# Determine rate law and rate constant for #2"I"^(-)(aq) + "S"_2"O"_8^(2-)(aq) -> "I"_2(aq) + 2"SO"_4^(2-)(aq)#?

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a. determine rate law

b. calculate rate constant

a. determine rate law

b. calculate rate constant

##### 1 Answer

You've given:

#2"I"^(-)(aq) + "S"_2"O"_8^(2-)(aq) -> "I"_2(aq) + 2"SO"_4^(2-)(aq)#

#ul("trial"" "["I"^(-)]_0("M")" "" "["S"_2"O"_8^(2-)]_0("M")" "" ""init rate"("M/s"))#

#1" "" "0.080" "" "" "" "0.040" "" "" "" "" "12.5 xx 10^(-6)#

#2" "" "0.040" "" "" "" "0.040" "" "" "" "" "6.25 xx 10^(-6)#

#3" "" "0.080" "" "" "" "0.020" "" "" "" "" "6.25 xx 10^(-6)#

#4" "" "0.032" "" "" "" "0.040" "" "" "" "" "5.00 xx 10^(-6)#

#5" "" "0.060" "" "" "" "0.030" "" "" "" "" "7.00 xx 10^(-6)#

Alright, so we know that we can pick

twotrials whereonereactant concentrationstayed constant, so that we can take the ratio of the remaining reactant concentration and compare it to the ratio of the rates.In other words, we take the

selective ratio of the rate laws:

#(r_b(t))/(r_a(t)) = (k["I"^(-)]_b^m["S"_2"O"_8^(2-)]_b^n)/(k["I"^(-)]_a^m["S"_2"O"_8^(2-)]_a^n)# where

#a# and#b# represent chosen trials withinitial rates#r_k(t)# , and#m# and#n# are to-be-determinedexponents(orders) for each reactant in the rate law.#k# is arate constantthat shouldnotchange with trials.

From here on out,subscriptsindicatetrial number, and it is implied that we are examininginitialconcentrations only.In this case, we can only choose trials

#bb3# over#bb1# for#"I"^(-)# (iodide) to stayconstantand#"S"_2"O"_8^(2-)# (dithionate) tochange.We'll write it the full way first, and the shortcut way second:

#(r_3(t))/(r_1(t)) = (cancelk["I"^(-)]_3^m["S"_2"O"_8^(2-)]_3^n)/(cancelk["I"^(-)]_1^m["S"_2"O"_8^(2-)]_1^n)#

#(6.25)/(12.5) = cancel((0.080)^m/(0.080)^m)^(1)(0.020)^n/(0.040)^n#

#0.50 = 0.50^n#

Therefore,#n = 1# , so the reaction is1st orderwith respect to dithionate only.Now we'll do it the shortcut way. We can choose trials that keep dithionate constant, and change iodide. That's a bit harder, but I would choose to divide

#bb2# over#bb1# (although#4# over#1# or#4# over#2# are also valid choices, those numbers aren't as nice).

#(6.25)/(12.5) = (0.040)^m/(0.080)^m#

#0.50 = 0.50^m#

So,#m = 1# as well, and the reaction is also1st orderwith respect to iodide only.Therefore, the

rate lawis:

#color(blue)(r(t)) = k[A]^m[B]^n = color(blue)(k["I"^(-)]["S"_2"O"_8^(2-)])#

Now, the

rate constantcan be calculated since we know the exponents. Choose any trial. I choose trial 1.

#12.5 xx 10^(-6) "M/s" = k("0.080 M")^1("0.040 M")^1#

#color(blue)(k) = (12.5 xx 10^(-6) "M/s")/(("0.080 M")^1("0.040 M")^1)#

#= color(blue)(3.91 xx 10^(-3) "M"^(-1)cdot"s"^(-1))# or

#1/("M"cdot"s")# , or#"L"/("mol"cdot"s")# .