<< Chapter < Page Chapter >> Page >

First-order reactions

We can derive an equation for determining the half-life of a first-order reaction from the alternate form of the integrated rate law as follows:

ln [ A ] 0 [ A ] = k t t = ln [ A ] 0 [ A ] × 1 k

If we set the time t equal to the half-life, t 1 / 2 , the corresponding concentration of A at this time is equal to one-half of its initial concentration. Hence, when t = t 1 / 2 , [ A ] = 1 2 [ A ] 0 .

Therefore:

t 1 / 2 = ln [ A ] 0 1 2 [ A ] 0 × 1 k = ln 2 × 1 k = 0.693 × 1 k

Thus:

t 1 / 2 = 0.693 k

We can see that the half-life of a first-order reaction is inversely proportional to the rate constant k . A fast reaction (shorter half-life) will have a larger k ; a slow reaction (longer half-life) will have a smaller k .

Calculation of a first-order rate constant using half-life

Calculate the rate constant for the first-order decomposition of hydrogen peroxide in water at 40 °C, using the data given in [link] .

A diagram of 5 beakers is shown, each approximately half-filled with colored substances. Beneath each beaker are three rows of text. The first beaker contains a bright green substance and is labeled below as, “1.000 M, 0 s, and ( 0 h ).” The second beaker contains a slightly lighter green substance and is labeled below as, “0.500 M, 2.16 times 10 superscript 4 s, and ( 6 h ).” The third beaker contains an even lighter green substance and is labeled below as, “0.250 M, 4.32 times 10 superscript 4 s, and ( 12 h ).” The fourth beaker contains a green tinted substance and is labeled below as, “0.125 M, 6.48 times 10 superscript 4 s, and ( 18 h ).” The fifth beaker contains a colorless substance and is labeled below as, “0.0625 M, 8.64 times 10 superscript 4 s, and ( 24 h ).”
The decomposition of H 2 O 2 ( 2H 2 O 2 2H 2 O + O 2 ) at 40 °C is illustrated. The intensity of the color symbolizes the concentration of H 2 O 2 at the indicated times; H 2 O 2 is actually colorless.

Solution

The half-life for the decomposition of H 2 O 2 is 2.16 × 10 4 s:

t 1 / 2 = 0.693 k k = 0.693 t 1 / 2 = 0.693 2.16 × 10 4 s = 3.21 × 10 −5 s −1

Check your learning

The first-order radioactive decay of iodine-131 exhibits a rate constant of 0.138 d −1 . What is the half-life for this decay?

Answer:

5.02 d.

Got questions? Get instant answers now!

Second-order reactions

We can derive the equation for calculating the half-life of a second order as follows:

1 [ A ] = k t + 1 [ A ] 0

or

1 [ A ] 1 [ A ] 0 = k t

If

t = t 1 / 2

then

[ A ] = 1 2 [ A ] 0

and we can write:

1 1 2 [ A ] 0 1 [ A ] 0 = k t 1 / 2 2 [ A ] 0 1 [ A ] 0 = k t 1 / 2 1 [ A ] 0 = k t 1 / 2

Thus:

t 1 / 2 = 1 k [ A ] 0

For a second-order reaction, t 1 / 2 is inversely proportional to the concentration of the reactant, and the half-life increases as the reaction proceeds because the concentration of reactant decreases. Consequently, we find the use of the half-life concept to be more complex for second-order reactions than for first-order reactions. Unlike with first-order reactions, the rate constant of a second-order reaction cannot be calculated directly from the half-life unless the initial concentration is known.

Zero-order reactions

We can derive an equation for calculating the half-life of a zero order reaction as follows:

[ A ] = k t + [ A ] 0

When half of the initial amount of reactant has been consumed t = t 1 / 2 and [ A ] = [ A ] 0 2 . Thus:

[ A ] 0 2 = k t 1 / 2 + [ A ] 0 k t 1 / 2 = [ A ] 0 2

and

t 1 / 2 = [ A ] 0 2 k

The half-life of a zero-order reaction increases as the initial concentration increases.

Equations for both differential and integrated rate laws and the corresponding half-lives for zero-, first-, and second-order reactions are summarized in [link] .

Summary of Rate Laws for Zero-, First-, and Second-Order Reactions
Zero-Order First-Order Second-Order
rate law rate = k rate = k [ A ] rate = k [ A ] 2
units of rate constant M s −1 s −1 M −1 s −1
integrated rate law [ A ] = − kt + [ A ] 0 ln[ A ] = − kt + ln[ A ] 0 1 [ A ] = k t + ( 1 [ A ] 0 )
plot needed for linear fit of rate data [ A ] vs. t ln[ A ] vs. t 1 [ A ] vs. t
relationship between slope of linear plot and rate constant k = −slope k = −slope k = +slope
half-life t 1 / 2 = [ A ] 0 2 k t 1 / 2 = 0.693 k t 1 / 2 = 1 [ A ] 0 k

Key concepts and summary

Differential rate laws can be determined by the method of initial rates or other methods. We measure values for the initial rates of a reaction at different concentrations of the reactants. From these measurements, we determine the order of the reaction in each reactant. Integrated rate laws are determined by integration of the corresponding differential rate laws. Rate constants for those rate laws are determined from measurements of concentration at various times during a reaction.

Questions & Answers

A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?
Aislinn Reply
cm
tijani
what is titration
John Reply
what is physics
Siyaka Reply
A mouse of mass 200 g falls 100 m down a vertical mine shaft and lands at the bottom with a speed of 8.0 m/s. During its fall, how much work is done on the mouse by air resistance
Jude Reply
Can you compute that for me. Ty
Jude
what is the dimension formula of energy?
David Reply
what is viscosity?
David
what is inorganic
emma Reply
what is chemistry
Youesf Reply
what is inorganic
emma
Chemistry is a branch of science that deals with the study of matter,it composition,it structure and the changes it undergoes
Adjei
please, I'm a physics student and I need help in physics
Adjanou
chemistry could also be understood like the sexual attraction/repulsion of the male and female elements. the reaction varies depending on the energy differences of each given gender. + masculine -female.
Pedro
A ball is thrown straight up.it passes a 2.0m high window 7.50 m off the ground on it path up and takes 1.30 s to go past the window.what was the ball initial velocity
Krampah Reply
2. A sled plus passenger with total mass 50 kg is pulled 20 m across the snow (0.20) at constant velocity by a force directed 25° above the horizontal. Calculate (a) the work of the applied force, (b) the work of friction, and (c) the total work.
Sahid Reply
you have been hired as an espert witness in a court case involving an automobile accident. the accident involved car A of mass 1500kg which crashed into stationary car B of mass 1100kg. the driver of car A applied his brakes 15 m before he skidded and crashed into car B. after the collision, car A s
Samuel Reply
can someone explain to me, an ignorant high school student, why the trend of the graph doesn't follow the fact that the higher frequency a sound wave is, the more power it is, hence, making me think the phons output would follow this general trend?
Joseph Reply
Nevermind i just realied that the graph is the phons output for a person with normal hearing and not just the phons output of the sound waves power, I should read the entire thing next time
Joseph
Follow up question, does anyone know where I can find a graph that accuretly depicts the actual relative "power" output of sound over its frequency instead of just humans hearing
Joseph
"Generation of electrical energy from sound energy | IEEE Conference Publication | IEEE Xplore" ***ieeexplore.ieee.org/document/7150687?reload=true
Ryan
what's motion
Maurice Reply
what are the types of wave
Maurice
answer
Magreth
progressive wave
Magreth
hello friend how are you
Muhammad Reply
fine, how about you?
Mohammed
hi
Mujahid
A string is 3.00 m long with a mass of 5.00 g. The string is held taut with a tension of 500.00 N applied to the string. A pulse is sent down the string. How long does it take the pulse to travel the 3.00 m of the string?
yasuo Reply
Who can show me the full solution in this problem?
Reofrir Reply
Got questions? Join the online conversation and get instant answers!
Jobilize.com Reply
Practice Key Terms 2

Get Jobilize Job Search Mobile App in your pocket Now!

Get it on Google Play Download on the App Store Now




Source:  OpenStax, Chemistry. OpenStax CNX. May 20, 2015 Download for free at http://legacy.cnx.org/content/col11760/1.9
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Chemistry' conversation and receive update notifications?

Ask