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?

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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?

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Md Ariful Islam · Answer 1 · Apr 02, 2024 09:48:30

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 $76%$.

A first-order reaction

$A -> B$

follows the :

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

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.

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?

1 Answers