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Half Life Period of a reaction is the time taken for half of reactions to finish or the time at which the concentration of the reactant is reduced to half of its original value. It is the time required for half of the reactants to be converted into products. The half-time provides valuable insights into the rate at which a reaction proceeds and helps scientists and engineers optimize processes in fields ranging from medicine to environmental science.
One essential concept in the realm of chemical kinetics is the half-time of a reaction.The half-time of a chemical reaction is denoted by the symbol "t1/2" and plays a central role in determining the reaction's rate and efficiency. This concept is particularly useful in understanding the progress of reactions over time and predicting the behavior of chemical systems.
Half Life Formula
The half-life of a first-order reaction is a constant that is related to the rate constant for the reaction t1/2 = 0.693/k.
Half time of Reaction: Importance
Quantifying Reaction Rates: The half-time provides a quantitative measure of how quickly a reaction occurs. By analyzing the rate of reaction at different time points, scientists can determine the specific half-life and gain insights into the overall efficiency of the process.
Optimizing Industrial Processes: In industrial settings, where efficient production processes are paramount, understanding the half-time of reactions is crucial. Engineers use this information to optimize reaction conditions, such as temperature, pressure, and concentration, to achieve desired yields in a cost-effective manner.
Drug Development and Medicine: The pharmaceutical industry relies heavily on the principles of chemical kinetics. Determining the half-time of drug metabolism and elimination is essential for establishing dosage regimens and ensuring therapeutic efficacy while minimizing side effects.
Environmental Chemistry: In environmental science, understanding the half-time of pollutants helps assess the persistence of contaminants in air, water, and soil. This information is vital for developing strategies to remediate polluted environments and mitigate the impact of human activities on ecosystems.
Safety in Chemical Reactions: The half-time also plays a role in ensuring the safety of chemical processes. Knowledge of reaction kinetics helps identify potential hazards, allowing for the implementation of safety measures and protocols to prevent accidents.
Examples of Half-Time in Real-world Reactions
Radioactive Decay: Radioactive elements decay over time, and their half-life is a crucial parameter in nuclear physics. For example, the half-life of carbon-14 is used in carbon dating to estimate the age of archaeological artifacts.
Enzyme Catalysis: Enzyme-catalyzed reactions in biological systems often exhibit specific half-times. Understanding these kinetics is essential for studying metabolic pathways and designing drugs that target specific enzymes.
FAQs on Half time of reaction
Q: Calculate the half-life of a first order reaction from their rate constants given below: (i) 200 s^–1 (ii) 2 min^–1 (iii) 4 years^–1
A: Half life of first order reaction is, t1/2 = ln2/K where t1/2 is half life of first order reaction, K is rate constant of First order reaction.
(i) t1/2 = ln2/200 s-1
⇒t1/2 = 0.693/200 s-1 (∵ln2 = 0.693)
⇒t1/2 = 0.003465 sec.
- (ii) t1/2 = ln2/2 min-1
⇒t1/2 = 0.693/2 min-1 (∵ln2 = 0.693)
⇒t1/2 = 0.3465 min
- (iii) t1/2 = ln2/4 year-1
⇒t1/2 = 0.693/4 years -1 (∵ln2 = 0.693)
⇒t1/2 = 0.17325 year.
Half life of 3 reactions are 0.003465 sec, 0.3465 min, 0.17325 year respectively.
Q: The half-life for radioactive decay of 14C is 5730 years. An archaeological artifact containing wood had only 80% of the 14C found in a living tree. Estimate the age of the sample.
A: Radio active decay occurs via first order rate law,
t1/2 = 5730 years. rate constant(k) of given decay is 0.693/t1/2
⇒0.693/5730 = 1.2 × 10-4 year-1
By first order integrated rate law age of sample will be,
T = ( 2.303 / 1.2 X 10-4 ) log (A0/At)
where T is the age of sample A0 is the initial activity of the sample. and At is the activity of sample at any time t
T = ( 2.303 / 1.2 X 10-4 ) log (A0/0.8 A0)
T = 0.18 × 104 years.
Age of given sample is 0.18 × 104 years.
Q: Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law, with t 1/2 = 3.00 hours. What fraction of sample of sucrose remains after 8 hours?
A: t1/2 = 3.00 hours
We know, t1/2 = 0.693/k
∴k = 0.693/3 k = 0.231 hrs-1
We know, time Where, k- rate constant
[R]° -Initial concentration [R]-Concentration at time ‘t’
Thus, substituting the values, log([R]0/[R]) = 0.8
log([R]/[R]0) = -0.8
[R]/[R]0 = 0.158
Hence, 0.158 fraction of sucrose remains.
Q: The rate constant for the first order decomposition of H2O2 is given by the following equation: log k = 14.34 – 1.25 × 104K/T. Calculate Ea for this reaction and at what temperature will its half-period be 256 minutes?
A: We know, The Arrhenius equation is given by k = Ae-Ea/RT Taking natural log on both sides,
Ln k = ln A-(Ea/RT)
Thus, log k = log A -(Ea/2.303RT)......................................... eqn 1
The given equation is log k = 14.34 – 1.25 × 104K/T.............. eqn 2
Comparing 2 equations, Ea/2.303R = 1.25 × 104K
Ea = 1.25 × 104K × 2.303 × 8.314
Ea = 239339.3 J mol-1 (approximately) Ea = 239.34 kJ mol-1
Also, when t1/2 = 256 minutes,
k = 0.693 / t1/2
= 0.693 / 256
= 2.707 × 10-3 min-1 k = 4.51 × 10-5s–1
Substitute k = 4.51 × 10-5s–1 in eqn 2,
log 4.51 × 10-5 s–1 = 14.34 – 1.25 × 104K/T
log(0.654-5) = 14.34– 1.25 × 104K/T T = 1.25 × 104/[ 14.34- log(0.654-5)] T = 668.9K or T = 669 K
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