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# Half-life of a second-order reaction

Deriving half-life equation of a second-order reaction starting from the integrated rate law.

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• How do you know which reaction order to use (first, second) for half-life, is there an equation to figure out which one to use?
• You have to know the order of the reaction first. Then you use the appropriate half-life equation.
Zero order: t½ = [A]₀/2k
1st order: t½ = ln2/k
2nd order: t½ = 1/(k[A]₀)
• What is a pseudo second order reaction?
• A pseudo second order reaction is a reaction for which the kinetics appear to be second order even though they are something else.
Consider a third order reaction with the rate law rate = k[A]²[B].
If B is in very large excess, the concentration of B will change very little as A gets used up.
[B] is almost a constant.
If we let k' = k[B], the rate law becomes rate = k'[A]².
The reaction is kinetically a pseudo second order reaction.
• sir, as per the definition the initial concentration gets half, but in th equation u have assigned 1/[A]t as [A]o/2 and the initial concentration 1/[A]o remains as it is.
• This is because [A]t is concentration at a specific time t . Here t = t1/2 ( half life) and as per the definition Half life is time at which the concentration of "A" is half of the initial concentration , so, [A]t = [A]o/2 . Thus, here, he assigned 1/[A]t as 1/[A]o/2, which can also be written as 2/[A]o. Hope this helps you.
• what is a half life?
• how did we got the graph of [A] VS t?
• We did an experiment and measured [A] at various times. Then we plotted a graph.
• why if less concentration makes time go slower in the first order reaction time remains constant?
(1 vote)
• Time doesn't go slower. It has the same half-life regardless of the amount.
For an analogy, imagine having 100 coins and you flip them all once. Approximately 50 will come up heads. If you then take the remaining tails and flip them once more, what do you expect the result will be? Do the coins have any way to know how many other coins there are being flipped?
• Well it makes sense when we're talking about the 2nd order reaction, but what about the zeroth order one, where the half life would be t1/2=[A]°/2K, meaning that as the concentration increases, the half life will increase too, and also in the 1st order reaction, how can the half life be independent of the intial concentration t1/2=0.693/K ? I find this really counterintuitive.
• In what way is the mathematics counter-intuitive? Its pretty simple stuff. And remember that experiment verifies the conclusions.
• I have a question I have to answer for my study but I don't have the initial concentration A. But I do have the concentration A. How do I know what the concentration A0 is if i want to calculate the half-life?