What will be the value of #t_(0.875)/(t_(0.500))# for #n#th order reaction?

The time it takes for #gamma# fraction of #A# to remain from the decomposition given by

#aA -> bB#

is:

#t_(gamma) = {(ln(1//gamma)/(ak), n=1),((1 - gamma^(n-1))/((n-1)gamma^(n-1) ak [A]_0^(n-1)), n = 0), (, n >= 2) :}#

If you wish take #a = 1# and it will reduce to the usual form you see in textbooks.

This decomposition time ratio is then given as:

#t_(0.875)/t_(0.500) = overbrace(0.250)^(n=0), overbrace(0.193)^(n=1), overbrace(0.143)^(n=2), overbrace(0.102)^(n=3), . . . #

(However, it is not worth pursuing #n >= 4#. Why?)

And this is independent of rate constant and initial concentration.

Well, if we took an arbitrary order #n# for some reaction

#aA -> bB#,

then

#-1/a(d[A])/(dt) = {(k, n = 0), (k[A]^n, n >= 1):}#

Separation of variables gives:

#-akdt = {(d[A], n = 0), ([A]^(-n) d[A], n >= 1):}#

Integrating both sides, we obtain:

#-akint_(0)^(t) dt = {(int_([A]_0)^([A]) d[A], n = 0), (int_([A]_0)^([A]) [A]^(-n) d[A], n >= 1):}#

#-akt = {([A] - [A]_0, n = 0), (ln\frac([A])([A]_0), n = 1),(-([A]^(1-n) - [A]_0^(1-n))/(n-1), n >= 2):}#

As a result, we get any 1-reactant integrated rate law...

#[A] = -akt + [A]_0#, #" "n = 0#

#ln[A] = -akt + ln[A]_0#, #" "n = 1#

#1/([A]) = akt + 1/([A]_0)#, #" "n = 2#

#1/(2[A]^2) = akt + 1/(2[A]_0^2)#, #" "n = 3#

#" "vdots" "" "" "" "vdots#

#1/((n-1)[A]^(n-1)) = akt + 1/((n-1)[A]_0^(n-1))#, #" "n = 0, 2, 3, 4, . . . #

• For a zero-order decomposition to leave #gamma# fraction of #A# leftover,

#gamma[A]_0 = -akt_(gamma) + [A]_0#

#akt_(gamma) = (1 - gamma)[A]_0#

#color(green)(t_(gamma) = ((1 - gamma)[A]_0)/(ak))#

So to get #t_(0.875)/(t_(0.500))#,

#color(blue)(t_(0.875)^((0))/(t_(0.500)^((0)))) = ((1 - 0.875)cancel([A]_0))/cancel(ak) cancel(ak)/((1 - 0.500)cancel([A]_0)) = color(blue)(0.250)#

• For a first-order decomposition to leave #gamma# fraction of #A# leftover,

#ln(gamma[A]_0) = -akt_(gamma) + ln([A]_0)#

#akt_(gamma) = ln(cancel([A]_0)/(gammacancel([A]_0))) = ln (1//gamma)#

#color(green)(t_(gamma) = ln(1//gamma)/(ak))#

So to get #t_(0.875)/(t_(0.500))#,

#color(blue)(t_(0.875)^((1))/(t_(0.500)^((1)))) = ln(1//0.875)/cancel(ak) cancel(ak)/ln(1//0.500) = color(blue)(0.193)#

• For a second-order decomposition to leave #gamma# fraction of #A# leftover,

#1/(gamma[A]_0) = akt_(gamma) + 1/([A]_0)#

#akt_(gamma) = (1 - gamma)/(gamma[A]_0)#

#color(green)(t_(gamma) = (1 - gamma)/(gamma ak[A]_0))#

So to get #t_(0.875)/(t_(0.500))#,

#color(blue)(t_(0.875)^((2))/(t_(0.500)^((2)))) = (1 - 0.875)/(0.875cancel(ak[A]_0)) (0.500cancel(ak[A]_0))/(1 - 0.500) = color(blue)(0.143)#

• For a third-order decomposition to leave #gamma# fraction of #A# leftover,

#1/(2(gamma[A]_0)^2) = akt_(gamma) + 1/(2[A]_0^2)#

#akt_(gamma) = (1 - gamma^2)/(2gamma^2[A]_0^2)#

#color(green)(t_(gamma) = (1 - gamma^2)/(2gamma^2 ak[A]_0^2))#

So to get #t_(0.875)/(t_(0.500))#,

#color(blue)(t_(0.875)^((3))/(t_(0.500)^((3)))) = (1 - 0.875^2)/(2 cdot 0.875^2cancel(ak[A]_0^2)) (2 cdot 0.500^2cancel(ak[A]_0^2))/(1 - 0.500^2) = color(blue)(0.102)#

• For an nth-order decomposition (#n = 0, 2, 3, 4, . . . #) to leave #gamma# fraction of #A# leftover,

#(gamma[A]_0)^(1-n) = a(n-1)kt_(gamma) + [A]_0^(1-n)#

#1/(gamma^(n-1)[A]_0^(n-1)) = a(n-1)kt_(gamma) + 1/([A]_0^(n-1))#

#a(n-1)kt_(gamma) = (1 - gamma^(n-1))/(gamma^(n-1)[A]_0^(n-1))#

#color(green)(t_(gamma) = (1 - gamma^(n-1))/((n-1)gamma^(n-1) ak [A]_0^(n-1)))#

So to get #t_(0.875)/(t_(0.500))#, #" "n = 0, 2, 3, 4, . . . #,

#color(blue)(t_(0.875)^((n))/(t_(0.500)^((n)))) = (1 - 0.875^(n-1))/(cancel((n-1))0.875^(n-1) cancel(ak [A]_0^(n-1))) (cancel((n-1))0.500^(n-1) cancel(ak [A]_0^(n-1)))/(1 - 0.500^(n-1))#

#= (1 - 0.875^(n-1))/(0.875^(n-1)) cdot (0.500^(n-1))/(1 - 0.500^(n-1))#

#= color(blue)(1/(1.75^(n-1))(1 - 0.875^(n-1))/(1 - 0.500^(n-1)))#