**4.30** The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy of activation of the reaction assuming that it does not change with temperature.

From the Arrhenius equation,
...................................(i)
it is given that
T1= 293 K
T2 = 313 K
Putting all these values in equation (i) we get,
Activation Energy = 52.86 KJ/mol
This is the required activation energy

**4.29 ** The time required for 10% completion of a first order reaction at 298K is equal to that required for its 25% completion at 308K. If the value of A is . Calculate k at 318K and Ea.

We know that,
for a first order reaction-
Case 1
At temp. = 298 K
= 0.1054/k
Case 2
At temp = 308 K
= 2.2877/k'
As per the question
K'/K = 2.7296
From Arrhenius equation,
= 76640.096 J /mol
=76.64 KJ/mol
k at 318 K
we have , T =318K
A=
Now
After putting the calue of given variable, we get
on takingantilog we get,
k =...

**4.28 ** The decomposition of A into product has value of k as at 10°Cand energy of activation 60 kJ mol–1. At what temperature would k be ?

The decomposition of A into a product has a value of k as at 10°C and energy of activation 60 kJ mol–1.
K1 =
K2 =
= 60 kJ mol–1
K2 =

**4.27 ** The rate constant for the first order decomposition of is given by the following equation:

.

Calculate for this reaction and at what temperature will its half-period be 256 minutes?

The Arrhenius equation is given by
taking log on both sides,
....................(i)
given equation,
.....................(ii)
On comparing both equation we get,
activation energy
half life () = 256 min
k = 0.693/256
With the help of equation (ii),
T =
= 669 (approx)

**4.26** The decomposition of hydrocarbon follows the equation . Calculate

The Arrhenius equation is given by
.................................(i)
given equation,
............................(ii)
by comparing equation (i) & (ii) we get,
A= 4.51011 per sec
Activation energy = 28000 (R = 8.314)
= 232.792 KJ/mol

**4.25 ** Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law, with hours. What fraction of sample of sucrose remains after 8 hours ?

For first order reaction,
given that half life = 3 hrs ()
Therefore k = 0.693/half-life
= 0.231 per hour
Now,
= antilog (0.8024)
= 6.3445
(approx)
Therefore fraction of sample of sucrose remains after 8 hrs is 0.157

**4.24** Consider a certain reaction A Products with . Calculatethe concentration of A remaining after 100 s if the initial concentration of A is 1.0 mol

Given that,
k =
t = 100 s
Here the unit of k is in per sec, it means it is a first-order reaction.
therefore,
Hence the concentration of rest test sample is 0.135 mol/L

**4.23 ** The rate constant for the decomposition of hydrocarbons is at 546 K. If the energy of activation is 179.9 kJ/mol, what will be the value of pre-exponential factor.

Given that,
k =
= 179.9 KJ/mol
T(temp) = 546K
According to Arrhenius equation,
taking log on both sides,
= (0.3835 - 5) + 17.2082
= 12.5917
Thus A = antilog (12.5917)
A = 3.9 per sec (approx)

**4.22 ** The rate constant for the decomposition of N2O5 at various temperatures

is given below:

Draw a graph between ln k and 1/T and calculate the values of A and

. Predict the rate constant at 30° and 50°C.

From the above data,
T/
0
20
40
60
80
T/K
273
293
313
333
353
()
3.66
3.41
3.19
3.0
2.83
0.0787
1.70
25.7
178
2140
-7.147
-4.075
-1.359
-0.577
3.063
Slope of line =
According to Arrhenius equations,
Slope =
12.30 8.314
= 102.27
Again,
When T = 30 +273 = 303 K and 1/T =0.0033K
k =
When T = 50 + 273 = 323 K and 1/T = 3.1 K
k = 0.607 per sec

**4.21 ** The following data were obtained during the first order thermal decomposition of at a constant volume.

** **

Calculate the rate of the reaction when total pressure is 0.65 atm.

The thermal decomposition of is shown here;
After t time, the total pressure =
So,
thus,
for first order reaction,
now putting the values of pressures, when t = 100s
when
= 0.65 - 0.5
= 0.15 atm
So,
= 0.5 - 0.15
= 0.35 atm
Thus, rate of reaction, when the...

**4.20 ** For the decomposition of azoisopropane to hexane and nitrogen at 543K, the following data are obtained.

Calculate the rate constant.

Decompostion is represented by equation-
After t time, the total pressure =
So,
thus,
for first order reaction,
now putting the values of pressures,
when t =360sec
when t = 270sec
So,

**4.19** A first order reaction takes 40 min for 30% decomposition. Calculate

For the first-order reaction,
(30% already decomposed and remaining is 70%)
therefore half life = 0.693/k
=
= 77.7 (approx)

**4.18 ** For a first order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction.

**4.17 ** During nuclear explosion, one of the products is with half-life of 28.1 years. If of was absorbed in the bones of a newly born baby instead of calcium, how much of it will remain after 10 years and 60 years if it is not lost metabolically.

Given,
half life = 21.8 years
= 0.693/21.8
and,
by putting the value we get,
taking antilog on both sides,
[R] = antilog(-0.1071)
= 0.781
Thus 0.781 of will remain after given 10 years of time.
Again,
Thus 0.2278 of will remain after 60 years.

**4.16 ** The rate constant for a first order reaction is . How much time will it take to reduce the initial concentration of the reactant to its 1/16th value?

**4.15 (6) ** The experimental data for decomposition of in gas phase at 318K are given below:

Calculate the half-life period from k and compare it with (ii).

**4.15 (5) ** The experimental data for decomposition of in gas phase at 318K are given below:

Calculate the rate constant.

From the log graph,
the slope of the graph is =
= -k/2.303 ..(from log equation)
On comparing both the equation we get,

**4.15 (4) ** The experimental data for decomposition of in gas phase at 318K are given below:

What is the rate law ?

Here, the reaction is in first order reaction because its log graph is linear.
Thus rate law can be expessed as

**4.15 (3)** The experimental data for decomposition of in gas phase at 318K are given below:

Draw a graph between and t.

0
1.63
-1.79
400
1.36
-1.87
800
1.14
-1.94
1200
0.93
-2.03
1600
0.78
-2.11
2000
0.64
-2.19
2400
0.53
-2.28
2800
0.43
-2.37
3200
0.35
-2.46

**4.15 (2)** The experimental data for decomposition of in gas phase at 318K are given below:

Find the half-life period for the reaction.

Half life of the reaction is-

time corresponding to the mol/ L = 81.5 mol /L is the half life of the reaction. From the graph, the answer shoul be in the range of 1400 s to 1500 s.

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