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August 18, 2010 11:36 9in x 6in b985-ch07 Elementary Physical Chemistry

Chemical Kinetics 67

c) After a time period, t, the concentration of N 2 O 5 is 10% of the initial
value. What is the value of t?

Solutions
1
a) r = d[NO 2]/dt = k[N 2O 5]
4
The rate of formation of N 2 O 5 is d[NO 2 ]/dt =4 k [N 2 O 5 ]
b) The rate, r, can be written also as r = −1/2d[N 2O 5]/dt,and so
−d[N 2O 5]/dt =2k[N 2O 5 ]= k A[N 2O 5]
showing that the rate of reaction is of first order. Consequently,
t 1/2 =ln 2/k A.Thus,
s
t 1/2 =ln 2/(2 × 1.73 × 10 −5 −1 )= 20.0s
c) ln[A] 0 /[A]= k A t
ln[A] 0 /(0.1)[A] 0 =ln 10 =(2 × 1.73 × 10 −5 )t
4
t =6.65 × 10 s

7.7. Other Reaction Orders
7.7.1. Zero-Order Reactions

The rate decomposition of a zero-order reaction is independent of concen-
tration,

−d[A]dt = k A. (7.21a)
Integration gives
[A] 0 − [A]= k At (7.21b)

and the half-life

t 1/2 =1/2[A] 0/k A. (7.21c)
Aplot of[A]vs. t yields a straight line with a negative slope (Fig. 7.3).

7.7.2. Third-Order Reactions
The rate of decomposition of a third-order reaction is

−d[A]/dt = k A[A] 3 (7.22a)
yielding
3
−d[A]/[A] = k A t. (7.22b)