The half-life of a zero order reaction A $\rightarrow$ products, is 0.5 hour. The initial concentration of A is 4 mol L$^{-1}$. How much time (in hr) does it take for its concentration to come from 2.0 mol L$^{-1}$ to 1.0 mol L$^{-1}$?
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Memorize the key equations for a zero-order reaction:
Rate Law: Rate = k
Integrated Rate Law: $[A]_t = [A]_0 - kt$
Half-life: $t_{1/2} = \frac{[A]_0}{2k}$
Unlike first-order reactions, the half-life of a zero-order reaction depends on the initial concentration.
For a zero-order reaction, the rate is independent of the concentration of the reactant.
Rate = $k$, where $k$ is the rate constant.
The integrated rate law for a zero-order reaction is:
$[A]_t = [A]_0 - kt$.
The half-life ($t_{1/2}$) is the time it takes for the concentration to drop to half of its initial value, $[A]_0/2$.
$[A]_0/2 = [A]_0 - kt_{1/2} \implies kt_{1/2} = [A]_0/2 \implies t_{1/2} = \frac{[A]_0}{2k}$.
We are given $t_{1/2} = 0.5$ hour and the initial concentration $[A]_0 = 4$ mol L$^{-1}$.
We can use this information to find the rate constant $k$.
$0.5 = \frac{4}{2k} \implies 0.5 = \frac{2}{k} \implies k = \frac{2}{0.5} = 4$ mol L$^{-1}$ hr$^{-1}$.
Now, we need to find the time it takes for the concentration to change from $[A]_1 = 2.0$ mol L$^{-1}$ to $[A]_2 = 1.0$ mol L$^{-1}$.
We can use the integrated rate law, setting the "initial" state to be when the concentration is 2.0.
Let $t$ be the time for this change. Then $[A]_t = 1.0$ and $[A]_0 = 2.0$.
$1.0 = 2.0 - kt$.
$kt = 2.0 - 1.0 = 1.0$.
$t = \frac{1.0}{k}$.
Substitute the value of $k=4$:
$t = \frac{1.0}{4} = 0.25$ hours, or $1/4$ hour.
