Multiple Choice Questions Kinetics (I):
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Which of the following following would NOT increase 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. 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
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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. 1.0x10-6 4.0x10-7 1.5x10-2 1.8x10-5
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1.2x10-3 The entropy of the surroundings surroundings goes up up when water vapor condenses. condenses. This is because:
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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. For the reaction reaction 2A + 3B + C = Products, Products, the rate equation is:
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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 For a certain first first order reaction, 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? 156 76 312 12.5 89.6 The following following initial rate-concentration rate-concentration data was collected collected for the the reaction 2A + B + 2C = D + E + F.
Initial Concentrations, M [A] [B] [C]
1.0 1.0
0.50 0.40
1.0 0.30 0.10 0.20 The rate equation for this reaction is:
Initial Rate, M/s [D]/t
0.40 0.40
1.8x10-4 1.8x10-4
0.80 0.40
9.0x10-5 1.8x10-5
R = k[A]2[B][C] 2 R = k[A][B]2[C] R =s k[A][C] R = k[A][C]-1 8
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R = k[B][C]2 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 2.15x10-4 7020 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 1.37x109 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 11 A catalyst functions by:
+4.42
-36200
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) + 2H 2 N2 (g) + 2H2O (g)
Rate = k[NO]2[H2]2/[N2][H2O]2 Rate = k[NO][H 2]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 NO is 0.75 M-1s-1. 2NO (g) 2 2
2NO (g) + O 2 (g) At 700 K the rate constant is 19.7 M -1s-1. The energy of activation for this reaction is _____ kJ/mole. 362 3.6 114 1.1 47 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 s 191 s 390 s 28 s 0.011 s 15 What is the half-life of phosphine (problem 14) at 680° C? 55 s 35 s 0.029 s 0.014 s 0.0056 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 H to answer this. Not enough information given. Must know E a to answer this. 17 The rate of the reaction :
BrO3- (aq) + 5Br - (aq) + 6H+ (aq) 3Br 2 (aq) + 3H2O was studied and the following information was obtained: Initial Rate{M/sec}
[BrO3-]0
[Br-]0
[H+]0
0.10 0.20
0.10 0.20
0.10 0.10
8.0x10-4 3.2x10-3
0.20 0.10 0.10 0.10 The rate law for this reaction is:
0.10 0.20
1.6x10-3 3.2x10-3
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) + O 2 (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 G° 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 H. 25 The half life for the first order decomposition of nitromethane, CH3 NO2, at 500K is 650 seconds. If the initial concentration of CH 3 NO2 is 0.500M, what will its concentration be (M) after 1300 seconds have elapsed? 0.125
0.140
0.250
0.425
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 0.125
0.500 0.500 0.125
2.04x10-2 4.08x10-2 2.55x10-3
0.125
0.250
5.09x10-3
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 [HI], M
0 1.00
2.0 0.50
4.0 0.33
27 What is the order of the reaction in HI?
one half minus one zero first 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
second
6.0 0.25
29 What is the average rate of the reaction, in moles HI/liter hour, over the first two hours?
0.25
0.35
0.50
2.0
4.0
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 + OHHOI + I- = I2 + OHH2O2 2OHH2O2 + 2I- I2 + 2OH31 The catalyst is:
IHOI OHI2 there is none 32 The reaction intermediate is: IHOI OHI2 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 2.5x10-4 Ms-1 when [A] is 0.20 M and [B] is 0.050 M. Calculate the numerical value of the specific rate constant.
0.025
0.125
0.50
0.050
0.20
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 + Br 2 CH3COCH2Br + H+ + Br -
Initial Rate Experiment
[CH3COCH3]0
[Br2]0
[H+]0
(M s-1)
1.
0.30 0.30 0.30
0.050 0.10 0.050
0.050 0.050 0.10
5.7x10-5 5.7x10-5 1.2x10-4
0.40 0.40
0.050 0.050
0.20 0.050
3.1x10-4 7.6x10-5
2. 3. 4. 5.
35 The rate law (or rate equation) is: (R = Rate)
R = k[CH3COCH3][Br 2] R = k[CH3COCH3][H+ ] R = k[CH3COCH3][Br 2]2 R = k[CH3COCH3]2[Br 2]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 N 2O5 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).
6.93x10-3 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 -2.20x10 4 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. 1.20x10-4 674 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?
first
second
zero
1/2
-1
41 What is the numerical value of the rate constant, k?
100
20
0.010
0.020
1.00
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 2 3
0.060 0.020 0.020
0.030 0.030 0.090
0.02480 0.00276 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. 0 1.000
time (seconds) [A]
70 0.600
140 ?
47 Determine the molar concentration of A after 140 seconds have elapsed.
0.500
0.300
0.360
0.200
0.000
48 Determine the value of the rate constant, k (in sec -1)
7.30x10-3 140 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:
8.314
6.00
49.9
15.0
22.4
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?
95
70
10
287
552
