ezto.mheducation com 0 Dashboard HWK Chapter 13 question 1 Chegg.c 4. .26 points

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ezto.mheducation com 0 Dashboard HWK Chapter 13 question 1 Chegg.c 4. .26 points Be sure to answer all parts. Report problem The reaction of peroxydisulfate ion (s,0s2 with iodide ion (I) is Would you like to try another version of this question? By doing so, you wil start from scratch and lose all points associated with this question. From the following data collected at a certain temperature, determine the rate law and calculate the rate constant. Experiment ISMIntial Rate IM/s] 0.0100 0.0100 0.0200 0.0360 0.0180 0.0180 7.60 x 10 3.80 x 104 7.60 x 10 Try another (a) Which of the following equations represents the rate law for this reaction? A. rate = k[S2082-1111 B, rate = kls,082-12[I-1 C. rate = k[S2082-ll 1-12 D. rate = kls,os2-1211-12 (b) What is the rate constant for the reaction? References eBook & Resources

Explanation / Answer

a)

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 points 1 and 2 so I- is left

(7.6*10^-4)/(3.8*10^-4) = (0.036/0.018){b

b = ln(2) / ln(2) = 1

now, hoose point 2 qand 3 so S2O8-2 is left

(3.8*10^-4)/(7.6*10^-4) = (0.01/0.02)^a

ln(0.5) / ln(0.5) = a

then

order is 1,1

Rate = k*[S2O8-2][I-]

b)

for K, choose any point

(7.6*10^-4) = k * (0.01*0.036)

k = (7.6*10^-4) / (0.01*0.036)

k = 2.111 1/(Ms)

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