The Foundations of Chemistry

A88nB
C

Plan

We use the equation given earlier for t1/2for a first-order reaction. The value of kis given in
the problem; the coefficient of reactant A is a1.

Solution

t1/215.4 s

After 15.4 seconds of reaction, half of the original reactant remains, so that [A]^12 [A] 0.

You should now work Exercise 39.

EXAMPLE 16-6 Concentration Versus Time: First-Order Reaction

The reaction 2N 2 O 5 (g) n2N 2 O 4 (g)
O 2 (g) obeys the rate law Ratek[N 2 O 5 ], in which the
specific rate constant is 0.00840 s^1 at a certain temperature. (a) If 2.50 moles of N 2 O 5 were
placed in a 5.00-liter container at that temperature, how many moles of N 2 O 5 would remain
after 1.00 minute? (b) How long would it take for 90% of the original N 2 O 5 to react?

Plan

We apply the integrated first-order rate equation.

ln


[

[

N

N

2

2

O

O

5

5

]

]

^0

akt

(a) First, we must determine [N 2 O 5 ] 0 , the original molar concentration of N 2 O 5. Then we
solve for [N 2 O 5 ], the molar concentration after 1.00 minute. We must remember to express k
and tusing the same time units. Finally, we convert molar concentration of N 2 O 5 to moles
remaining. (b) We solve the integrated first-order equation for the required time.

Solution

The original concentration of N 2 O 5 is

[N 2 O 5 ] 0 0.500 M

The other quantities are

a 2 k0.00840 s^1 t1.00 min60.0 s [N 2 O 5 ]__?

The only unknown in the integrated rate equation is [N 2 O 5 ] after 1.00 minute. Let us solve
for the unknown. Because ln x/yln xln y,

ln ln [N 2 O 5 ] 0 ln [N 2 O 5 ]akt

ln [N 2 O 5 ]ln [N 2 O 5 ] 0 akt

ln (0.500)(2)(0.00840 s^1 )(60.0 s)0.6931.008

ln [N 2 O 5 ]1.701

Taking the inverse natural logarithm of both sides gives

[N 2 O 5 ]1.82 10 ^1 M

[N 2 O 5 ] 0



[N 2 O 5 ]

2.50 mol



5.00 L

0.693



1(0.0450 s^1 )

ln 2



ak

inv ln xelnx

16-4 Concentration versus Time: The Integrated Rate Equation 665

See the Saunders Interactive

General Chemistry CD-ROM,

Screen 15.8, Half-Life: First-Order

Reactions.