Question

Question dealing with pseudo-first order reaction?

I really only want the answer to part a), but I'd appreciate the rest as well.

Answer 1

`(a)`

You only know that `C` is zero order, but that `B` is second order. Therefore, you do not assume that `Delta[B] = 0`, and you still don't assume that `Delta[C] = 0` either (only that `C` does not affect the rate).

However, we see that `[B]` and `[C]` are about `100` times larger than `[A]`, so we assume that the reaction is pseudo-first order with respect to `A`, i.e. that

`Delta[B]` `"<<"` `Delta[A]`
`Delta[C]` `"<<"` `Delta[A]`

Therefore, we can use the first-order integrated rate law for one reactant to get `k'`:

`ln[A] = -k't + ln[A]_0`

where `t` is the current time, `[A]` is concentration of `A` in `"M"`, and `[A]_0` is the starting concentration of `A`.

Solving for `k'`,

`ln(([A])/([A]_0)) = -k't`.

This gives

`color(blue)(k') = 1/tln(([A]_0)/([A]))`

`= 1/("8 s") ln((1.0 xx 10^(-2) "M")/(3.0 xx 10^(-3) "M"))`

`= color(blue)("0.15 s"^(-1))`

Since the reaction is supposed to be pseudo-first order,

`r(t) = k[A][B]^2~~ k'[A]`

What we actually got was `k' = "0.15 s"^(-1)`, so we can rewrite this as:

`color(blue)(k) ~~ (k')/([B]^2) ~~ ("0.15 s"^(-1))/("3.0 M")^2`

`= color(blue)("0.0167 M"^(-2)cdot"s"^(-1))`

for the overall reaction.

`(b)`

So then, as before, we assume that this is pseudo-first order with respect to `A`, giving a pseudo-first order half-life based on `A`.

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

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

`ln2 = kt_(1//2)`

Therefore, the half-life is:

`color(blue)(t_(1//2)) = (ln2)/k`

`= (ln2)/("0.15 s"^(-1)) = color(blue)("4.6 s")`

`(c)`

This can again be done using the first-order integrated rate law.

`ln[A] = -kt + ln[A]_0`

`color(blue)([A]) = [A]_0e^(-kt)`

`= 1.0 xx 10^(-2) "M" cdot e^(-"0.15 s"^(-1) cdot "13.0 s")`

`= color(blue)(1.4 xx 10^(-3) "M")`

Does this make physical sense? Well, it took longer than `8` seconds, and this concentration is indeed less than `3.0 xx 10^(-3) "M"` accomplished after `8` seconds.

`(d)`

For `C`, consider the stoichiometry of the problem. You will consume twice as much `C` as you do `A`.

(The reaction is zero order with respect to `C` though, so keep that in mind. If we didn't have `Delta[A]`, we wouldn't have any idea how to get `[C]` at any time `t` because `C` doesn't influence the rate.)

After `"13.0 s"`, `A` decreased by:

`|Delta[A]| = |1.4 xx 10^(-3) "M" - 1.0 xx 10^(-2) "M"|`

`= 8.6 xx 10^(-3) "M"`

`C` would decrease by twice as much, so

`|Delta[C]| = 2(8.6 xx 10^(-3) "M") = 1.7 xx 10^(-2) "M"`

Therefore,

`color(blue)([C]_(13)) = [C]_0 - |Delta[C]_(0->13)|`

`= "2.0 M" - 1.7 xx 10^(-2) "M"`

`=` `"1.98 M"`

`~~` `color(blue)("2.0 M")` to two sig figs.

(It was only a `0.86%` reduction in `[C]`.)

This proves that the assumption for a pseudo-first order reaction is good.