Question 1039c

Aug 22, 2017

Here's what I got.

Explanation:

The thing to remember about a zero-order reaction is that the half-life of the reaction actually depends on the initial concentration of the reactant.

You know that the integrated rate law for a zero-order reaction looks like this--I won't derive this here

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{{\left[\text{A"]_t = ["A}\right]}_{0} - k \cdot t}}}$

Here

• ${\left[\text{A}\right]}_{t}$ represents the concentration of the reactant after a time $t$
• ${\left[\text{A}\right]}_{0}$ represents the initial concentration of the reactant
• $k$ is the rate constant

Now, the half-life of a chemical reaction, ${t}_{\text{1/2}}$, tells you the time needed for half of the initial concentration of the reactant to be consumed by the reaction.

You can thus say that at half-life, you have

${\left[\text{A"]_ (t_"1/2") = ["A}\right]}_{0} / 2$

Plug this back into the integrated rate law

(["A"]_0)/2 = ["A"]_0 - k * t_"1/2"

and rearrange to get the expression for the half-life of zero-order reaction

t_"1/2" = (["A"]_0 - (["A"]_0)/2)/k

color(blue)(ul(color(black)(t_"1/2" = (["A"]_0)/(2k))))

Now, in your case, one reaction will have ${\left[\text{A}\right]}_{0}$ and ${t}_{\text{1/2}}$ and the other reaction will have

["A"]_ 0^' = (["A"]_0)/4 -> the initial concentration is reduced to $\frac{1}{4} \text{th}$ of its initial value

Keep in mind that you're actually performing the same reaction twice, meaning that the only parameter that is changing is the initial concentration of the reactant.

The half-life of the second reaction will be

t_ "1/2"^' = (["A"]_ 0^')/(2 k) = ((["A"]_ 0)/4)/(2k) = 1/4 * overbrace((["A"]_ 0)/(2k))^(color(blue)(t_"1/2"))#

This means that you have

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{{t}_{\text{1/2"^' = 1/4 * t_ "1/2}}}}}$

You can thus say that for a zero-order reaction, decreasing the initial concentration of the reactant to $\frac{1}{4} \text{th}$ of its value will cause the half-life to decrease to $\frac{1}{4} \text{th}$ of its value as well.