CY 102: Assignment III(Chemical Kinetics)
8th March 2004
1. The Half of uranium –235 is 7.1 x 106 years while that of the much shorter-lived
isotope sodium-24 is only 15.0 hours. Calculate the first order rate constants for the
radioactive decay of these isotopes in appropriate, and identical, units.
Ans: Ku-235 = 0.693/2.24 x 1016s = 3.09 x 10-17 s-1
KNa – 24 = 0.693/5.4 x 104s = 1.28 x 10-5 s-1.
2. The pseudo-first-order rate constant for the decomposition of NO Cl (g) to NO and
Cl2 is measured and found to be 1.96 x 106 m3mol-1s-1. The actual rate limiting step is
Cl + NOCl → Cl2 + NO which has rate constant k2. The reaction is pseudo-first-order
in chlorine atoms because the concentration of chlorine atoms is so much less than the
concentration of NOCl.
a. Calculate the value of the rate constant k2, with units, when p(NOCl) =
10.0 mol/m3.
b. Calculate the value of the reaction rate, with units, when p(NOCl) = 10.0
mol/m3 and p(Cl) = 1.0x103 mol/m3.
Ans: Rate = k2 [Cl.] [NOCl]
k2 = 1.96 x 106 m3 mol-1s-1 x 10 mol m3
= 1.96 x 107 s-1 (apparent 1st order)
Rate = 1.96 x 106 x 103 x 10 = 1.96 x 104 mol m-3 s-1.
3. For N2(g) + 3H2(g) → 2NH3(g), determine the rate expression using the given data.
Find k with units.
[N2] (M)
0.10
0.10
0.20
[H2] (M)
0.10
0.20
0.40
Initial rate (M/min)
0.0021
0.0084
0.0672
Ans: Rate = k[N2]a [H2]b
(0.0084/0.0021) = (0.20/0.10)b
∴b = 2
(0.0672/0.0084) = (0.20/0.10)a (0.40/0.20)2
∴a = 1
Rate = k[N2] [H2]2
K = 2.1 r-2 min-1.
4. The halogen astatine can only be obtained artificially through bombardment. It has
been found to be useful for the treatment of certain types of cancer of the thyroid
gland. One form of radioactive astatine is a particle emitter with a half-life of 7.21 hr.
If a sample containing 0.100 mg of astatine is given to a person at 9 a.m. one morning,
how much astatine will remain after about 14 hours?
Ans: ln 2 = k t1/2
k = 0.0961 h-1
ln ([At]/0.100) = -0.0961 m-1 x 14 hr = -1.35
∴[At] = 0.026 mg remaining.
5. N2O5 gas decomposes according to the reaction 2N2O5 → 4NO2 + O2. The rate of
reaction at 328 K is found to be 0.75 x 10-4 mol L-1 s-1. What are the values of
d [ N 2O5 ] d [ NO2 ]
d [O2 ]
,
and
?
dt
dt
dt
Ans: 2N2O5 → 4 NO2 + O2
d[N2O5]/dt = -1.5 x 10-4 mol l-1 s-1.
d[NO2]/dt = 3 x 10-4 mol l-1 s-1.
d[O2]/dt = 0.75 x 10-4 mol l-1 s-1.
6. The decomposition of ozone 2O3(g) → 3O2(g) is studied in a 2 L reaction vessel and
d [O3 ]
found
= −1.5 × 10−2 molL−1s −1.
dt
(a) What is the rate of reaction?
(b) What is the rate of conversion?
d [O2 ]
(c) What is the value of
?
dt
Ans: (a) rate of reaction = 0.75 x 10-2 mol L-1s-1
(b) Rate of conversion = 1.5 x 10-2 mol L-1 s-1
d [O2 ]
(c)
= 2.25 x 10-2 mol L-1 s-1.
dt
7. A first order reaction has an activation energy of 104,600 J mol-1 and a preexponential factor A = 5 x 1013 s-1, at what temperature will the reaction have a halflife of 30 days?
Ans: ln2/t1/2 = A exp-Ea/RT
∴T = 269.5 k.
8. Find the order and the value of the rate constant from the following kinetic data
obtained for a certain reaction.
C(M)
t(s)
8
0
6
42
4.5 3.4
97 169
Ans: Data fits into second order expression,
1/[A] = 1/[A]0 + kt
k = 1.004 x 10-3 m-1 s-1.
9. For an nth order reaction nA → P which goes to completion, derive the integrated rate
law and half life expressions.
Ans: A → product nth order in A.
Rate = d[A]/dt = -k[A]n
∫12[A]-n d[A] = -k∫12dt
[A]-n+1 – [A]0-n+1 / -n+1 = -kt for n≠1
or [[A]/[A]0]1-n = 1 + [A]0n-1 (n-1) kt
and t1/2 = 2n-1 -1/(n-1) [A]0n-1 k
10. A reaction follows the rate law [ A] − [ A]0 = −
kt
. Find the order and units for k.
2
Ans: Substitute n = ½ in the solution of prob. 9
kt
Which gives [ A] − [ A]0 = −
2
1/2
-1/2
-1
Units for k = mol litre sec .
11. T- butyl bromide is converted into t-butyl alcohol in a solvent containing 90%
acetone and 10% water. The reaction is
(CH3)3CBr + H2O → (CH3)3COH + HBr
The data for the concentration of t-butyl bromide versus time is given below
T(MIN)
0
9
18
24
40
54
72
105
(CH3)3CBR (MOL L-1)
0.1056
0.0961
0.0856
0.0767
0.0645
0.0536
0.0432
0.0270
Find the order, rate constant and half-life of the reaction?
Ans: Data fits into 1st order integrated rate expression.
K = 0.0125 min-1
t1/2 = 55.5 min.
12. Write down the differential rate equations for the removal of A, B, C and D species in
the following reaction scheme:
k1
→C + D
A + B 
k2
→ A+ B
C + D 
k3
→E + D
C + B 
k4
→F
2 D 
Ans: -d[A]/dt = k1 [A][B] – k2 [C][D]
-d[B]/dt = k1[A][B] – k2[C][D] + k2[C][B]
-d[C]/dt = k2[C][D] + k3[C][B] – k1[A][B]
-d[D]/dt = k2[C][D] – k1[A][B] – k3[C][B] + 2k4[D]2
k1
k2
→ B and A 
→ C , find the half-life of A.
13. For a parallel first order reactions A 
Show that the observed activation energy for the disappearance of A is given by
Ea =
k1 E1 + k2 E2
k1 + k2
Ans: [A] = [A]0 e-(k1+k2)t
[A] = [A]0/2 at t1/2
∴2 = e(k1 + k2)t1/2
t1/2 = ln2/(k1 + k2)
d[C]/dt = (k1 + k2) [A]
ln k = ln A – Ea/RT
Ea = RT2 d ln k/dT = RT2 d ln (k1 + k2)/dT
∴Eohs = RT2/(k1+k2) (k1E1/RT2 + k2E2/RT2)
= k1E1 + k2E2/k1 + k2

→ B , obtain expressions for [A] and [B] at time t and
14. For a reversible reaction A ←

k1
k2
also for the time required to reach half of [A]eq and [B]eq.
k1

→ B initially only A is present. i.e [A] = [A]0 & [B] = 0 at t=0.
Ans: A ←

k2
d[A]/dt = k2[A]0 – (k1 + k2) [A]
= -(k1 + k2) ([A] – [A]eq)
∵ [A]eq = k2/k1+k2 [A]0
Integration
ln [A]0 – [A]eq/[A] – [A]eq = (k1+k2)t
∴ [A] = k2[A]0/k1+k2 (1+k1/k2 e-(k1+k2)t)
llly [B] = k1[A]0/k1+k2 (1-e-(k1+k2)t)
and time for half [A]eq or [B]eq = ln2/k1+k2
k1
k2
→ B 
→ C , write down the rate
15. For a first order consecutive reaction A 
expressions and obtain concentrations [A], [B] and [C] at time t in terms of [A]0
k1 and k2. Obtain the same for B and C under steady state conditions. If [A]0 = 100 M,
k1 = 0.04 min-1, calculate the concentrations of A, B and C at 5 min.
Ans: See notes/text for the 1st part.
[A] = 81.87
[B] = 17.85
[C] = 0.278
k3

→ C 
→ D , find rate expression for the formation of D
16. For a reaction A + B ←

k1
k2
applying steady state approximation. C is a reactive intermediate.
Ans: Using SSA
d[C]/dt = k1[A][B] – k2[C] – k3[C] = 0
[C] = k1[A][B]/k2+k3
d[D]/dt = k3[c] = k1k3[A][B]/k2+k3.
17. The following mechanism is suggested for the decomposition of N2O5
2 N 2O5 ( g ) → 4 NO2 ( g ) + O2 ( g ) :
k1
N 2O5 
→ NO2 + NO3
k2
NO2 + NO3 
→ N 2O5
k3
NO2 + NO3 
→ NO2 + O2 + NO
k4
NO + N 2O5 
→ 3 NO2
Find the rate law consistent with this mechanism.
Ans: d[NO]/dt = k1[NO2][NO3] – k4[NO][N2O5] = 0
d[NO3]/dt = k1[N2O5] – k2[NO2][NO3] – k3[NO2][NO3] = 0
d[N2O5]/dt = -k1[N2O5] + k2[NO2][NO3] – k4[NO][N2O5]
∴d[N2O5]/dt = -2k1k3[N2O5]/k2+k3
18. Pyrolysis (thermal decomposition in the absence of air) of acetaldehyde
∆
CH 3CHO( g ) 
→ CH 4 ( g ) + CO( g ) is a chain reaction. The proposed Rice Herzfeld
mechanism is given below:
ka
→⋅CH 3 + ⋅CHO
CH 3CHO 
initiation
CH 3CHO + ⋅CH 3 → CH 4 + CH 3CO ⋅
kb
CH 3CO ⋅ →⋅CH 3 + CO
kc
⋅CH 3 + ⋅CH 3 → CH 3CH 3
kd
Show that this mechanism
3
d [CH 4 ]
= k[CH 3CHO] 2 .
dt
is
propagation
propagation
termination
consistent
with
the
experimental
Ans: intermediates: CH3 & CH3CO
d[CH3]/dt = ka[CH3CHO] – kb[CH3][CH3CHO] + kc[CH3CO] – 2kd[CH3]2 = 0
SSA
d[CH3CO]/dt = kb [CH3] [CH3CHO] – kc[CH3CO] = 0
∴[CH3] = [ka/2kd]1/2 [CH3CHO]1/2
d[CH4]/dt = kb[CH3][CH3CHO]
= kb (ka/2ka)1/2 [CH3CHO]3/2
d[CH4]/dt = kobs[CH3CHO]3/2
rate
law