# Question #d6345

##### 1 Answer

One hour.

#### Explanation:

Here's how you can answer this question by using an *intuitive approach*.

Notice that you have a **mole ratio** between **every mole** of **mole** of

Consequently, when **moles** of **moles** of

For the first trial, you start with

#"0.80 moles A " - overbrace("0.60 moles A")^(color(blue)("consumed to produce B")) = "0.20 moles A"#

This is equivalent to saying that in **one hour**, you end up with

#(0.20 color(red)(cancel(color(black)("moles A"))))/(0.80color(red)(cancel(color(black)("moles A")))) xx 100 = 25%#

of your **initial amount** of *unconsumed* in the second trial.

#"0.90 moles A " - overbrace("0.675 moles A")^(color(blue)("consumed to produce B")) = "0.225 moles A"#

Once again, you'll have

#(0.225 color(red)(cancel(color(black)("moles"))))/(0.90color(red)(cancel(color(black)("moles")))) xx 100 = 25%#

This can only mean that the time needed in the second trial is **equal** to the time needed in the first trial.

*Why is that the case?*

Your reaction is **first order**, which means that the *rate of the reaction* varies linearly with the concentration of

This means that the rate of the reaction depends exclusively on the **initial amount**, **final amount**, **integrated rate law** that describes a first-order reaction

#color(blue)(|bar(ul(color(white)(a/a)ln( (["A"]_t)/(["A"]_0)) = - k * tcolor(white)(a/a)|)))#

Here

**rate constant** for the reaction at a given temperature

So as long as the ratio between **the same** for two or more trials, the *time* needed to get from **must also be the same**.

**ALTERNATIVE APPROACH**

Another important thing to remember about **first-order reactions** is that the *half-life* of the reaction, **independent** of the initial concentration of the reactant.

Simply put, a first-order reaction **always** has the same *half-life*, which is the time needed for **half** of the reactant to be consumed, regardless of the initial concentration of said reactant.

Look at this first trial. You start with

#t_"1/2 " -> "0.80 moles A"/2 = "0.40 moles A"#

Notice that a *second half-life* must pass in order to get to

#t_"1/2 " -> "0.40 moles A"/2 = "0.20 moles A"#

So, it takes **one hour** for **two half-lives** to pass, which can only mean that the half-life is

#t_"1/2" = "1 h"/2 = "30 min"#

Now look at the second trial. You know that the **half-life is constant**, which means that it takes

#t_"1/2" = "30 min " -> "0.90 moles A"/2 = "0.45 moles A"#

#t_"1/2" = "30 min " -> "0.45 moles A"/2 = "0.225 moles A"#

Once again, **two half-lives** pass, so the *total time* needed will be

#t = 2 xx t_"1/2" = 2 xx "30 min" = "1 h"#