Chapter 5: Chemical Kinetics and Catalysis

Problem: 1, Page no: 149

In [2]:
# Variables
K = 3.5 * 10 ** - 2             # Rate constant

# Solution
print "First order reaction = 0.693 / K"
t_05 = 0.693 / K
print "Time taken for half the initial concentration to react", t_05, "minutes"

First order reaction = 0.693 / K
Time taken for half the initial concentration to react 19.8 minutes

Problem: 2, Page no: 150

In [5]:
# Variables
t05 = 40                    # minutes

# Solution
print "Rate constant = 0.693 / t05"
K = 0.693 / t05
print "Rate constant", "{:.4f}".format(K), "/ min"

Rate constant = 0.693 / t05
Rate constant 0.0173 / min

Problem: 3, Page no: 150

In [1]:
from math import log10

# Variables
t0 = 37.0                   # cm^3 of KMnO4
t5 = 29.8                   # cm^3 of KMnO4
t15 = 19.6                  # cm^3 of KMnO4
t25 = 12.3                  # cm^3 of KMnO4
t45 = 5.00                  # cm^3 of KMnO4

# Solution
K5 = 2.303 / 5 * log10(t0 / t5)
K15 = 2.303 / 15 * log10(t0 / t15)
K25 = 2.303 / 25 * log10(t0 / t25)
K45 = 2.303 / 45 * log10(t0 / t45)

print "At t = 5 min, K =", "{:.3e}".format(K5), "/min"
print "At t = 15 min, K =", "{:.3e}".format(K15), "/min"
print "At t = 25 min, K =", "{:.3e}".format(K25), "/min"
print "At t = 45 min, K =", "{:.3e}".format(K45), "/min"
print "As the different values of K are nearly same, so the reaction",
print "is of first-order"
K = (K45 + K25 + K5 + K15) / 4
print "The average value of K = ", "{:.3e}".format(K), "/min"

At t = 5 min, K = 4.329e-02 /min
At t = 15 min, K = 4.237e-02 /min
At t = 25 min, K = 4.406e-02 /min
At t = 45 min, K = 4.449e-02 /min
As the different values of K are nearly same, so the reaction is of first-order
The average value of K =  4.355e-02 /min

Problem : 4, Page no: 150

In [13]:
# Variables
t = 60                  # min
x = "0.5a"
K = 5.2 * 10 ** - 3     # / mol L / min

# Solution
a = 1 / (t * K)
print "Initial concentration", "{:.3f}".format(a), "mol / L"

Initial concentration 3.205 mol / L

Problem: 5, Page no: 151

In [18]:
from math import log10

# Solution
print "99.9 % / 50 % =",
times = round((2.303 * log10(100 / (100 - 99.9))) / (2.303 * log10(100 / (100 - 50))))
print times

99.9 % / 50 % = 10.0

Problem: 6, Page no: 151

In [22]:
from math import log10

# Constants
R = 1.987               # cal / K / mol

# Variables
T1 = 273.0              # K
T2 = 303.0              # K
K1 = 2.45 * 10 ** -5
K2 = 162 * 10 ** -5

# Solution
Ea = log10(K2 / K1) * R * 2.303 / ((T2 - T1) / (T1 * T2))
print "The activation energy of the reaction is", int(Ea), "cal / mol"

The activation energy of the reaction is 22968 cal / mol

Problem: 7, Page no: 151

In [23]:
# Variables
t05 = 30                # minutes
a = 0.1                 # M

# Solution
print "For second order reaction,"
print "t0.5 = 1 / Ka"
K = 1 / (a * t05)
print "The rate constant is", "{:.3f}".format(K), "mol / lit / min"

For second order reaction,
t0.5 = 1 / Ka
The rate constant is 0.333 mol / lit / min

Problem: 8, Page no: 151

In [4]:
from math import log

# Variables
T = 500                     # C
Pi = 350                    # torr
r1 = 1.07                   # torr / s
r2 = 0.76                   # torr / s

# Solution
print "1.07 = k(0.95a)^n"
print "0.76 = k(0.80a)^n"
n = log(r1 / r2) / log(0.95 / 0.80)
print "Hence, order of reaction is", round(n)

1.07 = k(0.95a)^n
0.76 = k(0.80a)^n
Hence, order of reaction is 2.0

Problem: 9, Page no: 152

In [5]:
# Solution
print "Applying steady state approximation to the concentration of NOCl2,",
print "we get"
print "d[NOCl2] / dt = 0 = k1 [NO] [Cl2] - k-1 [NOCl2] - k2 [NO] [NOCl2]"
print "[NOCl2] = k1 [NO] [Cl2] / (k-1 + k2 [NO])"
print "Now, the overall rate of reaction,"
print "d[NOCl2] / dt = k2 [NO] [NOCl2]"
print "From the above equations we get,"
print "(k1 k2 / k-1) * [NO]^2 * [Cl2], if k2[NO] << k-1"
print "k [NO]^2[Cl2], where k = k1 * k2 / k-1"

Applying steady state approximation to the concentration of NOCl2, we get
d[NOCl2] / dt = 0 = k1 [NO] [Cl2] - k-1 [NOCl2] - k2 [NO] [NOCl2]
[NOCl2] = k1 [NO] [Cl2] / (k-1 + k2 [NO])
Now, the overall rate of reaction,
d[NOCl2] / dt = k2 [NO] [NOCl2]
From the above equations we get,
(k1 k2 / k-1) * [NO]^2 * [Cl2], if k2[NO] << k-1
k [NO]^2[Cl2], where k = k1 * k2 / k-1

Problem: 10, Page no: 153

In [2]:
from math import log10

# Constant
R = 1.987               # cal / K / mol

# Variables
K2_K1 = 4               # factor increase
T1 = 27                 # C
T2 = 47                 # C

# Solution
T1 += 273.0
T2 += 273.0
Ea = log10(4) * 2.303 * R * (T1 * T2 / (T2 - T1))
print "The activation energy for the reaction is", "{:.2e}".format(Ea),
print "cal /mol"

The activation energy for the reaction is 1.32e+04 cal /mol

Problem: 11, Page no: 153

In [11]:
from math import log10

# Variables
a = 1                  # mole
x = 3 / 4.0            # reaction completed

# Solution
K = (2.303 / 6) * log10(1 / (1 - x))
print "The rate constant is", "{:.3f}".format(K), "/min"

The rate constant is 0.231 /min

Problem: 12, Page no: 153

In [15]:
from math import log10

# Solution
print "Let the initial concentration be 100, when x = 25",
print " t = 30 minutes"
a = 100
x = 25.0
t = 30          # minutes
K = 2.303 / t * log10(a / (a - x))
t05 = 0.683 / K
t = 2.303 / K * log10(a / x)
print "K =", "{:.2e}".format(K), "/ min"
print "T0.5 =", "{:.2f}".format(t05), "min"
print "t =", "{:.1f}".format(t), "min"

Let the initial concentration be 100, when x = 25  t = 30 minutes
K = 9.59e-03 / min
T0.5 = 71.21 min
t = 144.6 min