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Chemistry · General chemistry II · Worked example

Use the second-order integrated rate law

A reactant A is consumed by second-order kinetics with k=0.540 M−1 s−1. A run starts at [A]0=0.0100 M. Find [A] after 100 s and the first two half-lives.

Read the order from k

The first example was first order, with k in min−1. Here k is in M−1 s−1, and that M−1 marks a second-order rate constant. The given is the starting concentration.

[A]0=0.0100 M

Name the rate constant

k=0.540 M−1 s−1, in seconds, the same unit as the time.

k=0.540 M−1s−1

Name the time

The question asks about t=100 s.

t=100 s

Write the second-order law

For second order it is 1⁢/⁢[A] that rises in a straight line with time, with slope k. The law gives the reciprocal of the concentration, not the concentration.

1[A]=k⁢t+1[A]0

Substitute

k⁢t=(0.540 M−1 s−1)⁢(100 s) =54.0 M−1, and 1⁢/⁢[A]0 =1÷0.0100 M =100 M−1. Both terms are in M−1, so they can be added.

1[A]=(0.540)⁢(100) M−1+10.0100 M
1[A]=(0.540)⁢(100) M−1+10.0100 M

Add

54.0 M−1+100 M−1 =154 M−1.

1[A]=154 M−1

Invert

Take 1 over the reciprocal to get the concentration. 1÷154 M−1=0.00649 M, about two thirds of the start.

[A]=1154 M−1=0.00649 M

Find the first half-life

A second-order half-life depends on the starting concentration, so the 0.0100 M goes in.

t1⁢/2=1(0.540 M−1s−1)⁢(0.0100 M)=185 s
t1⁢/2=1k⁢[A]0=1(0.540)⁢(0.0100) s=185 s

Find the second half-life

The second half-life starts from 0.00500 M, half the original concentration, so it is twice as long, 1÷(0.540×0.00500) =370 s. Each second-order half-life is double the one before, where a first-order half-life would stay the same. That also explains [A] at 100 s, well short of one half-life and still at two thirds of the start.

Result

[A]=0.00649 M after 100 s. The first half-life is 185 s and the second is 370 s.

Your turn

1. One like the example

A second-order reaction has k=0.250 M−1 s−1 and starts at [A]0=0.200 M. What is [A] after 20.0 s?

Show the answer and explanation

0.100 M.

k⁢t=(0.250 M−1 s−1)⁢(20.0 s) =5.00 M−1 and 1⁢/⁢[A]0=5.00 M−1, so 1⁢/⁢[A]=10.0 M−1 and [A]=0.100 M. The answer is exactly half the start because 20.0 s is the first half-life, 1÷(0.250×0.200) =20.0 s.

1[A]=k⁢t+1[A]0[A]0=0.200Mk=0.250M−1s−1t=20.0s1[A]=(0.250M−1s−1)⁢(20.0s)+10.200M1[A]=10.0M−1110.0M−1=0.100M
1[A]=k⁢t+1[A]0[A]0=0.200Mk=0.250M−1s−1t=20.0s1[A]=(0.250M−1s−1)⁢(20.0s)+10.200M1[A]=10.0M−1110.0M−1=0.100M

2. A variation

A second-order reaction has k=0.500 M−1 s−1. How long does [A] take to fall from 0.0100 M to 0.00250 M?

Show the answer and explanation

600 s.

Solve 1⁢/⁢[A]=k⁢t+1⁢/⁢[A]0 for t=(1⁢/⁢[A]−1⁢/⁢[A]0)⁢/⁢k. 1⁢/0.00250 M=400 M−1 and 1⁢/0.0100 M=100 M−1, so t=300 M−1÷0.500 M−1 s−1 =600 s. Two half-lives took 200 s and then 400 s, which add to the same 600 s.

1[A]=k⁢t+1[A]0[A]0=0.0100M[A]=0.00250Mk=0.500M−1s−1t=10.00250M−10.0100M0.500M−1s−1=600.s
1[A]=k⁢t+1[A]0[A]0=0.0100M[A]=0.00250Mk=0.500M−1s−1t=10.00250M−10.0100M0.500M−1s−1=600.s

3. Somewhere new

A zero-order reaction has k=0.0025 M s−1 and starts at [A]0=0.100 M. Find [A] after 10 s, the first half-life, and the time at which A is used up.

Show the answer and explanation

[A]=0.075 M after 10 s. The first half-life is 20 s, and A is used up at 40 s.

For zero order the units of k are M s−1, and [A]=[A]0−k⁢t =0.100 M −(0.0025 M s−1)⁢(10 s) =0.075 M. The half-life is [A]0÷2⁢k =0.100÷0.0050 =20 s. [A] reaches zero when k⁢t=[A]0, at 0.100÷0.0025=40 s. The second half-life, starting from 0.050 M, is only 10 s, half the first.

[A]=[A]0−k⁢t[A]0=0.100Mk=0.0025Ms−1t=10s[A]=0.100M−(0.0025Ms−1)⁢(10s)=0.075Mt1⁢/2=[A]02⁢kt1⁢/2=0.100M2⁢(0.0025Ms−1)=20s
[A]=[A]0−k⁢t[A]0=0.100Mk=0.0025Ms−1t=10s[A]=0.100M−(0.0025Ms−1)⁢(10s)=0.075Mt1⁢/2=[A]02⁢kt1⁢/2=0.100M2⁢(0.0025Ms−1)=20s

Keep exploring

In Kinetics & Arrhenius, change the initial concentration to 0.005 mol⁢/⁢L. The half-life it reports doubles to 370 s, because a second-order half-life depends on where you start.

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