(10 pts) For thereaction Haseo3(aq) + 6 raq) + 4H+(ao Ses) + 213-(aq) + 3 HOah s

Question

(10 pts) For thereaction Haseo3(aq) + 6 raq) + 4H+(ao Ses) + 213-(aq) + 3 HOah several experiments were conducted and the initial rate of reaction was measured. 1. Experiment[H2SeOoJHlo Initial rate (mo/L (mol/L)(mol/L) (mol/L) 0.0010 0.0010 0.0010 0.10 0.10 0.20 0.20 0.010 0.020 0.010 3.0 x 10-6 2.4 x 10- 6.0 x 106 0.010 0.003 5.4x 105 a) Determine the order of the reaction with respect to each reactant and write the rate law. What is the overall order? (And no, I didn't make this one up-it's real!) calculating k for all experiments. increase, decrease, stay the same?) b) Determine the value of k and its units. Note the time units. Make sure to double check by e) What will happen to the rate of reaction in experiment 1 if the pH is increased? (Will it

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 H2S2O3 is left

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

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

now, choos points 3 and 4 so H+ is left

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

0.111 = 0.333^c

c = ln(1/3) = -1

then

rate is

Rate = k*[H2S2O3] [ I-]^3 [H+]^-1

b)

k can be calcualted via

(3*10^-6) = k * (0.1)(0.01^3)(0.001)^-1

k = (3*10^-6) / 0.0001 = 0.03

units must be

k = 0.03 1/(M2-s)

c)

if pH increases, [H+] decreases

since it is inversely proportional

the rate must increase

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