502 11 The Rates of Chemical Reactions
When the expressions forGandHare substituted into Eq. (11.3-10) and the resulting
expression is substituted into Eq. (11.3-9), a definite integration gives1
a[B] 0 −b[A] 0ln(
[B]t[A] 0
[A]t[B] 0)
kft (11.3-11)This integrated rate law can now be compared with experimental data.Exercise 11.14
Verify Eq. (11.3-11) by carrying out the integration.EXAMPLE11.9
Assume that for the reactionNO+O 3 −→NO 2 +O 2[NO] 0 1. 00 × 10 −^6 mol L−^1 and [O 3 ] 0 5. 00 × 10 −^7 mol L−^1. Find the concentration
of O 3 after 3.50 s.
Solution
Let O 3 be denoted by A and NO be denoted by B. Use of Eq. (11.3-11) withab1 gives,
with some manipulation:[B] 0 −x
[A] 0 −x
[B] 0
[A] 0
exp[kt([B] 0 −[A] 0 )]Using the values given, solution of the equation givesx 4. 73 × 10 −^7 and [O 3 ] 0. 27 ×
10 −^7 mol L−^1.Exercise 11.15
Carry out the mathematical steps to verify the solution of the preceding example.The Method of Initial Rates
Consider the hypothetical reactionaA+bB+fF−→products (11.3-12)We assume the reaction has definite orders so that the initial rate isrinitialkf[A]a 0 [B]β 0 [F]φ 0 ≈−1
a∆[A]
∆t(11.3-13)
Several experiments are carried out with the same values of [B] 0 and [F] 0 but
with different values of [A] 0. The initial rate is determined for each experiment. We
writeln(rinitial)ln(kf[B]β 0 [F]φ 0 )+αln([A] 0 ) (11.3-14)