Iodide ion reacts with iodate ion under acidic conditions: IO 3 - (aq) + 8 I - (

Question

Iodide ion reacts with iodate ion under acidic conditions: IO3- (aq) + 8 I- (aq) + 6 H + (aq) --> 3 I3- (aq) + 3 H2O (l)

The kinetics of the reaction were studied by the method of initial rates, neglecting any dependence of the rate on the concentration of products formed. The following data was obtained.

a) Determine the rate law, including the value of the rate constant (wit h units!).

b) What would the initial rate have been if [ IO3- ]0 = 0.300M, [ I-]0 = 0.400M and [ H+ ]0 = 0.200M?

                           Trial       [I- ]0 (M)          [IO3- ]o (M)         [H+ ]o (M)                rate (M/sec)

                            1            0.100                 0.200                    0.0100               0.530 x 10-6                                                         2             0.100                 0.500                    0.0100               1.325 x 10-6                                                                  3    0.200                 0.200                    0.0100               2.120 x 10-6                                                                                      4 0.200    0.200                    0.0050               1.50 x 10-6

a.) rate law:____________      k=_________________

b.) intial rate = __________________________

Explanation / Answer

The table:

In order to calculate the rate law expression for a A+B reaction, we need to apply Initial Rates Method.

Note that the generic formula goes as follows:

r = k [A]^a [B]^b

Note that if we got at least 3 sets of point, in which we have A and B constant, then we could use:

r1 / r2 = (k1 [A]1^a [B]1^b) / (k2 [A]2^a [B]2^b)

If we assume K1 and K2 are constant, then K1= K2 cancel each other

r1 / r2 = ([A]1^a [B]1^b) / ( [A]2^a [B]2^b)

Then, order according to [A] and [B]

r1 / r2 = ([A]1/[A2])^a * ([B]1/[B]2)^b

If we get two points in which A1 = A2, then we could get B, and vise versa for A...

From the data shown in YOUR table

Choose point 1 and 2...

r1 / r2 = ([A]1/[A2])^a * ([B]1/[B]2)^b * ([C]1/[C]2)^c

substitute

(0.53*10^-6) / (1.325*10^-6) = (0.1/0.1)^a * (0.2/0.5)^b * (0.1/0.1)^c

Cleary, the coefficient cancels:

0.4= (0.2/0.5)^b

ln(0.4) / ln(0.4) = b

b = 1

Choose now points3 and4:

r2 / r3 = ([A]2/[A]3)^a * ([B]2/[B]3)^b *   ([C]2/[C]3)^C

substitute

(2.12*10^-6)/(1.5*10^-6) = (0.2/0.2)^a * (0.20/0.20)^b * (0.01/0.005) ^c

Cleary, the coefficient cancels:

1.41 = 2^c

c = ln(1.41)/ln(2) =0.5

Choose now points 1 and 3

r1 / r3 = ([A]2/[A]3)^a * ([B]2/[B]3)^b *   ([C]2/[C]3)^C

((0.53*10^-6) / (2.12*10^-6)) = (0.1/0.2)^a * 0.2/0.2)^b 0.1/0.1)^c

0.25 = 0.5^a

a = ln(0.25)/ln(0.5) = 2

then

rate law = k*[I-]^2[IO3-][H+]^0.5

k

choose any point

rate law = k*[I-]^2[IO3-][H+]^0.5

0.53*10^-6 = k*(0.1^2)(0.2)(0.1^0.5)

k = (0.53*10^-6) / ((0.1^2)(0.2)(0.1^0.5)) = 0.0008380

initial rate...

rate law = k*[I-]^2[IO3-][H+]^0.5

rate law = 0.0008380*(0.3^2)(0.4)(0.2^0.5) = 0.000013491 M/s

Trial [I-]0 [IO3-]o [H+]o rate 1 0.1 0.2 0.01 0.000000530 2 0.1 0.5 0.01 0.000001325 3 0.2 0.2 0.01 0.000002120 4 0.2 0.2 0.005 0.000001500

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