The rate law for the reaction 2A → A2 is second order, with the rate equal to k[A]2. For a second-order reaction, the integrated rate law is ln([A]/[A]0) = -kt. The half-life is independent of the initial concentration and is equal to 0.693/k.

1. TOPIC 7 - THE RATE OF CHEMICAL REACTIONS
Examples of Solved Problems
1. The rate of formation of C in the reaction 2A + 3B → C + 4D is 2.7 mol dm-3 s-1. State
the rates of formation and consumption of the participants.
Answer:
d[C]
The rate of formation of C = = 2.7 mol dm −3 s −1
dt
1 d [ A] 1 d [ B] d [C ] 1 d [ D]
Rate = − =− = =
2 dt 3 dt dt 4 dt
The rate of consumption of A
d [ A] d [C ]
− =2 = 2(2.7 mol dm −3 s −1 ) = 5.4 mol dm −3 s −1
dt dt
d [ B]
− = 3(2.7 mol dm −3 s −1 ) = 8.1 mol dm −3 s −1
dt
The rate of formation of D,
d [ D]
= 4(2.7mol dm −3 s −1 ) = 10.8 mol dm −3 s −1
dt
2. The rate law for reaction A + 2B → 3C + 4D was reported as
d[C]/dt= kr[A][B][C]. Express the rate law in terms of the reaction rate; what is
the units for kr in each case?
Answer:
d [ A] 1 d [ B] 1 d [C ] 1 d [ D]
Rate = − =− = = = k r [ A][ B][C ]
dt 2 dt 3 dt 4 dt
Unit of k r ;
Rate mol dm −3 s −1
kr = = −3 3
= mol − 2 dm 6 s −1
[ A][ B][C ] (mol dm )
3. At 518 oC, the rate of decomposition of a sample of gaseous acetaldehyde,
initially at a pressure of 363 Torr, was 1.07 Torr s-1 when 5.0 percent had reacted
and 0.76 Torr s-1 when 20.0 per cent had reacted. Determine the order of the
reaction.
Answer:
Rate ∝ [ A]n ∝ (Pr essure A) n rate = k ( Po − x) n
rate1 1.07 Torr s −1 k ( Po − 5%) n k ( Po − 0.05Po ) n
= = =
rate 2 0.76 Torr s −1 k ( Po − 20%) n k ( Po − 0.20 Po ) n
n
 0.95  n
1.4079 =   = (1.1875) log1.4079 = n log1.1875
 0.80 
log1.4079 0.1486
n= = = 1.99 ≈ 2.00 second order reaction
log1.1875 0.0746
27

2. 4. At 518 oC, the half-life for the decomposition of a sample of gaseous
acetaldehyde, initially at 363 Torr was 410 s. When the pressure was 169 Torr,
the half-life was 880 s. Determine the order of the reaction.
Answer:
1 1
Half − life, t1 / 2 ∝ n −1
∝ n −1
∝ Po1−n
[ A]o P0
t1 / 2 (1) 410 s C (363)1− n
= = 0.4659 = 1− n
= (2.1479)1−n
t1 / 2 (2) 880 s C (169)
log 0.4659 = (1 − n) log 2.1479
log 0.4659 − 0.3317
(1 − n) = = = −0.999
log 2.1479 0.3320
n = 1 + 0.999 = 1.999 ≈ 2.00 second order reaction
5. A second-order reaction of the type A + B →P was carried out in a solution that
was initially 0.075 mol dm-3 in A and 0.060 mol dm-3 in B. After 1.0 h the
concentration of A had fallen to 0.020 mol dm-3. Calculate the rate constant.
For sec ond − order reaction type : A +B → P,
[ A] [ A]o − x [ A]o
The intergrated rate law : ln = ln = [ A]o − [ B]o kt + ln
[ B] [ B]o − x [ B]o
x = [ A]o − [ A] = [ B]o − [ B] = 0.075 − 0.020 = 0.055
0.020 0.075
ln = (0.075 − 0.060)k (1 × 60 × 60s ) + ln
0.060 − 0.055 0.060
1.3863 − 0.2231
1.3863 = 66326k + 0.2231 k = = 1.754 × 10 −5 dm 3 mol −1 s −1
66326
6. A reaction 2B →P has a second–order rate law with k = 4.30×10-4 dm3 mol-1 s-1.
Calculate the time required for the concentration of B to change from 0.210 mol
dm-3 to 0.010 mol dm-3.
d [ B]
For sec ond − order reaction type : 2 B → P, The rate law : − = 2k [ B ] 2
dt
[ B ]t [B] t t
d [ B]
− ∫ 2 [B]
[ B ]o [ B ]
= − ∫ [ B] −2 d [ B] = ∫ 2k dt
0
o
[ B ]t
− 2 +1
 [ B]0 2+1 
−
[ B]t−1 [ B]01−
1 1
− − −  =− + = − = 2kt − 2k (0 )
− 2 +1  − 2 +1 −1 −1 [ B ]t [ B ] o
1 1
−3
− −3
= (100 − 4.762) dm 3 mol −1 = 8.60 × 10 − 4 dm 3 mol −1 s −1t
0.010 mol dm 0.210 mol dm
t = 110742 s = 1.11 × 10 5 s
28

3. 7. Hydrogen peroxide reacts with thiosulfate ion in slightly acidic solution as
follows: H2O2 + 2S2O32- +2H+ → 2H2O + S4O62-
This reaction is second order, independent of the hydrogen ion concentration in
the pH range 4 to 6. The rate constant, k = 9.1×10-2 mol-1L min-1 were obtained at
25 o C and pH 5.0, and initial concentrations: [H2O2] = 0.036 mol L-1, [ S 2O3 − ] =
3
0.024 mol L-1. How long will it take for 50% of the thiosulfate ion, S 2O3 − to
3
react?
Answer:
d [ A]
For second − order reaction type : aA + bB → P, The rate law : − = k[ A][ B]
dt
([ A]o − ax) [ A]o
The interagrated rate equation : ln = (b[ A]o − a[ B]o )kt + ln
([ B]o − bx) [ B ]o
2−
[ A] = [ H 2 O2 ], [ B] = S 2 O3 , a = 1,.b = 2
bx = 2 x = 50%[ B]o , x = 0.25 × 0.024 = 6.0 × 10 −3
(0.036 − 6.0 × 10 −3 ) 0.036
ln −3
= (2(0.036) − 1(0.024))(9.1 × 10 − 2 )t + ln
(0.024 − 2(6.0 × 10 ) 0.024
0.9163 = 4.68 × 10 −3 t + 0.405 t = 117.06 s
Exercise 7a
1. If the reaction aA → Products is xth -order, where x = 0, x = 1, x = 2 and x = n (any
integer ≠1). Write the rate law, derive the integrated rate law, and also derive the
expression for half-life in terms of a, k, n and [A]o for each order.
2. The gas-phase formation of fosgene, CO + Cl2 → COCl2, is ½ order with respect to
CO. Derive the integrated rate equation for ½-order reaction and derive the
expression for the half-life.
3. The rate of the reaction A + 2 B → 3 C + D was reported as 1.0 mol dm−3 s−1. Which
of the following states of the rates is NOT TRUE.
A. –d[A]/dt =1.0 mol dm−3 s−1 B. –d[B]/dt =2.0 mol dm−3 s−1
−3 −1
C.-d[C]/dt=3.0 mol dm s D.- d[D]/dt= -1.0 mol dm−3 s−1
4. The rate law for the reaction A + 3 B → C + 2 D was found to be v = k[A][B].
Which of the following state(s) is/are true.
I The unit of the rate of reaction is mol-1 L s-1 II The unit of k is mol-1L s-1
III The reaction is second order reaction IV Rate = − d [C ] / dt = k[ A][ B ]
A. I and III B. II and III C. I, II , and III D. IV only
5. Nitrogen pentoxide gas decomposes according to: 2N2O5 (g) →4 NO2(g) + O2(g) .
At 328 K the rate of reaction v under certain conditions is 3.38 × 10−5 s−1 . Assuming
that none of the intermediates have appreciable concentrations. Which values is true.
A. -d[N2O5]/dt=3.38 × 10−5 s−1 B. –d[NO2]/dt=3.38 × 10−5 s−1
−5 −1
C.-d[O2]/dt= 3.38 × 10 s D. d[O2]/dt=3.38 × 10−5 s−1
29

4. 6. In studying the decomposition of ozone: 2O3(g) = 3O2(g)
in a 2-L reaction vessel, it is found that d[O3]/dt = -1.5×10-2 mol L-1 s-1. What is the
rate of reaction in mol L-1 s-1?
A.7.5×10−3 B.1.5×10-2 C. -1.5×10-2 D. - 7.5×10−3
7. Under certain conditions, it is found that ammonia is formed from its elements at a
rate of 0.10 mol L-1s-1. N2(g) + 3H2(g) = 2NH3(g) . Calculate the value of d[H2]/dt?
A. 0.10 mol L-1s-1 B. 0.20 mol L-1s-1 C. 0.30 mol L-1s-1 D. -0 .30 mol L-1s-1
8. The decomposition of N2O5: 2N2O5 = 4NO2 + O2 is studied by measuring the
concentration of O2 as a function of time, and it is found that
d[O2]/dt = (1.5 ×10-4 s-1)[N2O5] at constant temperature and pressure. Under these
conditions the reaction goes to completion to the right. Calculate the half-life of the
reaction under these conditions?
A. 4621s B. 2310 s C. 1155 s D. 577 s
9. At 400 K, the t1/2 for the decomposition of a sample of a gaseous compound initially at
55.5 kPa was 340 s. When the P was 28.9 kPa, the t1/2 was 178 s. The order of the
reaction is
A. n = 0 B. n =1 C. n = 2 D. n = 3
10. The composition of HI to H2 + I2 at 508 oC has a t1/2 of 135 min when the initial
pressure of HI is 0.1 atm and 13.5 min when the pressure is 1 atm. What is the order
of the reaction?
A. n = 0 B. n =1 C. n = 2 D. n = 3
11. The rate constant for the first-order decomposition of a compound A in the reaction
2A → P is k = 2.78 × 10−7 s−1at 25°C. Calculate the half-life of A?
A. 2.78 × 10−7 s−1 B. 1.25 ×10−6 s−1 C. 2.5 × 10 4 s D. 1.25 × 106 s−1
12. The first-order reaction 2A→ 2B + C is 35% complete after 325 s. How long will it
take for the reaction to be 70% complete?
A. 325 s B. 650 s C. 908 s D. 6.63 ×104 s
13. The t1/2 of a first-order reaction A → P is 12 min. What % of A remains after 50 min?
A. 5.6 % B. 6.5 % C. 9.08 % D. 16.63 %
14. The decomposition of ammonia at 1000 oC was a zero-order reaction with rate constant,
k = 4.30 mol dm-3s-1.What is the t1/2 for this reaction in which the initial concentration,
[NH3]o=0.0560 M?
A. 77 s B. 4.15 s C. 1.30×10−2s D. 1.61×10−1s
15. A solution of A is mixed with an equal volume of a solution of B containing the same
number of moles, and the reaction A + B = C occurs. At the end of 30 min. A is 75%
reacted. How much of A be left at the end of 50 min if the reaction is zero order in
both A and B?
A. 85% B. 60% C. 40% D. 15%
30

5. Exercise 7b
1. The rate law for the reaction 2A →A2 was found to be r = k [A]2. Which of the
following state(s) is / are FALSE.
A The unit of the rate of reaction is mol L-1 s-1 C. The unit of k is mol L-1 s-1
B The reaction is second order reaction D. - d [ A2 ] / dt = k A [ A] 2
2. A solution of A is mixed with an equal volume of a solution of B containing the same
number of moles, and the reaction A + B = C occurs. If the reaction is first order in
both A and B, what is the rate law?
A. Rate = k[A][B] C. d[B]/dt = k([A]o-x)([B]o-x)
B. d[C]/dt = k[A]2 = k[B]2 D. -d[B]/dt = k([A]o-x)([B]o-x)
3. The half-life, t1/2 of a reaction is the time required for half of the reactant to disappear.
Which of the following state(s) is / are true?
A. For a zero-order reaction, t1/2 is dependent of the initial concentration.
B. For a first-order reaction, t1/2 is independent of the initial concentration.
C. For a 2rd-order reaction, t1/2 is inversely proportional to the initial concentration
D. All of the above are true.
4. A dimer is formed in the solution reaction 2A →A2. The rate law is r = 2k[A]2,where k
= 0.015 M-1 s-1. What is the half-life of A when [A]o = 0.05 M?
A.1333 s B.667 s C. 1.5 s D. 1.5×10−2s
5. A reaction 2 A → P has a rate law with k = 3.50 ×10−2 dm3 mol−1 s−1. Calculate the
time required for the concentration of A to change from 0.26 mol L−1 to 0.11 mol L−1
A. 75 s B. 150 s C. 750 s D. 1333 s
6. A gas reaction 2A → B is second order in A and goes to completion in a reaction
vessel of constant volume and temperature with a half-life of 600 s. If the initial
pressure of A is 1 bar, what are the partial pressure of A at 800 s?
A. 0.328 bar B. 0.430 bar C. 0.504 bar D. 0.845 bar
7. A solution of A is mixed with an equal volume of a solution of B containing the same
number of moles, and the reaction A + B = C occurs. At the end of 30 min A is 65%
reacted. How much of A be left at the end of 50 min if the reaction is first order in
both A and B?
A. 84.8 % B. 75.6 % C. 48.8% D. 24.4%
8. Equal molar quantities of A and B are added to a liter of a suitable solvent. At the end
of 500 s one-half of A has reacted according to the reaction A + B = C. How much of
A will be reacted at the end of 800 s if the reaction is second order with respect to A
and zero order with respect to B?
A. 61.5 % B. 57.6 % C. 38.5% D. 26.0%
9. The second-order rate constant for an alkaline hydrolysis of ethyl formate in 85%
ethanol (aqueous) at 29.86 o C is 5.43 L mol-1s-1. If the reactants are both present at
0.01 mol L-1, calculate the t1/2.
A.0.13 s B. 5.43 s C. 18.42 s D. 85 s
31

6. 10. A second-order reaction of the type A + B → P was carried out in a solution that
was initially 0.050 mol dm−3 in A and 0.080 mol dm−3 in B. After1.0 h the
concentration of A had fallen to 0.020 mol dm−3. Calculate the rate constant?
A. 0.00413 mol-1Ls-1 B.0.0 413mol-1Ls-1 C. 0.248 mol-1 Ls-1 D. None of the above
11. A second-order reaction of the type A + 2 B → P was carried out in a solution that
was initially 0.75 mol dm−3 in A and 0.80 mol dm−3 in B. After1.0 h the concentration
of A had fallen to 0.65 mol dm−3. Calculate the rate constant.
A. 8.03×10-4 mol-1Ls-1 B. 5.74×10-5 mol-1Ls-1 C. 1.34×10-5 mol-1Ls-1 D. None of the above
12. The second-order rate constant for the reaction 2A + B → C is 0.21 dm3 mol−1 s−1.
What is the concentration of C after 10s when the reactants are mixed with initial
concentrations of [A] = 0.15 mol dm−3 and [B] = 0.25 mol dm−3?
A.0.184 mol dm−3 B. 0.046 mol dm−3 C. 0.021 mol dm−3 D. 0.07 mol dm−3
13. The reaction CH3CH2NO2 + OH- → H2O + CH3CHNO − 2
is second order, and k at 20o C is 9.1×10-2 mol-1 L min-1. An aqueous solution is
0.006 molar in nitroethane and 0.007 molar in NaOH. How long will it take for
10% of the nitroethane to react?
A. 1.44×10-3 s B. 15.82 s C. 15.82 min D. 9.1×102 min
14. The second-order rate constant for the reaction:
CH3COOC2H5 (aq) + OH− (aq) → CH3CO2− (aq) + CH3CH2OH(aq)
is 0.11 dm3 mol−1 s−1. What is the concentration of ester (ROOR) after 10 s when
ethyl acetate is added to NaOH so that the initial concentrations are [NaOH] = 0.050
mol dm−3 and [CH3COOC2H5] = 0.100 mol dm−3?
A. 0.100 mol dm−3 B. 0.095 mol L-1 C. 0.050 mol dm−3 D. 0.045 mol L-1
15. The rate constant, k, for second-order decomposition of HBr at 1000 K is 6.2×10-3
dm3mol-1 s-1. In a reaction in which the initial concentration, [HBr]o is 0.65 mol dm-3,
at what time [HBr]= 0.12 mol dm-3 ?
A. 18 .27 s B. 109 s C. 182 s D. 1.1×103 s
16. The following table gives kinetic data the following reaction at 25 oC: A+ B + C = D
[A] /molL-1 [B] /molL-1 [C] /molL-1 d[D]/dt /10-4mol L-1s-1
0.01 0.005 4.00 1.75
0.01 0.004 4.00 1.75
0.01 0.003 8.00 3.50
0.001 0.002 4.00 0.55
What is the rate law for the reaction?
A. r = k[ A]1 / 2 [C ] B. r = k[ A]1 / 2 [ B]−1 [C ] C. r = k[ A][C ] D. r = k[ A][ B ][C ]
17. A second-order reaction of the type A + B → P , the reaction is 60 % complete in
60 seconds when [A]o= 0.1 M and [B]o= 0.5 M. What is the rate constant for this
reaction?
A. 0.0041 mol-1Ls-1 B. 0.041mol-1s-1 C. 0.248 mol-1 L s-1 D. None of the above
32