Experiment 3: Iodide Persulphate Reaction

clock rxn

2 reactions occur together where a coloured product of one is a reactant of the second → can time how long it takes for colour to appear and thus reaction rate

why is starch used

facilitates seeing blue colour when mixed with iodine

what happens if starch is not used

iodine is only violet at high [ ] so when it starts to accumulate we would observe yellow → orange → red → violet

persulphate ion

S₂O₈^2- → think 2 sulphate ions (no charge change)

which reaction are we studying the rate of

persulphate with iodide ions to form iodine and sulphate

which reaction is the “clock”

iodine reacting with thiolsulphate

thiolsuplate

S₂O₃^2- → thiolsulphate so replace 1 oxygen with a sulphur

general rate law of studied reaction

-[ΔS₂O₈^2-] /Δt = k [S₂O₈²][I^-]

• rate of consuming persulphate = rate of overall rxn (1:1 ratio)

why is k not constant for the reaction of persulphate and Iodide ion

both are negatively charged → the repulse each other and don’t want to interact (threes low ionic strength)

ionic strength effect on reaction rate

high ionic strength → less repulsion → higher k → faster reaction

low ionic strength → high repulsion →lower k → slower reaction

k is ___ related to ionic strength

inversely

ionic strength: ½ ΣC_iZ_i² what does each component mean

C_i → concentration of species of interest

Z_i → charge of species of interest

what is ionic strength

measure of effective concentration of ions → high ionic strength means lots of same charges can interact

units of ionic strength

M

how is ionic strength kept constant

addition of inert electrolytes like NH4+ and K+

what happens if the trial turns blue immediately

no thiosulphate present

how can the rxn order of persulphate be determined

graph -log𝚫t vs log[S₂O₈] for first 3 reactions → order is the slope rounded to nearest whole number

how can the rxn order of Iodide be determined

graph -log𝚫t vs log[I^-] for trial 1, 4, and 5 → order is the slope rounded to the nearest whole number

how to find -Δ[S₂O₈]

equal to half [S₂O₃] initially