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?

$\endgroup$
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

| improve this answer | |
$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – user2351418 Feb 22 '19 at 5:42
  • $\begingroup$ "the reaction rate is constant because it is the slope of the logarithm of concentration verse time." $\text{the reaction rate constant is the slope of the logarithm of concentration verse time.}$ $\endgroup$ – Adnan AL-Amleh Feb 22 '19 at 5:42
  • 1
    $\begingroup$ yes Adnan, the reaction rate constant, is constant as all derivatives of linear functions are. $\endgroup$ – Charlie Crown Feb 22 '19 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. $\endgroup$ – Charlie Crown Feb 22 '19 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/… $\endgroup$ – user2351418 Feb 22 '19 at 5:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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