July 16, 2024, 03:18:55 AM
Forum Rules: Read This Before Posting


Topic: Effect of heat and orders in rate equations  (Read 7220 times)

0 Members and 1 Guest are viewing this topic.

Offline zilalti

  • Regular Member
  • ***
  • Posts: 46
  • Mole Snacks: +2/-2
  • Gender: Male
  • I'm a mole!
Effect of heat and orders in rate equations
« on: November 15, 2005, 09:23:03 AM »
If a chemical reaction has an order of zero it means the reaction rate can be independant of temperature. This got me thinking and the only reactions i could think of that this was true for was radioactive decay.
What I don't understand is how the rate of a chemical reaction can stay constant under different temperatures.
Surely all molecules gain kinetic energy from heat so therefore all reactions are speeded up by an increase of temperature.
Can someone explain this to me as i'm sure i'm missing out something really fundemental.
thanks.

Offline FeLiXe

  • Theoretical Biochemist
  • Chemist
  • Full Member
  • *
  • Posts: 462
  • Mole Snacks: +34/-7
  • Gender: Male
  • Excited?
    • Chemical Quantum Images
Re:Effect of heat and orders in rate equations
« Reply #1 on: November 16, 2005, 08:22:49 AM »
when you talk about reaction order you talk about how the rate of reaction changes with concentration not with temperature

a constant reaction order means that the rate of reaction does not depend on the concentration of your reagent. That would be for example if you have a heterogenious catalyst. Then the reaction rate almost only depends on the capacity of the catalyst.

the changing of reaction constants is calculated with the Arrhenius equation. (I think that's what it's called. And I don't know how it goes. some kind of e power)
« Last Edit: November 16, 2005, 08:23:32 AM by FeLiXe »
Math and alcohol don't mix, so... please, don't drink and derive!

Offline Borek

  • Mr. pH
  • Administrator
  • Deity Member
  • *
  • Posts: 27739
  • Mole Snacks: +1804/-411
  • Gender: Male
  • I am known to be occasionally wrong.
    • Chembuddy
Re:Effect of heat and orders in rate equations
« Reply #2 on: November 16, 2005, 09:32:41 AM »
k = k0 exp(-E/(RT))
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info

Offline Juan R.

  • Chemist
  • Full Member
  • *
  • Posts: 148
  • Mole Snacks: +24/-3
  • Gender: Male
    • The Center for CANONICAL |SCIENCE)
Re:Effect of heat and orders in rate equations
« Reply #3 on: November 20, 2005, 08:38:29 AM »
If a chemical reaction has an order of zero it means the reaction rate can be independant of temperature. This got me thinking and the only reactions i could think of that this was true for was radioactive decay.
What I don't understand is how the rate of a chemical reaction can stay constant under different temperatures.
Surely all molecules gain kinetic energy from heat so therefore all reactions are speeded up by an increase of temperature.
Can someone explain this to me as i'm sure i'm missing out something really fundemental.
thanks.

For A + B --> products

V = k [A]nm

zero order mean that m or n or both! are zero. Temperature dependence enter via rate constant k as explained by Borek.

Note: [C]0 = 1

It is true that is usually said that radioactive decay is independent of temperature. But what temperature? So far as i know only studies of k versus environmental temperature Tambient have been done. Nobody has measured the effect on radioative decay of spacetime temperature for example!

In fact, radioative decay may depend of temperature on some way because at 0 kelvin it cannot have any process on Nature (third law of thermodynamics).
« Last Edit: November 20, 2005, 08:44:04 AM by Juan R. »
The first canonical scientist.

Sponsored Links