# Question d6345

Aug 27, 2016

One hour.

#### Explanation:

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

Notice that you have a $1 : 1$ mole ratio between $\text{A}$ and $\text{B}$. This means that every mole of $\text{A}$ that reacts produces $1$ mole of $\text{B}$.

Consequently, when $0.60$ moles of $\text{B}$ are produced, $0.60$ moles of $\text{A}$ are consumed.

For the first trial, you start with $0.80$ moles of $\text{A}$ and end up with

$\text{0.80 moles A " - overbrace("0.60 moles A")^(color(blue)("consumed to produce B")) = "0.20 moles A}$

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 $\text{A}$. Now look what happens when you use the same approach to calculate the amount of $\text{A}$ that remains unconsumed in the second trial.

$\text{0.90 moles A " - overbrace("0.675 moles A")^(color(blue)("consumed to produce B")) = "0.225 moles A}$

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 $\text{A}$.

This means that the rate of the reaction depends exclusively on the initial amount, ${\left[\text{A}\right]}_{0}$, and the final amount, ${\left[\text{A}\right]}_{t}$, of the reactant, as shown by the integrated rate law that describes a first-order reaction

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \ln \left(\left({\left[\text{A"]_t)/(["A}\right]}_{0}\right)\right) = - k \cdot t \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Here

$k$ - the rate constant for the reaction at a given temperature

So as long as the ratio between ${\left[\text{A}\right]}_{t}$ and ${\left[\text{A}\right]}_{0}$ is the same for two or more trials, the time needed to get from ${\left[\text{A}\right]}_{0}$ to ${\left[\text{A}\right]}_{t}$ must also be the same.

$\textcolor{w h i t e}{a}$

ALTERNATIVE APPROACH

Another important thing to remember about first-order reactions is that the half-life of the reaction, ${t}_{\text{1/2}}$, is 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 $0.80$ moles of $\text{A}$. The half-life here would correspond to

${t}_{\text{1/2 " -> "0.80 moles A"/2 = "0.40 moles A}}$

Notice that a second half-life must pass in order to get to $0.20$ moles of $\text{A}$

${t}_{\text{1/2 " -> "0.40 moles A"/2 = "0.20 moles A}}$

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

${t}_{\text{1/2" = "1 h"/2 = "30 min}}$

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

${t}_{\text{1/2" = "30 min " -> "0.90 moles A"/2 = "0.45 moles A}}$

${t}_{\text{1/2" = "30 min " -> "0.45 moles A"/2 = "0.225 moles A}}$

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

$t = 2 \times {t}_{\text{1/2" = 2 xx "30 min" = "1 h}}$
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