Sapling Week 9/10 #13

[Scot Widjaja Dis 1J]

Can someone walk me through the steps of this mechanism problem?

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.

[Ranen_Chang_2G]

This one really stumped me too.

[Heather Costa 2C]

so for this problem first consider which is the slow step as it determines the rate of the reaction and create a rate law then see if there are any intermediates which in this case is HClO as it is a product in the first step and a reactant in the 2nd step so you are going to have to create an alternative expression for HClO using step 1 for the forward and reverse rate law then setequal to each other so you can isolate HClO . This then can be substituted into your original slow step rate law to create the rate law for the overall reaction

[Nan_Guan_1L]

Hi! I can help with this one.
so remember that the rate of the overall reaction is governed by the rate of the slowest step. so first you would need to identify the sow step with is step #2.

then, write out the rate law for step #2: rate=k2[HClO][I^-]

however, comparing step 1 and step 2 we know that HClO is an intermediate and thus can not be included in the rate law expression. Thus our mission here is to find a way to substitute HClO. we would be using the combination of the forward and reverse rate law for step 1.

rate (Forward)=k (forward) [ClO-][H2O]
rate (reverse)=k (reverse) [HClO][OH-]

make the above two equal to each other. since at equilibrium, the rates are equal and you get [HCLO]=[k (forward) [ClO-][H2O] / k (reverse) [OH-]] / [I^-]

since the concentration of H2O is constant, your final answer would be rate=k {[I^-][ClO-]} / [OH^-]

[Hannah Kim 3C]

This is a question for anyone, but how would you know that the rate of H2O is constant?

[Samantha Loc 1B]

Since we know the reaction rate is dependent on the slowest step, therefore, you can say that the rate = k(slow) [I-][HClO]; however, since HClO is an intermediate (does not show up in overall rxn equation), you need to replace it. By writing out the forward and reverse reaction rates for rxn 1 (since HClO is product in the equation), you can equate them to each other then solve for [HClO]. You can then substitute this answer for [HClO] in the rate of the slowest step. (H2O is omitted since it is a solvent and therefore constant)

[305561744]

This is a question for anyone, but how would you know that the rate of H2O is constant?

This is just a guess, but could it be because we start with water as a reactant and end with the same amount of water as a product?

[905744270]

Why can't we simply cross our the compounds that appear on both the reactant and product sides of the equations and go from there?

[105740916]

rate=k[I−][ClO−]/[OH−]

[705573518]

This is a question for anyone, but how would you know that the rate of H2O is constant?

Because H2O is the solvent, the concentration of H2O is a constant.

[Kaitlin Joya 1I]

This is a question for anyone, but how would you know that the rate of H2O is constant?

I believe the rate of H2O is constant when it is the solvent. I noticed when we would find equilibrium concentrations, we would exclude H2O if it was a liquid, but include it if it was a gas.

[Carlie_Dorostkar_1A]

Thank you so much to everyone! These response are really helpful!

[Akilah Beatty 2k]

Will H2O always be a constant?