In my Chemistry book, it states that sublimation is normally a zero-order reaction or process, as the concentration of molecules that have sufficient energy to sublime is limited by the solid's surface area. My book states that the concentration of these molecules does not change when the amount of subliming substance decreases.

However, I am struggling to understand one thing: if the amount of subliming substance decreases and the object gets smaller, wouldn't that have an effect on the surface area of the solid as well as a consequent effect on the rate of reaction?

Why exactly is sublimation zero-order when clearly the surface area of a solid is variable in different instances? According to me the only way for sublimation to be zero-order is if the solid is held in a sealed container and is under dynamic equilibrium with its vapor pressure. Am I correct about this? I would appreciate it greatly if someone could help me out about all this.

  • 2
    $\begingroup$ Well, there are about 1E15 at/cm2 atoms on a surface, and 6E22 at/cm3, so it takes quite a lot of atoms leaving a surface to change the volume or surface area much... $\endgroup$ – Jon Custer Dec 21 '18 at 4:00
  • $\begingroup$ Thank you for replying. That makes sense and provides me with some helpful insight, but what if you considered a case where on a macroscopic level you did notice a large difference in volume or surface area, such as a heap of snow reducing in size by a few factors after a period of a few days? Surely the rate of sublimation can't be considered approximately constant in that case? There would be a disparity between the initial and final rates, correct? Thanks for your help. $\endgroup$ – user596476 Dec 21 '18 at 4:22
  • $\begingroup$ I think that the rate of sublimation even on large time-scales can be taken as constant because even though it might change to some degree across that entire length of time, the rate of sublimation decreases at such a slow rate in general that we can simply consider it approximately constant. $\endgroup$ – user596476 Dec 21 '18 at 4:53
  • 3
    $\begingroup$ Zero'th order only means the rate does not depend on the concentration of the substance in question in the gas phase (or in solution). It's up to the experimentalist to keep everything else (temperature, catalyst concentration, active surface area, ...) constant. $\endgroup$ – Karl Dec 21 '18 at 7:54
  • $\begingroup$ In the case of your book, you just have to put the substance in an open box, and voilà, the surface doesn't change any more during sublimation. $\endgroup$ – Karl Dec 21 '18 at 8:09

Your Answer

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

Browse other questions tagged or ask your own question.