Question

Consider the reaction 3A + B + C ---> D + E where the rate law is defined as -D[A]/Dt=k[A]^2[B][C]?

The initial concentration of `B` and `C` are both `"1.00 M"`. The initial concentration of `A` is `1.00 xxx 10^(-4) "M"`, and after 3.00 min, it became `3.26 xx 10^(-5) "M"`.

`1)` Find the rate constant for the pseudo-second-order reaction in `"M"^(-3)cdot"s"^(-1)`.
`2)` Find the pseudo-second-order half-life in seconds.
`3)` Under what assumptions are `(1)` and `(2)` true?
`4)` Determine the concentration of `A` after 10.00 min have passed.

Answer 1

DISCLAIMER: LONG ANSWER!

Alright, first, let's write down what we were given here, and also what we can write out from what we know so far.

`A + B + C -> D + E`

(I assume the 3A is a typo, because the rate given, `-(Delta[A])/(Deltat)`, has no coefficient when it should if the stoichiometric cofficient of `A` were not `1`.)

• Initial concentration of `B`, `["B"]_0 = ["C"]_0 = "1.00 M"`
• Initial concentration of `A`, `["A"]_0 = 1.00 xx 10^(-4)` `"M"`
• Current concentration of `A`, `["A"]_"3.00 min." = 3.26 xx 10^(-5) "M"`

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

`= -1/1 (Delta[A])/(Deltat) = -1/1 (Delta[B])/(Deltat) = -1/1 (Delta[C])/(Deltat)`

`= +1/1 (Delta[D])/(Deltat) = +1/1 (Delta[E])/(Deltat)`

`1)`

The rate constant can be found by using the integrated rate law to track the fraction of just `A` leftover after some time `t`.

The integrated rate law with respect to ONLY `A` should be the second-order version as noted from the rate law, and fortunately we have tracked what the concentration of `A` had become in the context of `B` and `C` (which influence `A`!).

This version can be derived (assuming `B` and `C` don't interfere), but the derivation will be omitted. If you need to know it, it should be in your textbook, maybe off to the side? It requires a bit of calculus.

`bb(overbrace(1/([A]))^(y) = overbrace(k)^(m)overbrace(t)^(x) + overbrace(1/([A]_0))^(b))`

assuming `A` has a stoichiometric coefficient of `1`. Get to know this equation, because we'll be using it a lot.

Now, let's plot the two data points we have (and hope that that's enough).

Since the slope is `6891.6`, the rate constant is `k = "6891.6 M"^(-3)"min"^(-1)`. The units are this way because the reaction is fourth order overall, and the change in `[A]` was in the context of `B` and `C`:

`overbrace("M"/"min")^(r(t)) = overbrace(1/("M"^3 cdot "min"))^(k) xx overbrace("M"^2)^([A]^2) xx overbrace("M")^([B]) xx overbrace("M")^([C])`

In the units of your answer key though:

`color(blue)(k) = "6891.6 M"^(-3)cdot"min"^(-1) xx "1 min"/"60 s"`

`= color(blue)("114.86 M"^(-3)cdot"s"^(-1)) ~~ 115`

`2)`

Since we know the form of the integrated rate law for just `A` in the context of `B` and `C`, we can use the rate constant to find the half-life. The half-life is, again, when `[A] = 1/2[A]_0`, so:

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

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

Therefore, the half-life for the reaction based on only the second-order behavior of `A` in the context of `B` and `C` is:

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

`= 1/(114.86(1.00 xx 10^(-4))) " s"`

`=` `color(blue)("87.06 s")`

`3)`

As it turns out, we have been assuming in `(1)` and `(2)` to NOT consider the concentrations of `B` and `C`, and so, the half-life does not seem to be influenced by `B` or `C`...

That is not strictly true, unless the concentration of `B` and `C` do not change, or are held fixed at `color(blue)("1.00 M")` each.

`4)`

Using the second-order integrated rate law once more, denote `alpha` as the fraction of `A` leftover after `t = "10.0 min"`. This fraction will be less than `0.5` because the half-life was about `"1.45 min"`.

Then:

`1/([A]) = kt + 1/([A]_0)`

`1/(alpha[A]_0) - 1/([A]_0) = kt`

`(1/alpha - 1)/([A]_0) = kt`

`=> kt[A]_0 + 1 = 1/alpha`

So, the fraction of `bbA` leftover is:

`alpha = 1/(kt[A]_0 + 1)`

`= 1/(114.86 cdot 600 (1.00 xx 10^(-4)) + 1)`

`= 0.1267`

Therefore, the amount of `A` left is:

`color(blue)([A]_"10.00 min") = alpha[A]_0 = 0.1267 xx 1.00 xx 10^(-4)`

`= color(blue)(1.27 xx 10^(-5) "M")`