The rate of the following reaction in aqueous solution is monitored by measuring

Question

The rate of the following reaction in aqueous solution is monitored by measuring the rate of formation of I_3. Data obtained is listed in the table. S_2 O_8^2- + 3 I^- rightarrow 2 SO_4^2- + I_3^- Determine the rate law for the above reaction. What is the value of the rate constant, k? What would be the initial rate of the reaction if [S_2 O_8^2-] = 0.083 and [I^-] = 0.115 M? The tabulated data were collected for this reaction CH_3 Cl_(g) + 3 Cl_2(g) rightarrow CCl_4(g) + 3 HCl_(g) Write an expression for the reaction rate law and calculate the value of the rate constant, k. What is the overall order of the reaction?

Explanation / Answer

Q3.

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 2 and 3...

r2 / r3 = ([A]2/[A3])^a * ([B]2/[B]3)^b

substitute

(2.8*10^-5) / (5.6*10^-5) = (0.076/0.076)^a * (0.06/0.12)^b

Cleary, the coefficient cancels:

0.50= 1 * (0.5)^b

solve,

ln(0.50) / ln(0.5) = b

b = 1

Choose now points 1 and2:

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

substitute

(1.4*10^-5)/(2.8*10^-5) = (0.038/0.076)^a * (0.06/0.06)^b

Cleary, the coefficient cancels:

0.5= (0.5)^a * 1

solve,

a = 1

so...

a = 1, b = 1

then

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

so

r = k [A]^1 [B]^1

For "k" value... choose any point in your set of data, I will choose 1 for simplicity

substitute data

r = k*[A]*[B]

1.4*10^-5= k*(0.038)(0.06)

K = (1.4*10^-5)/((0.038)(0.06)) = 0.0061403 1/(M^-1 s^-1)

c)

Rate:

r = k [A]^1 [B]^1

r = 0.0061403 *(0.083)(0.115) = 0.0000586= 5.86*10^-5 M/s

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