERNAL ENERGY 97

Thirdly, we determine the amount of glucose consumed n. We obtain the value of n
−1
as ‘amount = mass ÷ molar mass’. The molar mass of glucose is 180 g mol ,sothe
number of moles is 5.88 × 10 −2 mol.
Finally, we calculate the value of U from Equation (3.9). Insert-
ing values: The minus sign is a
consequence of the
134 880 J K −1 × 1.224 K way Equation (3.9) is
U (combustion) =−
0.0588 mol written.
so
U (combustion) =−2.808 MJ mol −1 An energy change of MJ
mol −1 is exceptional.
Notice how this value of U is negative. As a good generalization, the Most changes in U are
change in internal energy U liberated during combustion is negative, smaller, of the order of
which helps explain why so many fires are self-sustaining (although kJ mol −1 .
see Chapter 4).
The value of U (combustion) for glucose is huge, but most values of
U are smaller, and are expressed in kilo joules per mole. We often calculate a
volume of U but
SAQ 3.4 A sample of anthracene (C 14 H 10 , II)was burnt cite the answer after
in a bomb calorimeter. A voltage of 10 V and a current adjusting for pres-
of 1.2 A were passed for exactly 15 min to achieve the sure–volume work;
same rise in temperature as that caused by the burning see p. 102.
of 0.40 g. Calculate the molar energy liberated by the
anthracene.

(II)

Aside

The large value of U in Worked Example 3.5 helps explain why sweets, meals and
drinks containing sugar are so fattening. If we say a single spoonful of sugar comprises
5 g of glucose, then the energy released by metabolizing it is the same as that needed
to raise a 3.5 kg weight from the ground to waist level 1000 times.
(We calculate the energy per lift with Equation (3.4), saying E = m × g × h,where
m is the mass, g is the acceleration due to gravity and h is the height through which
the weight is lifted.)