# Question c7286

Apr 4, 2017

In a 1st order reaction, the rate is directly proportional to the concentration of the reactants.

#### Explanation:

Hence, the rate is fast at the beginning.

As the reactants get used up the reaction goes slower and slower and slower
and …

We can see this happening in the first order plot shown below.

The blue curve is a plot of a first order reaction in which ${\left[A\right]}_{0} = \text{1 mol/L}$ and $k = 1 \textcolor{w h i t e}{l} \text{s"^"-1}$.

The instantaneous rate at a given time $t$ is a line tangent to the curve at that time.

The slope of the line is a measure of the rate ($\text{slope = -} k$).

Instantaneous rate at $t = 0$ .

The $\textcolor{b r o w n}{\text{brown}}$ line represents the instantaneous rate at $t = 0$

Since the reaction has just started, [A] is at its maximum and the rate is fastest
($\text{1 mol·L"^"-1""s"^"-1}$).

Instantaneous rate at $t \textcolor{w h i t e}{l} \text{= 20 s}$

At $t = \textcolor{w h i t e}{l} \text{20 s}$, the rate has slowed considerably because $\text{[A]}$ has decreased to 0.135 mol/L and the instantaneous rate (shown by the $\textcolor{g r e e n}{\text{green}}$ line) is 0.0135 mol/L.

Average rate

The average rate from 0 s to 20 s (shown by the color(purple)("purple" line) is a straight line joining the points, ("0,[A]"_text(0))#$\mathmr{and} \left({\text{20,[A]}}_{\textrm{20}}\right)$.

We see that the slope of this line ($\text{rate" = "0.43 mol·L"^"-1""s"^"-1}$) is between the values at the beginning and end of the 20 s interval.