Sapling Week 9/10 #13

[Lucy Wang 2J]

The mechanism proposed for the oxidation of iodide ion, I−, by the hypochlorite ion, ClO−, in aqueous solution is shown.

ClO−(aq)+H2O(l)⇌HClO(aq)+OH−(aq)fast in both directions
I−(aq)+HClO(aq)→HIO(aq)+Cl−(aq) slow
HIO(aq)+OH−(aq)→IO−(aq)+H2O(l) fast
Complete the rate law for the formation of IO− implied by this mechanism.

I'm a bit confused on how to solve this problem. I understand rate = k [I-][HClO]. and that the [HClO] must be replaced because it is an intermediate. But how do you solve for what it should be replaced with? and why is water included in the final answer?

[Melanie Lin 3E]

Hi! You basically use the equilibrium equation from the first step! So K=[HClO][OH-]/[ClO-] so you isolate [HClO] to get K[ClO-]/[OH-]. This would be used to replace [HClO] in the rate law. I'm pretty sure water isn't included in the answer (if I remember correctly). Hope this helps!

[jeffrey van 2I]

H2O is not included in the rate law. Since it is a solvent, its concentration will remain constant and be combined with the overall rate constant.

[Will Skinner]

In this problem the first elementary reaction is in equilibrium (the half arrows is notation for equilibrium). At equilibrium, the forward rate law is equal to the reverse. Once you set them equal to one another, you can rearrange the terms to get an expression for [HClO] and substitute that into the rate law for IO-.

Often when making expressions for intermediates we write the expression for the net rate of formation for the intermediate and set it equal to zero (assuming that the concentration of the intermediate does not change). With problem #13, we can instead use the equilibrium of the first step.

[Aditya Sundaram 3D]

The mechanism proposed for the oxidation of iodide ion, I−, by the hypochlorite ion, ClO−, in aqueous solution is shown.

ClO−(aq)+H2O(l)⇌HClO(aq)+OH−(aq)fast in both directions
I−(aq)+HClO(aq)→HIO(aq)+Cl−(aq) slow
HIO(aq)+OH−(aq)→IO−(aq)+H2O(l) fast
Complete the rate law for the formation of IO− implied by this mechanism.

I'm a bit confused on how to solve this problem. I understand rate = k [I-][HClO]. and that the [HClO] must be replaced because it is an intermediate. But how do you solve for what it should be replaced with? and why is water included in the final answer?

To replace [HClO] you can find the rate law from the first reaction in both directions. Since it is in equilibrium you can assume both k's are equal and cancel out giving you a solid [HClO] = [ClO-][H2O]

[David Liu 1E]

because we're looking at the slow reaction, we use that one and then replace hclo with the equation from the first as it's an intermediate product!

[Kyle Walsh 2J]

Hi! I believe that in order to solve for the unknown you just take the equilibrium reaction and isolate HClO. And as for the water, the solvent is never included in the rate law. I know I'm a bit late to this one, but I hope this helps anyways!