Multiple Choice Questions
Kinetics (I):

   
 1 Which of the following would NOT increase the rate of reaction. 

raising the temperature 
adding a catalyst 
increasing the concentration of reactants 
increasing the surface area of a solid reactant 
increasing the volume of the container for a gaseous reaction.

 2 The reaction A + B ® products is found to be second order in [A] and first order in [B]. The rate equation would be:

R = k[A][B] 
R = k[A]2[B] 
R = k[A][B]2
R = k[B] 
R = k[A]2

 3 For a certain reaction, it is found that the rate equation is R = 0.015 L/mol·s[A][B]2. Calculate the rate of the reaction, in mol/L·s, when [A] = 0.022 M and [B] = 0.055 M. 

1.0 x 10-6
4.0 x 10-7
1.5 x 10-2
1.8 x 10-5
1.2 x 10-3

 4 The entropy of the surroundings goes up when water vapor condenses. This is because:

Heat given off by the system increases the thermal motion of the surroundings.
The system is going to a more ordered state.
The system is going to a more random state.
Heat absorbed by the system increases the thermal motion of the surroundings.
Heat given off by the system decreases the thermal motion of the surroundings.

 5 For the reaction 2A + 3B + C = Products, the rate equation is:

R = k[A]2
R = k[A][B][C] 
R = k[A]2[B]3[C] 
R = k[B]3[C] 
Insufficient information to fix rate equation 

 6 For a certain first order reaction, it is found that it takes 156 seconds for the concentration of reactant to fall from 0.100 M to 0.0500 M. How much time would it take for the concentration of reactant to fall from 0.0500 M to 0.0250 M?

1567631212.589.6

 7 The following initial rate-concentration data was collected for the reaction 
2A + B + 2C = D + E + F. 
Initial Concentrations, M Initial Rate, M/s
[A] [B] [C] D[D]/Dt
1.0 0.50 0.40 1.8 x 10-4
1.0 0.40 0.40 1.8 x 10-4
1.0 0.30 0.80 9.0 x 10-5
0.10 0.20 0.40 1.8 x 10-5
The rate equation for this reaction is:

R = k[A]2[B][C]2
R = k[A][B]2[C] 
R =s k[A][C] 
R = k[A][C]-1
R = k[B][C]2

 8 A 1.000 g sample of live grass gives 116 decays per hour of carbon-14. A 1.000 g sample of grass found in an Egyptian tomb gives 34 decays per hour. The half-life of carbon-14 is 5720 years. How old, in years, is the ancient grass?

10100 
1680 
19500 
2.15 x 10-4
7020

 9 For a certain reaction, it is found that the equation relating the specific rate constant, k(M/s), and absolute temperature, T, is: 
lnk = -4420/T + 12.20. 
What is the value of the specific rate constant, k(M/s) at 500 K?

3.36 
28.8 
1.37 x 109
21.04
10.4

 10 What is the activation energy, in kJ/mole, for the process described in problem 9?

-4.42-36.7+36.7+4.42-36200

 11 A catalyst functions by:

lowering the energy of the reactants. 
lowering the energy of the products. 
providing a reaction path with a lower activation energy 
increasing the equilibrium constant. 
decreasing the rate of the reverse reaction.

 12 The following reaction is found to be first order in H2 (g) and second order in NO (g). The rate law for this reaction is:    2NO (g) + 2H2® N2 (g) + 2H2O (g)

Rate = k[NO]2[H2]2/[N2][H2O]2
Rate = k[NO][H2]2
Rate = k[NO]2[H2]2
Rate = k[NO]2[H2] 
Rate = k[N2][H2O]2/[NO]2[H2]2

 13 At 600 K the rate constant for the decomposition of NO2 is 0.75 M-1s-1.   2NO2 (g) ® 2NO (g) + O2 (g)
At 700 K the rate constant is 19.7 M-1s-1. The energy of activation for this reaction is _____ kJ/mole. 

3623.61141.147

 14 The thermal decomposition of phosphine (PH3) into phosphorous and molecular hydrogen is a first order process.    2PH3 (g) ® 2P (s) + 3H2 (g)
At 680° C the rate constant for this reaction is 0.0198 s-1. How long will it take for an initial concentration of 0.57 M to fall to 0.013 M?

75 s191 s390 s28 s0.011 s 

 15 What is the half-life of phosphine (problem 14) at 680° C?

55 s35 s0.029 s0.014 s0.0056 s

 16 You would expect the half-life of phosphine at room temperature to be _____ than at 680° C.

longer 
shorter 
the same 
Not enough information given. Must know DH to answer this. 
Not enough information given. Must know Ea to answer this. 

 17 The rate of the reaction :
                  BrO3- (aq) + 5Br- (aq) + 6H+ (aq) ® 3Br2 (aq) + 3H2O
was studied and the following information was obtained:
 

[BrO3-]0
[Br-]0
[H+]0		 Initial
Rate{M/sec}
0.10 0.10 0.10 8.0 x 10-4
0.20 0.20 0.10 3.2 x 10-3
0.20 0.10 0.10 1.6 x 10-3
0.10 0.10 0.20 3.2 x 10-3
The rate law for this reaction is:

Rate = k[BrO3-][Br-]5[H+]2
Rate = k[BrO3-][Br-]5[H+]6
Rate = k[BrO3-]2[Br-][H+]2
Rate = k[BrO3-][Br-]2[H+] 
Rate = k[BrO3-][Br-][H+]2

Questions 18 through 21 are based upon the following mechanism proposed for the 
decomposition of hydrogen peroxide using potassium bromide. 

    A.      H2O2 (aq) + Br- (aq) ® H2O (aq) + OBr- (aq)  ......Slow 
    B.      H2O2 (aq) + OBr- (aq) ® H2O (aq) + Br- (aq) + O2 (g)  ...... Fast
 18 The overall reaction is:

H2O2 (aq) + OBr- (aq) ® H2O (aq) + Br- (aq) + O2 (g) 
H2O2 (aq) + Br- (aq) ® H2O (aq) + OBr- (aq) 
2H2O2 (aq) + Br- (aq) ® 2H2O (aq) + OBr- (aq) + O2 (g) 
2H2O2 (aq) ® 2H2O (aq) + O2 (g) 
2H2O2 (aq) + Br- (aq) ® 2H2O (aq) + Br- (aq) + O2 (g

 19 The rate law for this reaction is:

Rate = k[H2O2][OBr-] 
Rate = k[H2O2]2
Rate = k[H2O2]2[Br-] 
Rate = k[H2O2][Br-] 
Rate = k[H2O2][OBr-]2

 20 Which of the following are reaction intermediates:

H2O 
Br-
OBr-
None are reaction intermediates 
Both (B) and (C) are reaction intermediates.

 21 Which of the following are catalysts:

H2O 
Br-
OBr-
None are catalysts. 
Both Br- and OBr- are catalysts.

 22 For the following reaction DG° is 2.60 kJ/mole at 25° C. The equilibrium constant for this reaction at 25° C is:
H2 (g) + I2 (g) « 2HI (g)

2.86 
0.999 
0.350 
1.05 x 10-3
1.05

 23 Reaction rates increase with temperature because as the temperature increases:

the equilibrium constant increases. 
the activation energy increases. 
the activation energy decreases. 
the rate constant increases. 
the rate constant decreases.

 24 A catalyst increases the rate of a reaction by:

increasing the temperature. 
decreasing the temperature. 
increasing the activation energy. 
decreasing the activation energy. 
decreasing DH. 

 25 The half life for the first order decomposition of nitromethane, CH3NO2, at 500K is 650 seconds. If the initial concentration of CH3NO2 is 0.500M, what will its concentration be (M) after 1300 seconds have elapsed?

0.1250.1400.2500.425

 26 Determine the rate law for the reaction, 2ICl + H2 ® I2 + 2HCl ,    from the following initial rate data:
 
[ICl]0 [H2]0 Initial Rate(Ms-1)
0.250 0.500 2.04x 10-2
0.500 0.500 4.08x 10-2
0.125 0.125 2.55 x 10-3
0.125 0.250 5.09 x 10-3

R = k[ICl]2
R = k[H2]2
R = k[ICl][H2]2
R = k[ICl][H2] 
R = k[ICl]2[H2]

Use the following data for the gas phase decomposition of 
hydrogen iodide to answer questions 27 through 29.


 
t, hours 0 2.0 4.0 6.0
[HI], M 1.00 0.50 0.33 0.25

 27 What is the order of the reaction in HI?

one halfminus onezerofirstsecond

 28 What is the rate constant, k (include units)?

0.25 h-1
0.25 Mh-1
0.50 h-1
0.50 M-1h-1
2.0 M-1h-1

 29 What is the average rate of the reaction, in moles HI/liter hour, over the first two hours?

0.250.350.502.04.0

Use the following information to answer questions 30 through 33

Hydrogen peroxide in basic solution oxidizes iodide ion to iodine. 
The proposed mechanism for the reaction is: 

        step 1.... H2O2 + I- ® HOI + OH- ......slow 
        step 2.... HOI + I- ® I2 + OH- ......fast :
 30 The equation for the overall reaction is: 

2I-® I2
H2O2 + I- ® HOI + OH-
HOI + I- ® I2 + OH-
H2O2® 2OH-
H2O2 + 2I- ® I2 + 2OH-

 31 The catalyst is:

I-
HOI 
OH-
I2
there is none

 32 The reaction intermediate is:

I-
HOI 
OH-
I2
there is none

 33 The rate law consistent with the mechanism is:

R = k[H2O2][I-]2
R = k[H2O2][I-] 
R = k[HOI][I-] 
R = k[I-]2
R = k[I2][OH-]2

 34 A certain reaction has the rate equation , R = k[A][B]2. The rate is 2.5 x 10-4 Ms-1 when [A] is 0.20 M and [B] is 0.050 M. Calculate the numerical value of the specific rate constant.

0.0250.1250.500.0500.20

Question 35 is based on the following information.

The following initial rates of the reaction were measured at various initial concentrations as 
recorded in the table below for:    CH3COCH3 + Br2® CH3COCH2Br + H+ + Br-


 

Experiment
[CH3COCH3]0
[Br2]0
[H+]0		 Initial Rate 
(M s-1)
1. 0.30 0.050 0.050 5.7 x 10-5
2. 0.30 0.10 0.050 5.7 x 10-5
3. 0.30 0.050 0.10 1.2 x 10-4
4. 0.40 0.050 0.20 3.1 x 10-4
5. 0.40 0.050 0.050 7.6 x 10-5

 35 The rate law (or rate equation) is: (R = Rate)

R = k[CH3COCH3][Br2] 
R = k[CH3COCH3][H+] 
R = k[CH3COCH3][Br2]2
R = k[CH3COCH3]2[Br2]2[H+] 
R = [CH3COCH3][H+]2

Questions 36 and 37 are based on the following information.

For a first order reaction the following data are obtained from experiment by measuring 
the concentration of N2O5 as a function of time, at constant temperature. 
The reaction is:     N2O5 (soln) ®  2NO2 (soln) + 1/2O2 (g)


 
time (seconds) 0 50 100 200 300 400
[N2O5] 0.100 0.0707 0.0500 0.0250 0.0125 0.00625

 36 Determine the value of the rate constant, k (in sec-1).

6.93 x 10-3
0.0707
0.0500
0.600
200

 37 Which of the following plots would be linear if the above data is plotted?

[N2O5] V time (in sec.) 
1/[N2O5] V time (in sec.) 
ln[N2O5] V time (in sec.) 
ln[N2O5] V 1/T(K) 
average rate V [N2O5]2

Question 38 is based on the following.

The rate constant, k for the decomposition of acetaldehyde 
CH3CHO (g) ® CH4 (g)+ CO (g) 
is measured at six different temperatures. The natural logarithm of k is plotted on the vertical axis (y-axis) and the reciprocal of the absolute temperature (1/T) is plotted on the horizontal axis (x-axis). The result is a linear plot with a slope equal to -2.20 x 104 K and a y-intercept equal to 27.0.

 38  The activation energy (Ea), in kJ/mole, is:

8.314
183
27.0
298
22.4

 39 The age of the Shroud of Turin in determined by radiocarbon dating. A Geiger counter measures 14.2 counts per minute (c.p.m.) in a sample of the shroud compared to 15.4 c.p.m. from carbon-14 in a living organism. If the half life is 5760 years, what is the age of the shroud in years? The radioactive decay of carbon-14 follows first order kinetics. 

1.20 x 10-4
674
5760
330
2880

Questions 40 and 41 are based on the following graph which represents experimental data obtained from a reaction of the type 

A ® products.

 40 What is the order with respect to A?

firstsecondzero1/2-1

 41 What is the numerical value of the rate constant, k?

100200.0100.0201.00

Questions 42 through 45 are based on the following mechanism for the destruction of ozone (O3) in the upper atmosphere.

        step 1..... O3(g) + NO(g) ® NO2(g) + O2(g) ......Slow 
        step 2...... NO2(g) + O(g) ® NO(g) + O2(g) ......Fast
 42 The net or overall reaction is:

O3(g) + NO(g) ® NO2(g) + O2(g) 
NO2(g) + O(g) ® NO(g) + O2(g) 
O3(g) + O2(g) ® O5(g) 
O3(g) + O(g) ® 2O2(g) 
2O2(g) ® O3(g) + O(g) 

 43 The catalyst in the above mechanism is:

O3
NO 
NO2
O2
O

 44 The intermediate in the above mechanism is:

O3
NO 
NO2
O2
O

 45 If the catalyst in question 42 is added to the reactants, the value of the rate constant, k, would be:

larger than that of the uncatalyzed reaction 
smaller than that of the uncatalyzed reaction 
the same as that of the uncatalyzed reaction 
impossible to determine

Question 46 is based on the following information. 

Chlorine Dioxide is a reddish-yellow gas that is soluble in water. 
In basic solution it reacts according to the following equation: 
2ClO2 (aq) + 2OH- (aq) ® ClO3- (aq) + ClO2- (aq) + H2O
The following initial rates of the reaction were measured at various initial concentrations 
as recorded in the table below.

Experiment [ClO2]0 [OH- ]0 Initial Rate (M s-1)
1 0.060 0.030 0.02480
2 0.020 0.030 0.00276
3 0.020 0.090 0.00828

 46 The rate law (or rate equation) is:

R = k[ClO2][OH-] 
R = k[ClO2]2[OH-] 
R = k[ClO2][OH-]2
R = k[ClO2]2[OH-]2
R = [ClO2][OH-]-1

Questions 47 and 48 are based on the following information. 

For a first order reaction the following data are obtained from experiment 
by measuring the concentration as a function of time. The reaction is of the type:   A ® products.

time (seconds) 0 70 140
[A] 1.000 0.600 ?

 47 Determine the molar concentration of A after 140 seconds have elapsed.

0.5000.3000.3600.2000.000

 48 Determine the value of the rate constant, k (in sec-1)

7.30 x 10-3
140
0.693
0.600
70

Question 49 and 50 are based on the following.

  • The rate constant, k, is measured at six different temperatures. 
  • The natural logarithm of k is plotted on the vertical axis (y-axis) and the reciprocal of the absolute temperature (1/T) is plotted on the horizontal axis (x-axis). 
  • The result is a linear plot with a slope equal to -6000 K and a y-intercept equal to 15.0
 49 The activation energy (Ea), in kJ/mole is:

8.3146.0049.915.022.4

 50 If the rate constant is 1.40 x 10-3 M-1s-1 at 276 K, to what temperature must the reaction be raised (in K) in order to double the rate of the reaction?

957010287552