Animated Solution for Chemistry - Chemical Kinetics: The time for half-life period of a certain reaction, $A⟶\text{products}$ is $1\text{ h}$. When the initial concentration of the reactant 'A' is $2.0{\text{ mol L}}^{-1}$, how much time does it take for its concentration to come from $0.50$ to $0.25{\text{ mol L}}^{-1}$, if it is a zero order reaction?

Decoding the Zero-Order Reaction

Imagine you are watching a chemical reaction unfold, where reactant $A$ transforms into products. The problem explicitly tells us that this is a zero-order reaction. This is a crucial piece of information! In a zero-order reaction, the rate at which the reactant disappears is completely independent of its concentration. It chugs along at a constant speed, much like a car driving on cruise control.

We are given a specific snapshot of this reaction: when the initial concentration $[A{]}_0$ is $2.0{\text{ mol L}}^{-1}$, the half-life $t_{1/2}$ is exactly $1\text{ h}$. Our ultimate mission is to find out how long it takes for the concentration to drop from $0.50{\text{ mol L}}^{-1}$ to $0.25{\text{ mol L}}^{-1}$.

The Heartbeat of the Reaction

Before we can predict the future of this reaction, we need to know its constant speed, which is the rate constant, $k$. For a zero-order reaction, the half-life formula is beautifully simple:

This equation tells us that the half-life is directly proportional to the initial concentration. A larger starting amount will take longer to halve. Let's rearrange this formula to solve for our unknown, $k$:

Now, we substitute the values provided in the first snapshot:

We've found the heartbeat! The reaction consumes $1.0{\text{ mol L}}^{-1}$ of reactant $A$ every single hour.

Calculating the Time Drop

Now we shift our focus to the second part of the problem. We want to know the time $t$ it takes for the concentration to go from an initial value of $0.50{\text{ mol L}}^{-1}$ to a final value of $0.25{\text{ mol L}}^{-1}$.

We bring in the integrated rate law for a zero-order reaction, which describes the straight-line decay of concentration over time:

We can rearrange this to solve directly for the time interval:

Let's plug in our new initial and final concentrations, along with the rate constant we just discovered:

And there is our answer! It takes exactly $0.25$ hours for the concentration to drop from $0.50$ to $0.25{\text{ mol L}}^{-1}$.

Zero vs. First Order

It is fascinating to note that going from $0.50$ to $0.25$ is exactly a halving of the concentration. If this had been a first-order reaction, the time required for this drop would simply be the half-life, which is constant and independent of the starting amount.

However, because this is a zero-order reaction, the time it takes to halve depends entirely on where you start. Since we started with a much smaller amount ($0.50$ compared to the original $2.0$), it took proportionally less time to halve ($0.25\text{ h}$ compared to the original $1\text{ h}$). Always let the order of the reaction guide your logic!