For a $1^{\text{st}}$ order reaction $ A \rightarrow B$, the corresponding rate law is: $$\text{rate} = k[A]^{1}$$ and the integrated rate law for this is $$ \ln \left(\frac{A}{A-x}\right) = kt$$
Going by this equation, every first-order reaction must get completed only at infinity. Is this true? If not, then what is the correct explanation?
Answer
As far as I can see, the expression you have come up with is not entirely correct. If we assume that the reaction is $\ce{A -> B}$ (i.e. saying that it is approximately irreversible, or that $k_\mathrm{reverse} \ll k_\mathrm{forward}$), we have:
$$ \mathrm{rate} = \frac{\mathrm{d}[\ce{A}]}{\mathrm{d}t} = -k[\ce{A}] $$
Separating:
$$ \frac{\mathrm{d}[\ce{A}]}{[\ce{A}]} = -k\,\mathrm{d}t $$
Integration gives:
$$ \ln\frac{[\ce{A}]}{[\ce{A}]_0} = -kt $$ $$ [\ce{A}] = [\ce{A}]_0 \mathrm e^{-kt} $$
Here, $[\ce{A}]_0$ is the initial concentration of $\ce{A}$. It is correct that mathematically, $[\ce{A}] > 0$ for all $t<\infty$, from the last equation. In practise, however, the amount of $[\ce{A}]$ after a certain time interval $t$ will be negligibly small. How long it takes depends on $k$; if $k$ is large, the time for the reaction to go essentially to completion will be short.
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