HaSe03 (aa) + 6 I-(aa) Se(s) + 2 13-(aa) + 3 H2O (l) Experiment [H2SeO3]o Initia

Question

HaSe03 (aa) + 6 I-(aa) Se(s) + 2 13-(aa) + 3 H2O (l) Experiment [H2SeO3]o Initial Rate mol/L min 3.0x106 2.4x10-5 6.0x106 5.4x10-5 2 3 4 mol/L 1 1 mol/L 01 02 01 01 mol/L 001 001 001 003 Find the order of the reaction with respect to each reactant and write the rate law. What's the overall order??? Calculate the K value and its units (notice the time units). Double checking K for every experiment. What happens to the experiment 1 rate of reaction if the Ph is increased? (increase, decrease stays the same) a. b. c.

Explanation / Answer

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, so I- is left

(3*10^-6)/(2.4*10^-5) = (0.01/0.02)^b

b = ln(0.125) / ln(0.5) = 3

now, choose point 1 and 3 so H2S2O4 is left

(3*10^-6)/(6*10^-6) = (0.1/0.2)^a

a = ln(0.5) / ln(0.5) = 1

now, choose points: 3 and 4, so H+ remains

(6*10^-6)/(5.4*10^-5) =(0.001/0.003)^c

ln(0.111) / ln(0.3333) = c

c = 2

then

Rate = k*[H2S2O3]^a[I-]^b[H+]^c

Rate = k*[H2S2O3][I-]^3[H+]^2

b)

for k

(3*10^-6) = k * (0.1)(0.01^3)(0.001^2)

k = (3*10^-6)/ ((0.1)(0.01^3)(0.001^2)) = 30000000

choose another point

(6*10^-6) = k * (0.2)(0.01^3)(0.001^2)

k = (6*10^-6)/ ((0.2)(0.01^3)(0.001^2)) = 30000000

c)

if we increase pH --> [H+] decreases

since [H+] depends to ^2

then the rate should decreae by x^2 factor

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