The amount of substance produced by a chemical reaction
The amount of substance produced by a chemical reaction is $x = A\bigl(1-e^{-kt}\bigr)$, with $A,k>0$. Find the rate of the reaction.
Explanation of Method
The rate turns out to depend on how much is left, not on the clock.
Step 1: Differentiate
The constant $A$ disappears from the bracket and the chain rule supplies $+k$: $\frac{dx}{dt} = A\cdot ke^{-kt} = Ake^{-kt}$.
Step 2: Re-express through $x$
From the given law, $e^{-kt} = 1-\dfrac{x}{A} = \dfrac{A-x}{A}$, so
\[ \frac{dx}{dt} = Ak\cdot\frac{A-x}{A} = k\bigl(A-x\bigr). \]
Step 3: What the second form says
The rate is proportional to the amount of substance still unreacted, $A-x$. This is the first-order rate law, and the given $x(t)$ is precisely its solution with $x(0) = 0$. The reaction begins fastest, at $\dfrac{dx}{dt}\big|_{0} = Ak$, and slows as it approaches completion.
Step 4: Verification
\[t = 0:\ x = 0 \text{ and } \frac{dx}{dt} = Ak \text{ by both formulas} \;\;\checkmark\]
\[t\to\infty:\ x\to A,\ \frac{dx}{dt}\to0 \ \text{ (the reaction runs to completion but never overshoots)} \;\;\checkmark\]
\[\text{Half-completion at } t = \frac{\ln2}{k}, \text{ where the rate is } \frac{Ak}{2} \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,\frac{dx}{dt} = Ake^{-kt} = k\bigl(A-x\bigr)\,}\]
Remark
Remark. The quantity $A-x$ decays exponentially with the same constant $k$, so the half-life $\dfrac{\ln2}{k}$ is independent of how much has already reacted — the same structure as the barometric law of Problem 348, and as radioactive decay. Only the sign of the exponent distinguishes a quantity that grows to a ceiling from one that falls to zero.