CHAP. 9: CHEMICAL KINETICS [CONTENTS] 288
Solution
We first try whether this is a first-order reaction. From equation (9.24) we express the rate
constant as
k=
1
τ
ln
cA0
cA
.
The initial concentration is given by the first item in the table:cA0= 8. 46 × 10 −^3 mol dm−^3. We
calculate the constant for the other 5 items; it systematically falls with time:
τ[s] 240 660 1200 1800 2400
k· 104 [s−^1 ] 6.72 6.39 6.16 5.67 5.23
Hence the reaction is not of first order. We then try the second order. In this case we have from
relation (9.32)
k=
1
τ
( 1
cA
−
1
cA0
)
.
The constant increases systematically with time:
τ[s] 240 660 1200 1800 2400
k· 105 [s−^1 dm^3 mol−^1 ] 8.62 9.39 10.78 11.65 12.36
We can thus expect the reaction order to be lower than 2 but higher than 1. We now try one
and half order. From relation (9.72) we have forn= 1.5
k=
2
τ
[(
1
cA
) 1 / 2
−
(
1
cA0
) 1 / 2 ]
.
The constant does not change systematically with time:
τ[s] 240 660 1200 1800 2400
k· 104 [s−^1 (dm^3 mol−^1 )(1/2)] 2.41 2.44 2.56 2.54 2.50
Hence the order of the reaction is one and half. The average value of the rate constant is
k= 2. 49 × 10 −^4 s−^1 (dm^3 mol−^1 )(1/2).
