The data below were obtained in a study on how the rate of a reaction was affect
ID: 1062524 • Letter: T
Question
The data below were obtained in a study on how the rate of a reaction was affected by the concentration of its reactants.
7. The data below were obtained in a study on how the rate of a reaction was affected by the concentration of its reactants CCI Rate (mol L i hr 1) Experiment [A] LB] 0.200 0.100 0.600 5.0 80.0 0.200 0.400 0.400 15.0 0.600 0.100 0.200 0.200 0.100 0.200 5.0 0.200 0.200 0.400 20.0 a. From this data, calculate the reactant orders of A, B and C b. Determine the overall order of the reaction. c. Calculate the rate constant.Explanation / Answer
Answer:
a)
i) By observin Exp-1 and 4 it's clear that when [A] and [B] kept constant and [C] is varied rate of rection remains constatant.
Hence order of C is 0.
ii) By observation of Exp-1 and 2 it's clear that when [A] is kept constant (As C has order 0 no thought given to it's concentration change) and that of [B] is quadrapled rate of reaction increases 16 fold ( i.e. from 5 to 80 mole.L-1.hr-1.). 16 = 42 (4 taken as concentration is quadrrapled) it means,
Order of B is 2
iii) By observation of Exp -1 and 3 it's clear that as [A] is tripled by keeping [B] constant (As C has order 0 no thought given to it's concentration change). as 3 = 31.
Hence order of A is 1.
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b) Overall order or reaction = sum of orders of each reactant involved.
Hence,
Order of given reaction = order of A + order of B + order of C = 1 + 2 + 0 = 3
Hence overall order of reaction is 3.
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c) Rate expression for this reaction is,
Rate = k [A][B]2[C]0.
Or
Rate = k [A][B]2.
k = Rate / ([A][B]2)
Where, k = rate constant.
Using Exp-1 we have,
[A] = 0.2, [B] = 0.1, [C] = 0.6 (not involved) and Rate = 5 mol.L-1.hr-1.
Using these values in above equation 1 we get,
k = 5 / (0.2)(0.1)2.
k = 5 / (0.2 x 0.01)
k = 5 / 0.002
k = 2.5 x 103.L2 mol-2 hr-1.
Rate constant for given reaction is 2.5 x 103.L2 mol-2 hr-1.
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