Animated Solution for Chemistry - Chemical Kinetics: For a first order reaction, $A⟶\text{products}$ the concentration of $A$ changes from $0.1\text{ M}$ to $0.025\text{ M}$ in $40\text{ min}$. The rate of reaction when the concentration of $A$ is $0.01\text{ M}$ is

Imagine you are watching a water tank empty out through a small hole at the bottom. At first, when the tank is full, the water gushes out rapidly. But as the water level drops, the pressure decreases, and the flow slows down. This is exactly how a first-order chemical reaction behaves! The rate at which the reactant disappears is directly proportional to how much reactant is currently present.

In this problem, we are given a first-order reaction $A⟶\text{products}$. We are told that the concentration of $A$ drops from $0.1\text{ M}$ to $0.025\text{ M}$ in exactly $40\text{ minutes}$. Our mission is to find the instantaneous rate of the reaction at a later time when the concentration has dwindled down to $0.01\text{ M}$.

To predict the rate at any given moment, we first need to find the unique fingerprint of this reaction: its rate constant ($k$). For a first-order reaction, the integrated rate law is our master key:

Let's plug in the data we have. The initial concentration $[A{]}_0$ is $0.1\text{ M}$, and the concentration at time $t=40\text{ min}$ is $[A{]}_t=0.025\text{ M}$.

Simplifying the fraction inside the logarithm, we get exactly $4$. Since $\log (4)\approx 0.602$, we can compute $k$:

Ninja Technique: Did you notice that $0.1$ to $0.025$ is exactly two half-lives? ($0.1\to 0.05\to 0.025$). This means $2t_{1/2}=40\text{ min}$, so $t_{1/2}=20\text{ min}$. Using the half-life formula $k=\frac{0.693}{t_{1/2}}$, we get $k=\frac{0.693}{20}=0.03465{\text{ min}}^{-1}$. It's much faster and gives the exact same result!

Now that we have the rate constant $k$, we can find the rate at any concentration using the differential rate law. For a first-order reaction, the rate depends linearly on the concentration of $A$:

We want to find the rate when $[A]=0.01\text{ M}$. Let's substitute our values:

Geometrically, this value represents the negative slope of the tangent line drawn on the concentration vs. time curve exactly at the point where the concentration is $0.01\text{ M}$. The math perfectly mirrors the physical reality of the slowing reaction!