Half Life for a second order reaction

[preet]

Homework Statement

Homework Equations

HalfLife = 1 / k*[A]i

where k is the rate constant and [A]i is the initial concentration of a reactant A.

The Attempt at a Solution

I don't have enough information to attempt this problem. I don't know what to do with the partial pressure of O2 (is it part of the reactants?) If it is, then does the reaction become third-order?

I also don't see the point of having HClO4 in this problem... does it change anything?

By using the given values of k and [Fe] and plugging into the equation, you get 2702.7 atm hours... I think this is wrong because I didnt use the partial pressure of O2 at all...

One final thing, the units by just using the half life equation give atm*hours... and if you divide by the pressure you get the correct unit, just hours. Is this the correct thing to do (the result is 13 513 hours)...

I just want to know whether I'm in the right direction.


Thanks
Preet

 

[GCT]

Fe2+ does react with HClO4, assuming you can ignore the formation
of Fe(OH)2, you need to incorporate the rate dyanmics of this reaction and that of the Fe2+ reacting with oxygen gas. You need to put everything in terms of how the Fe2+ is depleted.

 

[Blueshift5]

i

Hello Preeti,

Thank you for your question. It seems like you are on the right track, but there are a few things that need to be clarified.

Firstly, the half-life equation that you have provided is correct for a second-order reaction. However, the units are incorrect. The correct units for the half-life of a second-order reaction are M^-1*s (or mol^-1*L*s depending on the units of the rate constant).

Secondly, the partial pressure of O2 is not part of the reactants, it is simply a measurement of the reactant's concentration in the gas phase. In this case, it is used to calculate the initial concentration of Fe in the reaction.

Thirdly, the presence of HClO4 does not change anything in this problem. It is simply a spectator ion and does not participate in the reaction.

Lastly, your final result of 13,513 hours is correct. Dividing by the pressure is the correct way to convert the units to hours.

I hope this helps clarify any confusion you may have had. Keep up the good work!