Question

If it takes `"3 min"` for `40%` of sucrose to dissolve in a certain amount of water, how long does it take for all but `10%` of the sugar to dissolve?

Answer 1

Not exactly a half-life problem... but it seems we'll have to assume that it works that way (even though it is more random than that).

Here are three ways to do it.

CONCEPTUAL WAY

A half-life is the amount of time needed to see the concentration drop to one half its starting amount. We could represent this as:

`[suc] = (1/2)^(n)[suc]_0`

where `n` is the number of "half-lives" (the number of times we've multiplied the starting concentration by `0.5`).

Extending the idea of a half-life, the "0.4-life" can be represented as:

`bb([suc] = (0.4)^(n)[suc]_0)`

where `n` is the number of "0.4-lives" (the number of times we've multiplied the starting concentration by `0.4`). Each "0.4-life" is `"3 min"`, as defined in the question.

If we then let `[suc] = 0.1[suc]_0`, we have "all but `10%` of the sugar" left.

`=> 0.1cancel([suc]_0) = 0.4^n cancel([suc]_0)`

`=> 0.1 = 0.4^n`

`=> ln0.1 = ln 0.4^n`

Use the property that `lna^b = blna` to get:

`=> ln0.1 = nln0.4`

`=> n = (ln0.1)/(ln0.4) = 2.513`

That means about `2.5` "0.4-lives" passed. Since one "0.4-life" is `"3 min"`, it means that `3 xx 2.513 = color(blue)("7.54 min")`.

USING EXCEL?

If you wanted to try doing this with excel, what I did was choose `n` from `1` to `3` in increments of `0.1`, and multiply `t_(0.4)` by `n` to find the total time passed.

Then, I used the equation above, `[suc] = 0.4^n [suc]_0` to graph `[suc]` vs. `t`.

And the amount of time needed to reach `10%` of the sugar left looks to be about `"7.5 min"`, in agreement with the first method used.

USING THE INTEGRATED RATE LAW

Assuming sugar solvation follows first-order half-life kinetics, another way is to use the first-order integrated rate law:

`bb(ln[suc] = -kt + ln[suc]_0)`

Rearrange to get:

`ln(([suc])/([suc]_0)) = -kt`

In the question, it states that `[suc]_0 -> 0.4[suc]_0` for `t = 3`. Therefore:

`ln((0.4cancel([suc]_0))/(cancel([suc]_0))) = -k(3)`

`ln0.4 = -3k`

And the rate constant for this solvation process would be:

`=> k = -ln0.4/3 = "0.3054 min"^(-1)`

As a result, if we want all but `10%` of the sugar to be gone, we want `90%` solvation, i.e. we want `[suc]_0 -> 0.1[suc]_0`. Therefore, we can reuse the integrated rate law from before:

`ln((0.1cancel([suc]_0))/(cancel([suc]_0))) = -("0.3504 min"^(-1))t`

`=> ln 0.1 = -0.3504t`

`=> color(blue)(t) = -ln0.1/0.3054 "min"`

`=` `color(blue)("7.54 min")`

You can see we got the same thing as we got above.