The Foundations of Chemistry

Thus, after 1.00 minute of reaction, the concentration of N 2 O 5 is 0.182 M.The number of

moles of N 2 O 5 left in the 5.00-L container is

__? mol N 2 O 5 5.00 L0.910 mol N 2 O 5

(b) Because the integrated first-orderrate equation involves a ratio of concentrations, we do

not need to obtain the numerical value of the required concentration. When 90.0% of the

original N 2 O 5 has reacted, 10.0% remains, or

[N 2 O 5 ](0.100)[N 2 O 5 ] 0

We make this substitution into the integrated rate equation and solve for the elapsed time, t.

ln akt

ln (2)(0.00840 s^1 )t

ln (10.0)(0.0168 s^1 )t

2.302(0.0168 s^1 )t or t137 seconds

You should now work Exercises 36 and 40.

2.302



0.0168 s^1

[N 2 O 5 ] 0



(0.100)[N 2 O 5 ] 0

[N 2 O 5 ] 0



[N 2 O 5 ]

0.182 mol



L

666 CHAPTER 16: Chemical Kinetics

Problem-Solving Tip:Does Your Answer Make Sense?

We know that the amount of N 2 O 5 in Example 16-6 must be decreasing. The calculated

result, 0.910 mol N 2 O 5 after 1.00 minute, is less than the initial amount, 2.50 mol N 2 O 5 ,

which is a reasonable result. If our solution had given a result that was largerthan the

original, we should recognize that we must have made some error. For example, if we

had incorrectly written the equation as

ln

[

[

N

N

2

2

O

O

5

5

]

]

0

akt

we would have obtained [N 2 O 5 ]1.37 M,corresponding to 6.85 mol N 2 O 5. This would

be more N 2 O 5 than was originally present, which we should immediately recognize as

an impossible answer.

Second-Order Reactions

For reactions involving aA n products that are second order with respect toA and second

order overall,the integrated rate equation is

akt 


second order in A,

second order overall

1



[A] 0

1



[A]