Question

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

Answer 1

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)))`