Chemistry SECOND ORDER AND `n^(th)` ORDER REACTION AND THEIR HALF-LIVES

Second Order Reaction :

`text(Type I)` :

`A->` Product

Initial concentration `a` `0` Concentration after time `t` `(a-x)` `x`

Differential rate law:

Integrated rate law (on integration of above equation):

`(d(a-x))/(dt)=+(dx)/(dt)=k_2(a-x)^2`

Integrated rate law (on integration of above equation):

`k_2=1/t x/(a(a-x))`

`text(Type II)` :

(i) Reactants `A` and `B` have the same initial concentration

`A + B ->` Products

Initial concentration `a` `0`

Concentration after time, `t` `(a-x)` `(a-x)` `x`

Differential Rate Law :

`-(d(a-x))/(dt)=+(dx)/(dt)-k_2(a-x)(a-x)-k_2(a-x)^2`

`k_2=1/t * x/(a(a-x))` (Same as for `A->` products)

(ii) Reactants `A` and `B` have different initial concentrations.

`A + B ->` products

Initial concentration `a` `b` `0`

Concentration after time, `t` `(a -x)` `(b-x)` `x`

Differential Rate Law:

`-(d(a-x))/(dt)=-(d(b-x))/(dt)=+(dx)/(dt)=k_2(a-x)(b-x)`

Integrated Rate Law :

When `a>b` : `k_2 = 2.303/(t(a-b)) log (b(a-x))/(a(b-x))`

when `b> a` : `k_2=2.303/(t(a-b)) log (a(b-x))/(b(a-x))`

The unit of the rate constant, `k_2` is `text(concentration) ^(-1) text(time)^(-1)`.

Pseudo-first-order reaction :

Measuring a second-order reaction rate with reactants `A` and `B` can be problematic: The concentrations of the two reactants must be
followed simultaneously, which is more difficult then to measure one of them and calculate the other as a difference, which is less precise. A common solution for that problem is the pseudo-first-order approximation

If either `[A]` or `[B]` remains constant as the reaction proceeds, then the reaction can be considered pseudo-first-order because, in fact,
it depends on the concentration of only one reactant. If, for example, `[B]` remains constant, then:

`r=k[A][B]=k'[A]`

Where `k' = k[B]_0`, (`k'` or `k_(obs)` with units `s^(-1)`) and an expression is obtained identical to the first order expression above.
One way to obtain a pseudo-first-order reaction is to use a large excess of one of the reactants `([B]`>> `[A])` so that, as the reaction progresses, only a small amount of the reactant is consumed, and its concentration can be considered to stay constant.

Half life for Second-Order Reaction :

For the second order reaction of the type

`A ->` Product

or

`A + B ->` Product, in which `A` and `B` have the same initial concentrations, we have

`k_2=1/t xx x/(a(a-x))`

for `t` to be `r_(1//2),x=a/2`

`k_(1//2)=1/k_2 xx 1/a`

or `t_(1//2) prop 1/a`

Thus `t_(1//2)` for a second-order reaction is inversely proportional to `'a'`.
Note that in reactions with more than one reactant, `t_(1//2)` of the reaction is calculated from the limiting reactant.

`n^(th)` order reaction :

Assume `A->` Product

is a `n^(th)` order reaction

Rate law is `(dx)/(dt)=k[A]^n=k(a-x)^n`

on integration , `int (dx)/(a-x)^n=k_(A) int dx`

`k_(A)=1/((n-1))[1/((a-x)^(n-1))-1/a^(n-1)]`

But `n > 1`

Half-Life of a `n^(th)` Order Reaction :

Let us venture on to find out the `t_(1//2)`, for a `n^(th)` order reaction where `n ne 1`.

`(-d[A])/(dt)=k[A]^n => (-d[A])/([A]^n)=kdt`

`-int_([A]_0)^([A]_0//2) (d[A])/([A]^n) =k int_0^ (t_(1//2)) dt`

`int_([A]_0//2)^([A]_0) [A]^(-n)d[A] = k t_(1//2) => [([A]^(1-n))/(1-n)]_(A_0//2)^(A_0)= kt_(1//2)`

`=>1/(1-n) ([A]_0^(1-n)-[([A]_0)/2]^(1-n)) = kt_(1//2)`

`=>([A]_0^(1-n))/(1-n)[1-(1/2)^(1-n)] = kt_(1//2)`

`1/((1-n)[A_0]^(n-1))[1-2^(n-1)]=kt_(1//2)`

`=> (2^(n-1)-1)/((n-1)[A_0]^(n-1))=kt_(1//2)` (order `n ne 1`)

Therefore for a `n^(th)` order reaction, the half life is inversely related to the initial concentration raised to the power of `(n - 1 )`. i.e.

`t_(1//2) prop 1/([A_o]^(n-1))`

 
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