Question

If the rate constant is `3.5 xx 10^(-3) "s"^(-1)` for a first order decay of reactant `A`, how long does it take the reaction to get to `24%` completion?

`A)` `"285.7 s"`
`B)` `"407.7 s"`
`C)` `7.84 xx 10^(-5) "s"`
`D)` `4.08 xx 10^(-4) "s"`
`E)` `"78.41 s"`

Answer 1

I got about `"78 s"`.

`A)` is incorrect because it is merely one over `k`.
`B)` is incorrect because it results from taking `24%` completion to mean `24%` of reactant `A` leftover instead of `76%` of `A` leftover.
`C)` is incorrect because it results from an exponent error due to not using parentheses around `k` in evaluating the time, even though the rest was correct.
`D)` is incorrect for the same reasons as `C)`, except that it also takes the reaction to have `24%` of `A` leftover instead of `76%`.

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A first-order reaction

`A -> B`

follows the rate law:

`r(t) = k[A] = -(Delta[A])/(Deltat) = (Delta[B])/(Deltat)`

For an infinitesimally small time `dt`, we can rewrite this as:

`r(t) = k[A] = -(d[A])/(dt) = (d[B])/(dt)`

Rearranging, we get:

`-kdt = 1/([A])d[A]`

What we can do is "integrate" from time `t = 0` to time `t` on the left and from initial concentration `[A]_0` to current concentration `[A]` on the right.

In simple words, this describes the evolution of time as compared to how the concentration changes, in a "smooth" manner (rather than "stepwise").

`-int_(0)^(t) kdt = int_([A]_0)^([A])1/([A])d[A]`

`-(k*t - k*0) = ln[A] - ln[A]_0`

Therefore, we have the integrated rate law for first-order kinetics:

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

Manipulating this rate law, we can determine at what time this reaction is at `24%` completion, i.e. when `24%` of `A` is gone, or when `76%` of `A` is left. In other words, `[A] = 0.76[A]_0`.

This means:

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

`ln\frac([A])([A]_0) = -kt`

`ln\frac(0.76cancel([A]_0))(cancel([A]_0)) = -kt`

`ln(0.76) = -kt`

Therefore:

`color(blue)(t) = -ln(0.76)/k`

`= -ln(0.76)/(3.5 xx 10^(-3) "s"^(-1))`

`=` `color(blue)("78.41 s")`

This is closest to answer E.