What is the rate law for the following reaction and data?

#rate = k [A]^x [B]^y#
Solve for the Order of each reaction (the exponents) by taking ratios of any two of them. For example:

1) 9.85e-7 0.0025 0.015 0.015
3) 3.94e-6 0.010 0.015 0.015 (change in I)

#(rate3)/(rate1) = (3.94 X 10^(-6) M/s) / (9.85 X 10^(-7) M/s) = 4#

#(k (0.0025M)^x (0.015M)^y)/(k (0.010M)^x (0.015M)^y# #= 4#

#(0.0025M)^x = (0.010M)^x xx 4#
#xlog(0.0025M) = xlog(0.010M) + log4#
#x xx(-2.60) = x xx(-2) + 0.602#
#x xx(-0.60) = 0.602#
#x = 1#, first order.

Calculate the 'y' exponent from a set of experiments where H2SeO3 is changed and I is constant. Then, using the data from experiment 2 and the proper complete rate equation, find the k.

#k = (rate)/([A]^1[B]^y) = (rate_2)/([A_2][B_2]^y#

#r(t) = k["H"_2"SeO"_3]["H"^(+)]^2["I"^(-)]^3#

Reaction:

#"H"_2"SeO"_3(aq) + 6 "I"^(−) (aq) + 4 "H"^(+)(aq) -> "Se"(s) + 2"I"_3^(−)(aq) + 3 "H"_2"O"(l)#

First of all, let's clean up the data...

#ul(r(t)("M/s")" "" "["H"_2"SeO"_3]("M")" "" "["I"^(-)]("M")" "" "["H"^(+)]("M"))#
#9.85 xx 10^(-7)" "" "0.0025" "" "" "" "0.015" "" "" "0.015#
#7.88 xx 10^(-6)" "" "0.0025" "" "" "" "0.030" "" "" "0.015#
#3.94 xx 10^(-6)" "" "0.010color(white)(.)" "" "" "" "0.015" "" "" "0.015#
#3.15 xx 10^(-5)" "" "0.0025" "" "" "" "0.030" "" "" "0.030#

For the rate law, we know that it must involve potentially all three reactants, so we first write:

#r(t) = k["H"_2"SeO"_3]^m["H"^(+)]^n["I"^(-)]^q#

where:

• #k# is the rate constant in the appropriate units.
• #[" "]# is the molar concentration of whatever it is.
• #m#, #n#, and #q# are unknown orders.

It is convenient to take the ratio of two sets of trial concentrations while holding the second and third reactant concentrations constant. That way, those two reactant concentrations go to #1^"finite number" = 1#.

[Otherwise, it just makes it so you have to plug in what you already know, which you don't really know anything until you start.]

ORDER FOR SELENOUS ACID

We can choose trials 1 and 3 here.

#(3.94 xx 10^(-6))/(9.85 xx 10^(-7)) = ((0.010)/(0.0025))^(m)((0.015)/(0.015))^(n)((0.015)/(0.015))^(q)#

#4 = 4^m * 1^n * 1^q#

Therefore, #ul(m = 1)#.

ORDER FOR HYDROGEN CATION

We can choose trials 2 and 4 here.

#(3.15 xx 10^(-5))/(7.88 xx 10^(-6)) = ((0.0025)/(0.0025))^(1)((0.030)/(0.015))^(n)((0.030)/(0.030))^(q)#

In fact, we already know the order for selenous acid, so we plugged it in anyway.

#4 = 1^1 * 2^n * 1^q#

We know that #2*2 = 4#. Therefore, #ul(n = 2)#.

ORDER FOR IODIDE ANION

We can choose trials 1 and 2 here. Or we could now pick any two trials. Let's do it the hard way and pick 1 and 4 for no reason.

#(3.15 xx 10^(-5))/(9.85 xx 10^(-7)) = ((0.0025)/(0.0025))^(1)((0.030)/(0.015))^(2)((0.030)/(0.015))^(q)#

In fact, we already know the order for selenous acid and the proton, so we plugged them in anyway.

#32 = 1^1 * 2^2 * 2^q#

#8 = 2^q#

Thus, #ul(q = 3)#. Convince yourself that doing this for trials 1 and 2, which is easier, also gives you third order in #"I"^(-)#.

RATE LAW

Thus, the rate law is:

#r(t) = k["H"_2"SeO"_3]["H"^(+)]^2["I"^(-)]^3#

What is the overall order of the reaction? (How do you get the sum of the orders?)

RATE CONSTANT

Choose any trial and plug in your concentrations and rate for the known rate law. Solve for #k#.

#k = (r_j(t))/(["H"_2"SeO"_3]_j["H"^(+)]_j^2["I"^(-)]_j^3)#

I'll leave this as an exercise for you to convince yourself that any trial SHOULD give you the same rate constant at the same temperature.

And furthermore, you should be able to show that the units of #k# become #ul("M"^(-5)"s"^(-1))#.