
Education Journalist | Study Abroad Lead
The half-life of a reaction, t1/2, is defined as the time it takes for a reactant concentration to drop by half when compared to its original concentration. In chemistry, the application half-life is used to anticipate the concentration of a chemical over time. Half-life is used to determine how rapidly a drug decreases in the target once it has been absorbed over a period (sec, minute, day) or the elimination rate constant ke (minute-1, hour-1, day-1). It's important to remember that the half-life varies from reaction to reaction.
| Table of Content |
Key Takeaways: Radioactivity, Zero order reaction, First-order reactions, Second order Reaction, Half-life, Half-life reaction
Half-Life Formula
[Click Here for Sample Questions]
It's important to remember that a reaction's half-life formula changes depending on the order of the reactions. The formula for the half-life of different reactions is given below.
- The half-life of a zero-order reaction, the formula is given as t1/2 = R0/2k
- The half-life of a first-order reaction is given as t1/2 = 0.693/k.
- The half-life of a second-order reaction is given by the formula 1/kR0.
Where,
The half-life of a reaction is referred to as t1/2 (unit - seconds)
The initial reactant concentration is referred to as R0 (in mol.L-1 or M) is [R0], and
The reaction's rate constant is given as k (unit - M(1-n)s-1, where 'n' is the reaction's order)
Also Read:
| Related Articles | ||
|---|---|---|
| Chemical Kinetics | Collision Theory of Chemical Reactions | Ionization enthalpy |
| Emulsions | Integrated Rate Equations | Surface Chemistry |
Derivation of Half-Life Formula for Zero Order Reaction
[Click Here for Sample Questions]
The rate constant for a zero-order reaction is measured in mol.L-1.s-1.
The following is the formula for a zero-order rate constant:
k=[R]0−[R ]/ t
By substituting t = t1/2, [R] = [R]0/2 is obtained (at the half-life of a reaction, reactant concentration is half of the initial concentration):
k=[R]0−[R]0/2 / t1/2
The half-life of a zero-order reaction can be calculated by rearranging the equation:
t1/2 = [R]0 / 2k
Derivation of Half-Life Formula for First Order Reaction
[Click Here for Sample Questions]
The following is the formula for a first-order rate constant expressed mathematically:
k = 2.303/t log [R]0 / [R]
At t = t1/2, [R] = [R]0/2, according to the reaction half-life definition. The following equation is obtained by substituting these values in the expression for the first-order rate constant:
k=2.303/ t1/2 log[R]0 / [R]0/2
Rearranging the expression:
t1/2 = 2.303 log(2) / k = 0.693/k
Thus, the half-life of a first-order reaction is obtained as 0.693/k.
Derivation of Half-Life Formula for Second-Order Reactions
[Click Here for Sample Questions]
An equation relating the half-life of a second-order reaction to its rate constant and starting concentration may be obtained from its integrated rate law using the same method as for first-order reactions:
\(\frac{1}{[A]_t} = kt + \frac{1}{[A]_0}\) Or \(\frac{1}{[A]_t} - \frac{1}{[A]_0} = kt \)
The integrated rate law is simplified when \(t = {t_\frac{1}{2}}\) and \({[A]_t} = {12[A]_0}\) are substituted:
\({kt_{\frac{1}{2}}} = {\frac{1}{\frac{1}{2}[A]_0}} - {\frac{1}{[A]_0}} = {\frac{2}{[A]_0}} - {\frac{1}{[A]_0}}\)
\({kt_{\frac{1}{2}}} = {\frac{1}{[A]_0}}\)
\({t_{\frac{1}{2}}} = {\frac{1}{k[A]_0}}\)
The half-life of a second-order reaction is inversely proportional to the reactant concentration, and the half-life rises as the reaction progresses since the reactant concentration falls. A second-order reaction's rate constant cannot be computed simply from the half-life, unlike first-order reactions, unless the initial concentration is known.
Effect of Half Life
[Click Here for Sample Questions]
- The radioisotope component has a physical Half-life, which means that the amount of radioactivity drops by half with each Half-life. Tc99m, a common radioisotope with a half-life of 6 hours, is an example of a common radioisotope. If it holds 100 MBq of this isotope right now, it will contain 50 MBq in 6 hours and 25 MBq in the next 6 hours.
- The patient can also excrete a little amount of the radiopharmaceutical, usually through the urine and kidneys, thus there is a certain amount that the body gets rid of, and so the radiation can be considered outside the body. If this occurs, that portion of the patient's radiation dose is no longer affected. The biological half-life is the term for this.
Things to Remember
- The half-life chemistry or reaction half-life, t1/2, is defined as the time it takes for a reactant concentration to decrease by half when compared to its initial concentration.
- Half-life is used to determine how quickly a chemical decreases in the target once it has been absorbed over a period of time (sec, minute, day) or the elimination rate constant 'k' (minute-1, hour-1, day-1).
- A reaction's half-life formula changes depending on the order of the reactions.
- The half-life of a zero-order reaction, the formula is given as t1/2 = R0/2k, The half-life of a first-order reaction is given as t1/2 = 0.693/k, The half-life of a second-order reaction is given by the formula 1/kR0.
- The rate constant for a zero-order reaction is measured in mol.L-1.s-1.
Also Read:
Sample Questions
Ques. Half life period of a radioactive element is 10 years. Calculate its disintegration constant & average life? (2 marks)
Ans: Given, t1/2 =10 years
k = 0.693 / t1/2 = 0.693 /10
K= 0.0693
Average life = 1.44 × t1/2=1.44×10
Average life = 14.4 years
Ques. The Half life period of U234 is 2.5×105 years. In how many years it will remain 25% of its original amount? (3 marks)
Ans: Given t1/2 =2.5×105 years
No = 100 gm
N = 25 grams
N= N0(1/2)n
25 = 100 (1/2)n
25 / 100 =(1/2)n
(1/4 ) =(1/2)n
(1/2)2 =(1/2)n
n=2
T=n×t1/2 = 2×2.5×105
T=5×105 years
Ques. 10-gram thorium remains 5 grams in 24 days. How much of thorium remains in 48 days? (2 marks)
Ans: t1/2=24 days
T=n×t1/2
n =48/24 =2
N= N0(1/2)n
N= 10 (1/2)2
N = 2.5 grams
Ques. Half life of 83 I125 is 60 days.How much of its radioactivity remains after 180 days? (2 marks)
Ans: Given,
t1/2 = 60 days
T = 180 days
Let initial reactivity ‘No’ = 100 gm
T=n×t1/2
n = T / t1/2
n =180 / 60 = 3
N= N0(1/2)n
N = 100 (1/2)3 =100 (1/8) = 12.5
N = 12.5 %
Ques. The activity of a radioactive element remains 12.5% in 90 days. Calculate the disintegration constant of the element? (3 marks)
Ans: Given,
T = 90 days
No = 100 gm
N = 12.5 grams
N= N0(1/2)n
12.5 = 100 (1/2)n
12.5 / 100 =(1/2)n
(1/8 ) =(1/2)n
(1/2)3 =(1/2)n
n=3
T=n×t1/2
90 = 3 × t1/2
t1/2 =90/3 = 30 days
K =0.693/t1/2 = 0.693 / 30
K = 0.0231 days-1
Ques. Starting with 16 atoms of a radioactive element , how many atoms are left after 4 half lives? (2 marks)
Ans: No = 16 atoms
N = ?
n = 4
N= N0(1/2)n
N = 16 (1/2)4 = 16 (1/16) = 1
N = 1 atom
Ques. Half life of 1H3 is 13.3 years. How much of 1H3 should be taken initially so that it remains 4 Kg after 26.6 years ? (3 marks)
Ans: Given,
T = 26.6 years
t1/2 = 13.3 years
T=n×t1/2
n = T / t1/2 = 26.6 / 13.3 = 2 half lives
N = 4 Kg
N= N0(1/2)n
4 = N0 (1/2)2
4 = N0 /4
N0 = 16 kg
Ques. How much of a radioactive element is left after 5 half lives? (2 marks)
Ans: Initial amount = N0
n = 5
N= N0(1/2)n
N = N0 (1/2)5 = N0 (1/32) = N0 / 32
N = N0 / 32
The amount left after 5 half lives is 1/32 of the initial amount.
Ques. At 318 K, the starting concentration of N2O5 was 1.24 10-2 mol L-1 in the first order reaction N2O5 (g)= 2 NO2(g) + 1/2O2 (g). After 60 minutes, the concentration of N2O5 was 0.20 10-2 mol L-1. Calculate the reaction's rate constant at 318 K. (3 marks)
Ans: In first order reaction
Log [R]1/[R]2 = k(t2-t1)/2.303
k= 2.303/(t2-t1) log [R]1/[R]2
k = 2.303/(60 min-0min) log (1.24x10-2 mol L-1)/ 0.20×10-2 mol L-1)
k=2.303/60 log 6.2 min-1
k= 0.0304 min-1
Ques. The rate constant of a first order reaction is determined to be k = 5.5 10-14 s-1. Determine the reaction's half-life. (2 marks)
Ans: In first order reaction,
t1/2 = 0.693/k
t1/2 = 0.693/(5.5 10-14 s-1)
= 1.26 x 1013s
Ques. Demonstrate that the time necessary to complete a first-order reaction at 99.9% is 10 times the reaction's half-life (t1/2). (3 marks)
Ans: 99.9% of complete reaction, [R]n = [R]0 -0.999[R]0
k = 2.303/t log [R]0/[R]
=2.303/t log [R]0/ [R]0 -0.999[R]0
=2.303/t log103
t= 6.909/k
Half life of the reaction
t1/2 = 0.693/k
t/ t1/2 = 6.909/k x k/0.693 = 10
Ques. What exactly do you mean by "order of a reaction"? From each of the following units of reaction rate constant, determine the reaction order: I mol L-1 s-1 (ii) L mol-1 s-1 mol L-1 s-1 mol L-1 s-1 mol L-1 s-1 mol L-1 s-1. (Delhi, 2012, 2 marks)
Ans: The order of the reaction is defined as the sum of the powers of the concentration of the reactants in the rate law statement.
r = K[A]x[B]y, Order = x + y
mol L-1 s-1 is zero-order reaction
L mol-1 s0-1 is second-order reaction
Ques. In reactant A, a reaction is first order, but in reactant B, it is second order. When I the concentration of B alone is increased to three times (ii) the concentrations of A and B are both doubled, how does the pace of this reaction change? (Delhi 2012, 2 marks)
Ans: r = K[A]1 [B]2
When the concentration of B increases 3 times
r= KA(3B)2
r = 9KAB2 = 9 times
When the concentration of bot A and B is doubled
r = K(2A) (2B)2
r = 8KAB2 = 8 times
Ques. Differentiate between a reaction's 'rate expression' and 'rate constant.' (Delhi 2011, 2 marks)
Ans: Rate expression: In a balanced chemical equation, the rate of reaction is stated in terms of molar concentrations of the reactants, with each term raised to its power, which may or may not be the same as the reactant's stoichiometric coefficient.
Rate Constant: The rate of reaction is defined as the rate of change in the concentration of either the reactants or the products per unit time.
Ques. What is the order of the reaction whose rate constant and rate of reaction have the same units? Give the order of the reaction A + H2O → B; Rate ∝ [A] (Comptt. All India 2017, 2 marks)
Ans: The reaction will have zero order if the rate constant has the same units as the rate of reaction.
Because the rate of reaction is only dependent on the concentration of A, the order of the given reaction will be a pseudo first order reaction.
For Latest Updates on Upcoming Board Exams, Click Here: https://t.me/class_10_12_board_updates
Check-Out:






Comments