1 (568)

Prévia do material em texto

560 16MOLECULES INMOTION
�is expression is compared with that in the question by taking T∗ = (20 +
273) = 293 K and rewriting the relationship as
log η/η20 =
1.3272[20 − (T/K − 273)] − 0.001053[20 − (T/K − 273)]2
(T/K − 273) + 105
where θ/○C = (T/K − 273) is used. One approach is to generate values of
log η/η20 in the range 20 ○C to 100 ○C and then plot these data against 1/T ;
according to eqn 16.2 the slope of such a graph is Ea/R ln 10. Such a plot is
shown in Fig. 16.1.
0.0026 0.0028 0.0030 0.0032 0.0034
−0.6
−0.4
−0.2
0.0
(1/T/K)
lo
g(
η/
η 2
0)
Figure 16.1
�e data are a modest �t to a straight line with equation
log (η/η20) = (7.4667 × 102) × 1/(T/K) − 2.5655
�e activation energy is computed from the slope as
Ea = R ln 10 × (slope)
= (8.3145 JK−1mol−1) × (ln 10) × (7.4667 × 102 K) = 14.3 kJmol−1
P16B.4 �e conductance G is the reciprocal of the resistance R, G = 1/R. �e con-
ductivity of the solution in the cell depends on the measured conductance and
the physical dimensions of the cell. However, because in these measurements
the same cell is being used, it follows that the conductivity of the solution is
given by κ = A/R, where A is a constant. If the resistance of two solutions are
measured in the cell the ratio of the measured resistances is the inverse of the
ratio of their conductivities: R1/R2 = κ2/κ1.
Both solutions are aqueous and hence their conductivities consist of contribu-
tions from the water component and acid component.�erefore
κ(acid solution)
κ(KCl solution)
= κ(acid) + κ(water)
κ(KCl) + κ(water)
= R(KCl solution)
R(acid solution)