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Rate Equation for Peroxodisulphate and Iodide Ions, with a Iron(III) catalyst

For my chemistry coursework I am investigating the kinetics of the reaction between Peroxodisulphate ions and Iodide ions. I have worked out the rate equation without the involvement of a catalyst and have discovered that both reactants are of order one.

I am now going to introduce Iron(III) ions as a catalyst and was wondering what affect this would have on the Rate Equation. I think iron(III) is involved in the rate determining step and therefore should be in the Rate equation:


2Fe2+(aq) + S2O82-(aq) → 2Fe3+(aq) + 2SO42-(aq)
Then: 2Fe3+(aq) + 2I-(aq) → 2Fe2+(aq) + I2(aq)

Any input would be greatly appreciated, and if someone knows the orders of the reactants and the catalyst that would be great so I can compare it to my results.
Thanks :smile:
Original post by Dayve
For my chemistry coursework I am investigating the kinetics of the reaction between Peroxodisulphate ions and Iodide ions. I have worked out the rate equation without the involvement of a catalyst and have discovered that both reactants are of order one.

I am now going to introduce Iron(III) ions as a catalyst and was wondering what affect this would have on the Rate Equation. I think iron(III) is involved in the rate determining step and therefore should be in the Rate equation:


2Fe2+(aq) + S2O82-(aq) → 2Fe3+(aq) + 2SO42-(aq)
Then: 2Fe3+(aq) + 2I-(aq) → 2Fe2+(aq) + I2(aq)

Any input would be greatly appreciated, and if someone knows the orders of the reactants and the catalyst that would be great so I can compare it to my results.
Thanks :smile:


Well you can cancel out the Fe2+ and Fe3+ ions, as they balance out:

2 Fe2+(aq) + S2O82-(aq) → 2 Fe3+(aq) + 2 SO42-(aq)

2 Fe3+(aq) + 2 I-(aq) → 2 Fe2+(aq) + I2(aq)

Then combine the equations:

S2O82-(aq) + 2 I-(aq) → 2 SO42-(aq) + I2(aq).

You now have your rate equation:

Rate = k[S2O82-(aq)][I-(aq)]2
Reply 2
Avatar for Dayve
Dayve
OP
Then combine the equations:

S2O82-(aq) + 2 I-(aq) → 2 SO42-(aq) + I2(aq).

You now have your rate equation:

Rate = k[S2O82-(aq)][I-(aq)]2

I'm fairly sure you need to do experiments to work out order, so I'm not convinced that this is correct.

Also is this for the catalyzed reaction? Because the Rate Equation for the uncatalyzed reaction is definitely just order 1 for both reactants:

Rate = k[S2O82-(aq)][I-(aq)]
Original post by Dayve
Then combine the equations:

S2O82-(aq) + 2 I-(aq) → 2 SO42-(aq) + I2(aq).

You now have your rate equation:

Rate = k[S2O82-(aq)][I-(aq)]2

I'm fairly sure you need to do experiments to work out order, so I'm not convinced that this is correct.

Also is this for the catalyzed reaction? Because the Rate Equation for the uncatalyzed reaction is definitely just order 1 for both reactants:

Rate = k[S2O82-(aq)][I-(aq)]

But the stoichiometry is that 1 mole of S2O82- ions reacts with 2 moles of I- ions, so if you were to double the concentration of I-, you would see a four-fold increase in rate of reaction, so the order (with respect to iodide ions) is 2.
Reply 4
Avatar for t.1996
t.1996
Hey, im doing this as part of my coursework, so i was wondering could you send it so i can compare results and use it like a guideline cos im really stuck.....

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