Table 1: Summary information of kinetic run Question 1 a) What is the order of t
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Question
Table 1: Summary information of kinetic run
Question 1
a) What is the order of the reaction with respect to dye? (Use zero, first, second, third, fourth, etc.)
b) What is the value of K_obs?
c) What is the units for K_obs?
d) What is the order of reaction with respect to the bleach? (Use zero, first, second, third, fourth, etc.)
e) What is the numeric part of the rate constant for the reaction?
f) What is the units for the rate constant?
2) In the experiment we wanted one of the reactants in large excess. Why was the bleach chosen to be in excess and not the dye?
Question 3
The rate law for the reaction between A and B was determined to be:
Reaction rate = [A]3 [B]2
Reagent B was decreased from 0.14 M to 0.07 M
Select items a to c to indicate what effect this change will have on the reaction rate. Select d to j to indicate the magnitude of this change.
a) Increase
b) Decrease
c) Remains the same
d) no change in reaction rate
e) Increase 2 times the initial rate
f) increase to 4 ties the initial rate
g) Increase to 8 times the initial rate
h) Decrease to 1/2 times the initial rate
i) Decrease to 1/4 times the initial rate
j) Decrease to 1/8 times the initial rate
Trial Concentration of bleach (M) measured in run Kobs 1 0.125 ? 2 0.250 0.0303 y -0.0017x +0.6413 R2 0.8586 Time (s) 0.0596x 4.39 Rar 0.84 Time (s) y -0.00076x-0.0504 R2 0.9954 Time (s)Explanation / Answer
Question 1)
a) The above plots for the first run is a plot that relates to the first order kinetics.the data points fits a line in ln vs t plot only
ln[A]=ln[A]o-kt [integrated rate law for first order rxn] ,ln [A] vs time,t (where [A]= concentration of reactant A after time t.) is linear and follows linear equation of the form y=mx+c ,m=slope,c=intercept
Also ,as the concentration of bleach dye is changing ,so bleach conc must be constant throughout the experiment(by using huge excess of bleach).
The rate law is,
rate=K[bleach]^m [dye]^n =K''[dye]^m [ with K''=K[bleach]^n that makes the pseudo order with respect to dye]
b)Kobs=K[bleach] [eqn relates apparant and actual rate constant]
kobs=K*(0.125M)^m [m=1 ,first order]
0.0303=K[0.250M]
kobs/0.303=(1/2)
kobs=(1/2)*0.0303=0.01515
c)unit for kobs=s-1 [first order]
d)first order with respect to bleach as well
e) from part b)
Kobs=K[bleach] [eqn relates apparant and actual rate constant]
0.0303=K[0.250M]^m ,m=1
0.0303=K[0.250M]
k=0.303/0.250=0.1212
f) rate law=K[bleach]^m [dye]^n ,as both m=n=1 the rxn follows first order kinetics with respect to both dye and bleach
so overall order=m+n=1+1=2
overrall rate constant for second order=M^-1 s^-1
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