Find the overall order of the reaction, the rate constant, and the overall rate expression for the following set of experiments by using the kinetics method of initial rates.[A]0.174 M[B]0.968 M Exptvolume Avolume Bvolume H20time (minutes:seconds)1667 uL250 uL19.083 mL2:452333 uL250 uL19.417 mL5:263667 uL500 uL18.883 mL0:40

"The initial rate of reaction is not given to find the rate constant." According to the rate law, the rate of the reaction (R) is related to the concentration of reactants as follows: Rate=kAaBb where k is the rate constant, [A], [B] are the concentration of the reactants and a, b are their respective orders in the reaction. The concentration of A and B in the given reaction are 0.174 M and 0.968 M respectively. The volume of the solution remains constant in all the trials.The concentration of the reactants is constant in all the trials while the their volume being used differs which determines the end point for product formation. The end point indicates the amount of product consumed to give a dramatic change in color in a given time. Thus, rate of the reaction can be written inversely proportional to the time taken since the amount of product consumed will be the same in order to show change in color. Further, since the concentration of reactants used is also the same in all the trials performed, the rate of the reaction will depend upon the number of moles of reactant of each type consumed to show the end point. The rate law of the given reaction can be expressed as: dndt=k[nA]a[nB]b where dn is the change in number of moles of the product which is same for all the trials, dt is the time taken to show the corresponding change, nA and nB are the number of moles of A and B respectively participating in the given reaction and a and b are their respective orders.To determine the order for a particular reactant, select two trials in which the amount of the other reactant is invariable. The rate law equations for the trials 1, 2 and 3 can be represented as: dn2.75=k[0.174×667]a[0.968×250]b……(1)dn5.43=k[0.174×333]a[0.968×250]b……(2)dn0.67=k[0.174×667]a[0.968×500]b……(3) As the concentration of A and B are constant in all the trials, consider only the volume of A and B in the rate law equations.To determine the order of A, select trial 1 and 3 in which the volume of B is equal and thus, their number of moles (product of concentration and volume) are also equal and then write their ratio. dn2.75dn5.43=k667a250bk[333]a[250]b5.432.75=k[667]ak[333]a1.975=2a a=1 To determine the order of B, select trial 1 and 3 in which the volume of A is equal and thus, their number of moles (product of concentration and volume) are also equal and then write their ratio. dn2.75dn0.67=k250b667ak[500]b[667]a0.672.75=k[250]ak[500]a4.1=2b b=2The order of A and B thus, calculated are 2 and 1 respectively and thus, the overall rate order is the sum of order of A and B that is, 3. The overall rate expression can be written by substituting the order of each reactant in the rate law equation as follows: Rate=kA1B2 Knowing the initial rate of reaction, the rate constant can be calculated as: k=Rate0.17410.9682

Answered: Find the overall order of the reaction,…

Chemistry

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Chapter1: Chemical Foundations

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Expt	volume A	volume B	volume H20	time (minutes:seconds)
1	667 uL	250 uL	19.083 mL	2:45
2	333 uL	250 uL	19.417 mL	5:26
3	667 uL	500 uL	18.883 mL	0:40

[A]	0.174 M
[B]	0.968 M