Question

In first order reaction 60%of the reactant decomposes in 45min.Calculate the half life for the reaction.

Answer 1

Answer:

the half life for the reaction is 34 min.

Answer 2

The reaction has a half-life of 34 minutes.

As per the question,

Half-life is the time required for a quantity to reduce to half of its initial value. The term is commonly used in nuclear physics to describe how quickly unstable atoms undergo radioactive decay or how long stable atoms survive. The term is also used more generally to characterize any type of exponential decay.

$Given: [A] 0 =100% [A]t = 100 - 60 =40\% t= 45 minTo find: Half-life of reaction (t_{1/2})Formula: k= 2.303/t log 10 [A] 0 [A] t .Calculation: Substitution of these in abovek = 2.303/t * log_10(100/40)= 2.303/(45min) * log_10(2.5)= 2.303/(45min) * 0.3979 ...(Using log table)$

$= Antilog_{10} (log 10 2.303+log 10 0.3979-log_10(45) )= Antilog_{10} (0.3623+ overline 1 .5998-1.6532)= Antilog_{10} ( overline 2 .3089)= 0.0203 min t_{\frac{1}{2}} = 0.693/k = 0.693/(0.0203mi * n ^ - 1) = 34minThe half-life of the reaction is 34 min.$

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