0
$\begingroup$

Does the rate of change of the reaction rate for a 1st order reaction change linearly over time? My teacher says it does, but my textbook seems to contradict this. Could anyone clarify?

$\endgroup$

1 Answer 1

0
$\begingroup$

The logarithm of the concentration of one of the components will vary linearly. However, the concentration does not vary linearly with time, only its logarithm does.

Reaction Rate $= -\frac{d[A]}{dt} = kA = $ constant * [concentration] for a first order reaction.

If you integrate this using separation of variables you end up with

$\ln \frac{[A]_t}{[A]_0} = -kt$

if you plot $\ln[A]_t$ verse time the slope will be constant.

If you plot concentration as a function of time it will not be constant because we can rearrange the above equation to get

$[A]_t = [A]_0 \exp(-kt)$

which is clearly exponential and not linear.

So yes, for a first order reaction, the reaction rate is constant because it depends linearly on the reaction rate constant which is... constant because it is the slope of the logarithm of concentration verse time. Here is a useful link:

www.chem.libretexts.org/reaction_rates

Here is a short answer to a similar question that is useful: chemistry.stackexchange.com

$\endgroup$
6
  • $\begingroup$ Thanks Charlie. I a still a little confused, how can it be that "for a first order reaction, the reaction rate is constant" when the rate changes with concentration? $\endgroup$ Feb 22, 2019 at 5:42
  • $\begingroup$ "the reaction rate is constant because it is the slope of the logarithm of concentration verse time." $\text{the reaction rate constant is the slope of the logarithm of concentration verse time.}$ $\endgroup$ Feb 22, 2019 at 5:42
  • 1
    $\begingroup$ yes Adnan, the reaction rate constant, is constant as all derivatives of linear functions are. $\endgroup$
    – B. Kelly
    Feb 22, 2019 at 5:45
  • $\begingroup$ what I mean @user2351418 is that after plotting the logarithm of concentration verse time THEN you calculate the slope, which is the reaction rate constant. It is a single number. $\endgroup$
    – B. Kelly
    Feb 22, 2019 at 5:53
  • $\begingroup$ The rate constant does not change with concentration, but surely the reaction rate must? I thought the definition of a 0th order was that rate does not change with concentration, and of a 1st order is that rate changes linearly with concentration, and of a 2nd order that rate changes with the square of concentration. As depicted here chem.libretexts.org/@api/deki/files/15966/… $\endgroup$ Feb 22, 2019 at 5:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.