60 of a first order reaction was completed in 12 minutes calculate the time taken For 84 completion

Nuclear half-life expresses the time required for half of a sample to undergo radioactive decay. Exponential decay can be expressed mathematically like this:

#A(t) = A_0 * (1/2)^(t/t_("1/2"))# (1), where

#A(t)# - the amount left after t years;
#A_0# - the initial quantity of the substance that will undergo decay;
#t_("1/2")# - the half-life of the decaying quantity.

So, if a problem asks you to calculate an element's half-life, it must provide information about the initial mass, the quantity left after radioactive decay, and the time it took that sample to reach its post-decay value.

Let's say you have a radioactive isotope that undergoes radioactive decay. It started from a mass of 67.0 g and it took 98 years for it to reach 0.01 g. Here's how you would determine its half-life:

Starting from (1), we know that

#0.01 = 67.0 * (1/2)^(98.0/t_("1/2")) -> 0.01/67.0 = 0.000149 = (1/2)^(98.0/(t_("1/2"))#

#98.0/t_("1/2") = log_(0.5)(0.000149) = 12.7#

Therefore, its half-life is #t_("1/2") = 98.0/(12.7) = 7.72# #"years"#.

So, the initial mass gets halved every 7.72 years.

Sometimes, if the numbers allow it, you can work backwards to determine an element's half-life. Let's say you started with 100 g and ended up with 25 g after 1,000 years.

In this case, since 25 represents 1/4th of 100, two hal-life cycles must have passed in 1,000 years, since

#100.0/2 = 50.0# #"g"# after the first #t_("1/2")#,

#50.0/2 = 25.0# #"g"# after another #t_("1/2")#.

So, # 2 * t_("1/2") = 1000 -> t_("1/2") = 1000/2 = 500# #"years"#.

Answer

Verified

Hint: In this question we will use the 1st order reaction equation to establish a relationship between the two different stages of reaction i.e. at 60% and 84%. We know that in 1st order reaction $k = \dfrac{{2.303}}{t}\log \dfrac{a}{{a - x}}$, wherea = initial stage of reactionx= final stage of reaction t = total time

Complete step by step answer:

Let a = 100 Therefore according to the given question the reaction is 60% complete in 20 minutes. \[ \Rightarrow k = \dfrac{{2.303}}{{20}}\log \dfrac{{100}}{{100 - 60}}\] ..... (i)Now as per the question we need to find out that how long it will take to complete 84% of reactionTherefore, \[\; \Rightarrow k = \dfrac{{2.303}}{t}\log \dfrac{{100}}{{100 - 84}}\]..... (ii)But as we all know $k$remains same throughout the reaction, therefore (i) = (ii),\[ \Rightarrow \dfrac{{2.303}}{{20}}\log \dfrac{{100}}{{100 - 60}} = \;\dfrac{{2.303}}{t}\log \dfrac{{100}}{{100 - 84}}\]\[ \Rightarrow t = 20 \times \dfrac{{0.795}}{{0.397}}\]$ \Rightarrow t = 40{\text{ mins}}$

Therefore, the correct option is C.

Additional information:

Reaction lifetime is defined as the time taken by the reaction to proceed to 98% of completion. The shorter the life time, the faster the reaction. Reaction life times are used to compare the various reactions. Reactions are also compared with half life periods. The half life period is defined as the time during which the concentration of a reactant is reduced to one-half of its initial value or the time in which half of the reaction is completed.

Note: There are several factors which affect the rate of reaction and time period, few of them are listed below:

1.Concentration of reactants2.Effect of temperature3.Presence of a positive catalyst4.Presence of a negative catalyst5.Physical state of reactantsThese need to be taken care while solving such problems.

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