The Foundations of Chemistry

Plan

As long as some B is present, only the concentration of A affects the rate. The reaction is

second order in [A] and second order overall, so we use the appropriate equation for the half-

life.

Solution

t1/239.2 min

EXAMPLE 16-8 Concentration Versus Time: Second-Order Reaction

The gas-phase decomposition of NOBr is second order in [NOBr], with k0.810 M^1 s^1

at 10°C. We start with 4.00 10 ^3 MNOBr in a flask at 10°C. How many seconds does it

take to use up 1.50 10 ^3 Mof this NOBr?

2NOBr(g)88n2NO(g)
Br 2 (g) ratek[NOBr]^2

Plan

We first determine the concentration of NOBr that remains after 1.50 10 ^3 Mis used up.

Then we use the second-order integrated rate equation to determine the time required to

reach that concentration.

Solution

__?MNOBr remaining(0.004000.00150) M0.00250 M[NOBr]

We solve the integrated rate equation 

[NO

1

Br]



[NO

1

Br] 0

aktfor t.

t
 
 

 (400 M^1  250 M^1 )

 92.6 s

You should now work Exercise 34.

EXAMPLE 16-9 Concentration Versus Time: Second-Order Reaction

Consider the reaction of Example 16-8 at 10°C. If we start with 2.40 10 ^3 MNOBr, what

concentration of NOBr will remain after 5.00 minutes of reaction?

Plan

We use the integrated second-order rate equation to solve for the concentration of NOBr

remaining at t5.00 minutes.

Solution

Again, we start with the expression akt.Then we put in the known

values and solve for [NOBr].

(2)(0.810 M^1 s^1 )(5.00 min)


60 s



1 min

1



2.40 10 ^3 M

1



[NOBr]

1



[NOBr] 0

1



[NOBr]

1



1.62 M^1 s^1

1



0.00400 M

1



0.00250 M

1



(2)(0.810 M^1 s^1

1



[NOBr] 0

1



[NOBr]

1



ak

1



(1)(0.622 M^1 min^1 )(4.10 10 ^2 M)

1



ak[A] 0

The coefficient of NOBr is a2.

668 CHAPTER 16: Chemical Kinetics