Print

Please login or register.
1 Hour 1 Day 1 Week 1 Month Forever
Login with username, password and session length

[zacko94]

Consider the decomposition of hydrogen peroxide.

2H₂O₂ 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.

[zacko94]

Sorry, for time, I should say per second - not per minute

[Borek]

Your logic is not broken down, you are just referring to two different cases - easy to interconvert.