Question #c7286

1 Answer
Apr 4, 2017

Answer:

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 #[A]_0 = "1 mol/L"# and #k = 1color(white)(l) "s"^"-1"#.

1st Order

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 (#"slope = -"k#).

Instantaneous rate at #t =0# .

The #color(brown)("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
(#"1 mol·L"^"-1""s"^"-1"#).

Instantaneous rate at #t color(white)(l)"= 20 s"#

At #t = color(white)(l)"20 s"#, the rate has slowed considerably because #"[A]"# has decreased to 0.135 mol/L and the instantaneous rate (shown by the #color(green)("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))## and ("20,[A]"_text(20))#.

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