January 22, 2021, 11:11:01 AM
Forum Rules: Read This Before Posting


Topic: Clarify, please - finding rate constants from a rate against concentration graph  (Read 1404 times)

0 Members and 1 Guest are viewing this topic.

Offline zacko94

  • Very New Member
  • *
  • Posts: 2
  • Mole Snacks: +0/-0
Consider the decomposition of hydrogen peroxide.

2H2O2 :rarrow: 2H2O + O2

I will follow the reaction by measuring the volume of oxygen given off at regular intervals. The first graph that I will draw will be Volume of O2/cm3 against Time/min. I will repeat the experiment a number of times for different initial concentrations of hydrogen peroxide, plotting the results of each on the same axes. I will obtain, from this graph, the initial rate of appearance of oxygen (in cm3 min-1) for each initial concentration of hydrogen peroxide, and, after converting these rates from cm3 min-1 into mol dm-3 min-1 (I know this isn't necessary, but you'll see why soon), will plot a new graph of Rate of Formation of O2/mol dm-3 min-1 against [H2O2]/mol dm-3. Now, since the reaction is first order (overall and with respect to hydrogen peroxide), the graph will show a straight line through the origin (in theory), and, since there are no other reactants to affect the rate, the gradient of that straight line will be the rate constant, k.

I hope I am right so far. Let's get to the bit I'm struggling with.

Now, according to the equation above, two moles of hydrogen peroxide will disappear in the time that one mole of oxygen will appear. That must mean the rate (that being the change in concentration... per unit time) of the disappearance of hydrogen peroxide will be twice the rate of the appearance of oxygen.

Suppose, then, if I had, instead of drawing the graph of Rate of Formation of O2/mol dm-3 min-1 against [H2O2]/moldm-3, first multiplied each value for the rate of formation of oxygen by two (to get the rate of disappearance of hydrogen peroxide) for each initial concentration of hydrogen peroxide and then drawn a graph of Rate of Disappearance of H2O2/mol dm-3 min-1 against [H2O2]/mol dm-3. I would get a straight line again, of course, but this time the gradient of that line would be twice the gradient of the other one. Thus, there would be two different values for k.

Of course, the second way is a lot more convoluted, but, still, could someone explain where my logic has broken down, please.

Thanks in advance.



Offline zacko94

  • Very New Member
  • *
  • Posts: 2
  • Mole Snacks: +0/-0
Sorry, for time, I should say per second - not per minute

Offline Borek

  • Mr. pH
  • Administrator
  • Deity Member
  • *
  • Posts: 26247
  • Mole Snacks: +1706/-402
  • Gender: Male
  • I am known to be occasionally wrong.
    • Chembuddy
Your logic is not broken down, you are just referring to two different cases - easy to interconvert.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

Sponsored Links