Chem 201(Chemical Kinetics) Final Exam 2022/2023.Well Elaborate And Detailed Questions &Answers for Distinction Pass

Chem 201(Chemical Kinetics) Final Exam 2022/2023.Well Elaborate
And Detailed Questions &Answers for Distinction Pass.
(ANSWERS FROM PAGE 51)


1. The average rate of disappearance of ozone in the reaction is found to be 9.04 10–3 atm over a certain interval
of time. What is the rate of appearance of O2 during this interval?
A. 13.6 10–3 atm/s
B. 9.04 10–3atm/s
C. 6.03 10–3atm/s
D. 369 10–3atm/s
E. 27.2 10–3atm/s



2. The balanced equation for the reaction of bromate ion with bromide ion in acidic solution is given by:

At a particular instant in time, the value of –[Br–]/t is 2.6 10–3 mol/L s. What is the value of [Br2]/t in the same
units?
A. 1.6 10–3

B. 2.6 10–3

C. 4.3 10–3

D. 1.3 10–3

E. 2.2 10–3




3. Consider the reaction 2H2 + O2 2H2O
What is the ratio of the initial rate of the appearance of water to the initial rate of disappearance of oxygen?
A. 1 : 1
B. 2 : 1
C. 1 : 2
D. 2 : 2
E. 3 : 2

,4. Consider the reaction: 4NH3 + 7O2 4NO2 + 6H2O
At a certain instant the initial rate of disappearance of the oxygen gas is X. What is the value of the appearance
of water at the same instant?
A. 1.2 X
B. 1.1 X
C. 0.86 X
D. 0.58 X
E. cannot be determined from the data



5. For the reaction 5A + 4B 2C + 3D, at a particular instant in time, the rate of the reaction is 0.0211 M/s. What
is the rate of change of B?
A. –0.0211 M/s
B. 0.0844 M/s
C. –0.0844 M/s
D. –0.00528 M/s
E. 0.00528 M/s



6. Consider the reaction X Y + Z
Which of the following is a possible rate law?
A. Rate = k[X]
B. Rate = k[Y]
C. Rate = k[Y][Z]
D. Rate = k[X][Y]
E. Rate = k[Z]



7. Consider the following rate law: Rate = k[A]n[B]m
How are the exponents n and m determined?
A. by using the balanced chemical equation
B. by using the subscripts for the chemical formulas
C. by using the coefficients of the chemical formulas
D. by educated guess
E. by experiment

,8. The following data were obtained for the reaction of NO with O2. Concentrations are in molecules/cm3 and
rates are in molecules/cm3s.

[NO]0 [O2]0 Initial Rate
1 1018 1 1018 2.0 1016
2 1018 1 1018 8.0 1016
3 1018 1 1018 18.0 1016
1 1018 2 1018 4.0 1016
1 1018 3 1018 6.0 1016
What is the rate law?
A. Rate = k[NO][O2]
B. Rate = k[NO][O2]2

C. Rate = k[NO]2[O2]
D. Rate = k[NO]2

E. Rate = k[NO]2[O2]2



9. The reaction of (CH3)3CBr with hydroxide ion proceeds with the formation of (CH3)3COH.
(CH3)3CBr(aq) + OH–(aq) (CH3)3COH(aq) + Br–(aq)
The following data were obtained at 55C.

[(CH3)3CBr]0 [OH–]0 Initial Rate
Exp. (mol/L) (mol/L) (mol/L)
1 0.10 0.10 1.0 10–3
2 0.20 0.10 2.0 10–3
3 0.10 0.20 1.0 10–3
4 0.30 0.20 ?
What will the initial rate (in mol/L·s) be in Experiment 4?
A. 3.0 10–3

B. 6.0 10–3

C. 9.0 10–3

D. 18 10–3

E. none of these

, 10. For a reaction in which A and B react to form C, the following initial rate data were obtained:

[A] [B] Initial Rate of Formation of C
(mol/L) (mol/L) (mol/L·s)
0.10 0.10 1.00
0.10 0.20 4.00
0.20 0.20 8.00
What is the rate law?
A. Rate = k[A][B]
B. Rate = k[A]2[B]
C. Rate = k[A][B]2

D. Rate = k[A]2[B]2

E. Rate = k[A]3



11. Tabulated below are initial rate data for the reaction
2Fe(CN)63– + 2I– 2Fe(CN) 4– 6 + I 2



Initial
Run [Fe(CN)63–]0 [I–]0 [Fe(CN)64–]0 [I2]0 Rate (M/s)
1 0.01 0.01 0.01 0.01 1 10–5
2 0.01 0.02 0.01 0.01 2 10–5
3 0.02 0.02 0.01 0.01 8 10–5
4 0.02 0.02 0.02 0.01 8 10–5
5 0.02 0.02 0.02 0.02 8 10–5
The experimental rate law is:
A. = k[Fe(CN)63–]2[I–]2[Fe(CN)64–]2[I2]
B. = k[Fe(CN)63–]2[I–][Fe(CN)64–][I2]
C. = k[Fe(CN)63–)]2[I–]
D. = k[Fe(CN)63–][I–]2

E. = k[Fe(CN)63–][I–] [Fe(CN)64–]