Question

`A` and `B` react to form `C` and `D` in a reaction that was found to be second order in `A`. The rate constant at `30^@ "C"` is `"0.622 M"^(-1)cdot"min"^(-1)`. What is the half-life of A if `4.10xx10^(-2) "M"` is mixed with excess `B`?

Answer 1

From the expression (derived below) for the second-order half-life, assuming the stoichiometric coefficient of `A` is `a = 1`,

`t_"1/2" = 1/(k[A]_0)`

And so, if `4.10 xx 10^(-2) "M"` of `A` is added, that is `[A]_0` and:

`color(blue)(t_"1/2") = 1/(0.622 cancel("M"^(-1))cdot"min"^(-1) cdot 4.10 xx 10^(-2) cancel"M")`

`=` `color(blue)ul("39.2 min")`

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The idea here is that since `A` is mixed with excess `B`, the reaction is primarily controlled by the concentration of `A` (pseudo-second order).

So, the rate law is of the form:

`r(t) ~~ k[A]^2[B]^0 = k[A]^2`

where `r(t)` is the rate in `"M/min"`, `k` is the rate constant in `"M"^(-1)cdot"min"^(-1)`, and `[A]` is the molar concentration of `A` at `t = "0 min"`.

The rate is also related to the change in concentration over time:

`aA + bB -> cC + dD`

`r(t) = -1/a(Delta[A])/(Deltat) = -1/b(Delta[B])/(Deltat)`

`= +1/c(Delta[C])/(Deltat) = +1/d(Delta[D])/(Deltat)`

You aren't told what the coefficient `a` of `A` is, but since this is all general, we don't assume `a = 1`. So we keep that in.

Equating the two expressions for the rate:

`k[A]^2 = -1/a(Delta[A])/(Deltat)`

We mainly care about the rate of reaction near the initial time, so we can look at infinitesimally small changes in concentration such that

`(Delta[A])/(Deltat) ~~ (d[A])/(dt)`.

This then becomes an equation to rearrange and integrate over an interval.

`k[A]^2 = -1/a(d[A])/(dt)`

Rearrange to get:

`akdt = -1/([A]^2)d[A]`

Integrate the left side over time `0->t` and the right side over `[A]_0 -> [A]`, where `[A]_0` is the initial concentration and `[A]` is the current concentration.

`a` and `k` are constants which float outside the integral sign.

`=> akint_(0)^(t)dt = -int_([A]_0)^([A])1/([A]^2)d[A]`

The integral of `-1/x^2` is `1/x`, so:

`akt = 1/([A]) - 1/([A]_0)`

This would rearrange to give a general integrated rate law for any stoichiometric coefficient `a`:

`barul(|stackrel(" ")(" "1/([A]) = akt + 1/([A]_0)" ")|)`

For a half-life, the initial concentration `[A]_0` drops to half, so we call `t = t_"1/2"` at the point where `[A] = 1/2[A]_0`. Thus,

`2/([A]_0) = akt_"1/2" + 1/([A]_0)`

Solve for the half-life to get:

`color(blue)barul(|stackrel(" ")(" "t_"1/2" = 1/(ak[A]_0)" ")|)`

And if `a = 1` for the stoichiometric coefficient of `A`, then

`t_"1/2" = 1/(k[A]_0)`