Question

6. Calculate the rate constant for the decay of C-14 (in units of 1/year).

Also could if you could explain how kinetics and radioactivity relate, that would be great. thank you

Answer 1

Following is the - complete Answer -&- Explanation: for the given: Question: in......typed format...

$\Rightarrow$Answer:

Rate constant for decay of C-14, is: k = 1.2096 x 10 -4 year -1 ( i.e. per year.... )

$\Rightarrow$Explanation:

• Step - 1:

We know the following equation: for radio-active , beta-decay, of carbon-14 ( C-14 ), producing nitrogen-14 ( N-14 ), and beta ( $\beta$ ) particle in the process:

$\Rightarrow$i.e. C-14   $\rightarrow$ N-14 +   $\beta$- ----------------------------------Eq. ( 1 )

• Step - 2:

​​​​​​​We know for the above Eq. ( 1 ), and for the radio-active beta decay , of carbon-14 ( C-14 ), the following is the rate equation:

$\Rightarrow$   ln ( [A]o / [A] ) =  k x t   -----------------------------------------Eq. ( 2 )

$\Rightarrow$ Where:

1. [A]o =   Initial concentration of C-14
2. [A] =  Final concentration of C-14
3. k = Rate constant of the reaction Eq. ( 2 )
4. t =   Time required to chage the concentration from [A]o to  [A]

• Step - 3:

​​​​​​​We know, half life ( t1/2 ), is the time required to make the concentration, half of its initial concentration...

Then using Eq. ( 2 ), we can form the following equation:

$\Rightarrow$   ln ( [A]o / [A]o /2 ) =  k x t1/2   --------------------------------------Eq. ( 3 )

[ t1/2 = half life ; k = rate constant [=] yr -1 ]

$\Rightarrow$ ln ( 2 ) = k x t1/2

$\Rightarrow$0.693 = k x  t1/2

$\Rightarrow$   k =   ( 0.693 ) /  t1/2 ---------------------------------- Eq. ( 4 )

• Step - 4:

​​​​​​​We know for carbon-14, ( C-14 ), following is the value of Half-Life (   t1/2 ) , is the following:

$\Rightarrow$   t1/2   = 5730 yr. ( years )

Plugging in values in Eq. ( 4 ), we would get:

$\Rightarrow$ k =   ( 0.693 ) /  t1/2

$\Rightarrow$   k =   ( 0.693 ) / ( 5730.0 year )

$\Rightarrow$ k = 1.2096 x 10 -4 year -1 ( i.e. per year.... )

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