#### Page No 98:

#### Question 4.1:

For the reaction R → P, the concentration of a reactant changes from 0.03 M to 0.02 M in 25 minutes. Calculate the average rate of reaction using units of time both in minutes and seconds.

#### Answer:

Average rate of reaction

= 6.67 × 10^{−6} M s^{−1}

#### Question 4.2:

In a
reaction, 2A →
Products, the concentration of A decreases from 0.5 mol L^{−1} to 0.4 mol L^{−1} in 10 minutes. Calculate the rate during this interval?

#### Answer:

Average rate

=
0.005 mol L^{−1 }min^{−1}

= 5 **×** 10^{−3} M min^{−1}

#### Page No 103:

#### Question 4.3:

For a reaction, A + B → Product; the rate law is given by,. What is the order of the reaction?

#### Answer:

The order of the reaction

= 2.5

#### Question 4.4:

The conversion of molecules X to Y follows second order kinetics. If concentration of X is increased to three times how will it affect the rate of formation of Y?

#### Answer:

The reaction X → Y follows second order kinetics.

Therefore, the rate equation for this reaction will be:

Rate = *k*[X]^{2} (1)

Let [X] = *a* mol L^{−1}, then equation (1) can
be written as:

Rate_{1} = *k* .(*a*)^{2}

= *ka*^{2}

If the concentration of X is increased to three times, then [X] =
3*a* mol L^{−1}

Now, the rate equation will be:

Rate = *k* (3*a*)^{2}

= 9(*ka*^{2})

Hence, the rate of formation will increase by 9 times.

#### Page No 111:

#### Question 4.5:

A
first order reaction has a rate constant 1.15 10^{−3} s^{−1}.
How long will 5 g of this reactant take to reduce to 3 g?

#### Answer:

From the question, we can write down the following information:

Initial amount = 5 g

Final concentration = 3 g

Rate
constant = 1.15 10^{−3} s^{−1}

We
know that for a 1^{st} order reaction,

= 444.38 s

= 444 s (approx)

#### Question 4.6:

Time
required to decompose SO_{2}Cl_{2} to half of its initial amount is 60 minutes. If the decomposition is
a first order reaction, calculate the rate constant of the reaction.

#### Answer:

We
know that for a 1^{st} order reaction,

It
is given that t_{1/2} = 60 min

#### Page No 116:

#### Question 4.7:

What will be the effect of temperature on rate constant?

#### Answer:

The rate constant of a reaction is nearly doubled with a 10° rise in temperature. However, the exact dependence of the rate of a chemical reaction on temperature is given by Arrhenius equation,

Where,

A is the Arrhenius factor or the frequency factor

*T* is the temperature

*R* is the gas constant

*E*_{a} is the activation energy

#### Question 4.8:

The rate of the chemical reaction doubles for an increase of 10 K
in absolute temperature from 298 K. Calculate *E*_{a}.

#### Answer:

It is given that *T*_{1} = 298 K

∴*T*_{2} = (298 + 10) K

= 308 K

We also know that the rate of the reaction doubles when temperature is increased by 10°.

Therefore, let us take the value of *k*_{1} = *k* and that of *k*_{2} = 2*k*

Also, *R* = 8.314 J K^{−1} mol^{−1}

Now, substituting these values in the equation:

We get:

= 52897.78 J mol^{−1}

= 52.9 kJ mol^{−1}

**Note:** There is
a slight variation in this answer and the one given in the NCERT
textbook.

#### Question 4.9:

The activation energy for the reaction

2HI_{(}_{g}_{)} → H_{2} + I_{2}_{(}_{g}_{)}

is 209.5 kJ mol^{−1} at 581K. Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy?

#### Answer:

In the given case:

*E*_{a} = 209.5 kJ mol^{−1} = 209500 J mol^{−1}

*T* = 581 K

*R* = 8.314 JK^{−1} mol^{−1}

Now, the fraction of molecules of reactants having energy equal to or greater than activation energy is given as:

x=e-Ea/RT⇒In x=-EaRT⇒log x=-Ea2.303RT⇒log x=-209500 J mol-12.303×8.314 JK-1mol-1×581=-18.8323Now, x =Antilog -18.8323 = 1.471×10-19

#### Page No 117:

#### Question 4.1:

From the rate expression for the following reactions, determine their order of reaction and the dimensions of the rate constants.

**(i) ** 3 NO(g) →
N_{2}O_{ }(g) Rate
= *k*[NO]^{2}

**(ii) ** H_{2}O_{2 }(aq) + 3
I^{− }(aq) + 2
H^{+} → 2 H_{2}O
(l) + Rate = *k*[H_{2}O_{2}][I^{−}]

**(iii) ** CH_{3}CHO(g)
→ CH_{4}(g)
+ CO(g) Rate = *k *[CH_{3}CHO]^{3/2}

**(iv) ** C_{2}H_{5}Cl(g)
→ C_{2}H_{4}(g)
+ HCl(g) Rate = *k *[C_{2}H_{5}Cl]

#### Answer:

**(i)** Given
rate = *k *[NO]^{2}

Therefore, order of the reaction = 2

Dimension of

**(ii) **Given
rate = *k *[H_{2}O_{2}]
[I^{−}]

Therefore, order of the reaction = 2

Dimension of

**(iii)** Given
rate = *k *[CH_{3}CHO]^{3/2}

Therefore, order of reaction =

Dimension of

**(iv)** Given
rate = *k *[C_{2}H_{5}Cl]

Therefore, order of the reaction = 1

Dimension of

#### Question 4.2:

For the reaction:

2A
+ B → A_{2}B

the
rate = *k*[A][B]^{2} with *k* = 2.0 × 10^{−6} mol^{−2} L^{2} s^{−1}.
Calculate the initial rate of the reaction when [A] = 0.1 mol L^{−1},
[B] = 0.2 mol L^{−1}.
Calculate the rate of reaction after [A] is reduced to 0.06 mol L^{−1}.

#### Answer:

The initial rate of the reaction is

Rate
= *k *[A][B]^{2}

= (2.0 ×
10^{−6} mol^{−2} L^{2} s^{−1})
(0.1 mol L^{−1})
(0.2 mol L^{−1})^{2}

= 8.0 ×
10^{−9} mol^{−2} L^{2} s^{−1}

When
[A] is reduced from 0.1 mol L^{−1} to 0.06 mol^{−1},
the concentration of A reacted = (0.1 − 0.06) mol L^{−1 }= 0.04
mol L^{−1}

Therefore,
concentration of B reacted =
0.02 mol L^{−1}

Then,
concentration of B available, [B] = (0.2 − 0.02) mol L^{−1}

=
0.18 mol L^{−1}

After
[A] is reduced to 0.06 mol L^{−1},
the rate of the reaction is given by,

Rate
= *k *[A][B]^{2}

= (2.0 × 10^{−6} mol^{−2} L^{2} s^{−1})
(0.06 mol L^{−1})
(0.18 mol L^{−1})^{2}

= 3.89 mol L^{−1} s^{−1}

#### Question 4.3:

The
decomposition of NH_{3} on platinum surface is zero order reaction. What are the rates of
production of N_{2} and H_{2} if *k *=
2.5 × 10^{−4} mol^{−1} L s^{−1}?

#### Answer:

The
decomposition of NH_{3} on platinum surface is represented by the following equation.

Therefore,

However, it is given that the reaction is of zero order.

Therefore,

Therefore,
the rate of production of N_{2} is

And,
the rate of production of H_{2} is

=
7.5 × 10^{−4} mol L^{−1} s^{−1}

#### Question 4.4:

The
decomposition of dimethyl ether leads to the formation of CH_{4},
H_{2} and CO and the reaction rate is given by

Rate
= *k *[CH_{3}OCH_{3}]^{3/2}

The rate of reaction is followed by increase in pressure in a closed vessel, so the rate can also be expressed in terms of the partial pressure of dimethyl ether, i.e.,

If the pressure is measured in bar andtime in minutes, then what are the units of rate and rate constants?

#### Answer:

If pressure is measured in bar and time in minutes, then

Unit
of rate = bar min^{−1}

Therefore, unit of rate constants

#### Question 4.5:

Mention the factors that affect the rate of a chemical reaction.

#### Answer:

The factors that affect the rate of a reaction are as follows.

**(i)** Concentration
of reactants (pressure in case of gases)

**(ii)** Temperature

**(iii)** Presence
of a catalyst

#### Page No 118:

#### Question 4.6:

A reaction is second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant is

**(i)** doubled **(ii)** reduced to half?

#### Answer:

Letthe concentration of the reactant be [A] = *a*

Rate of reaction, R = *k *[A]^{2}

= *ka*^{2}

**(i)**If the concentration of the reactant is doubled, i.e. [A] = 2*a*, then the rate of the reaction would be

= 4*ka*^{2}

= 4 R

Therefore, the rate of the reaction would increase by 4 times.

**(ii)** If the concentration of the reactant is reduced to half, i.e. , then the rate of the reaction would be

Therefore, the rate of the reaction would be reduced to

#### Question 4.7:

What is the effect of temperature on the rate constant of a reaction? How can this temperature effect on rate constant be represented quantitatively?

#### Answer:

The rate constant is nearly doubled with a rise in temperature by 10° for a chemical reaction.

The temperature effect on the rate constant can be represented quantitatively by Arrhenius equation,

where, *k* is the rate constant,

A is the Arrhenius factor or the frequency factor,

R is the gas constant,

*T* is the temperature, and

*E*_{a} is the energy of activation for the reaction

#### Question 4.8:

In a pseudo first order hydrolysis of ester in water, the following results were obtained:

t/s | 0 | 30 | 60 | 90 |

[Ester]mol
L | 0.55 | 0.31 | 0.17 | 0.085 |

**(i) ** Calculate
the average rate of reaction between the time interval 30 to 60
seconds.

**(ii) ** Calculate
the pseudo first order rate constant for the hydrolysis of ester.

#### Answer:

**(i) **Average
rate of reaction between the time interval, 30 to 60 seconds,

=
4.67 × 10^{−3} mol L^{−1} s^{−1}

**(ii) **For
a pseudo first order reaction,

For *t* = 30 s,

=
1.911 × 10^{−2} s^{−1}

For *t* = 60 s,

=
1.957 × 10^{−2} s^{−1}

For *t* = 90 s,

=
2.075 × 10^{−2} s^{−1}

Then, average rate constant,

#### Question 4.9:

A reaction is first order in A and second order in B.

**(i) **Write the
differential rate equation.

**(ii) **How is
the rate affected on increasing the concentration of B three times?

**(iii) ** How is
the rate affected when the concentrations of both A and B are
doubled?

#### Answer:

**(i) **The
differential rate equation will be

**(ii) **If
the concentration of B is increased three times, then

Therefore, the rate of reaction will increase 9 times.

**(iii) **When
the concentrations of both A and B are doubled,

Therefore, the rate of reaction will increase 8 times.

#### Question 4.10:

In
a reaction between A and B, the initial rate of reaction (r_{0})
was measured for different initial concentrations of A and B as given
below:

A/ mol L

^{−1}0.20

0.20

0.40

B/ mol L

^{−1}0.30

0.10

0.05

r

_{0}/ mol L^{−1}s^{−1}5.07 × 10

^{−5}5.07 × 10

^{−5}1.43 × 10

^{−4}

What is the order of the reaction with respect to A and B?

#### Answer:

Let
the order of the reaction with respect to A be *x* and with respect to B be *y*.

Therefore,

Dividing equation (i) by (ii), we obtain

Dividing equation (iii) by (ii), we obtain

= 1.496

= 1.5 (approximately)

Hence, the order of the reaction with respect to A is 1.5 and with respect to B is zero.

#### Question 4.11:

The following results have been obtained during the kinetic studies of the reaction:

2A + B → C + D

Experiment

A/ mol L

^{−1}B/ mol L

^{−1}Initial rate of formation of D/mol L

^{−1}min^{−1}I

0.1

0.1

6.0 × 10

^{−3}II

0.3

0.2

7.2 × 10

^{−2}III

0.3

0.4

2.88 × 10

^{−1}IV

0.4

0.1

2.40 × 10

^{−2}

Determine the rate law and the rate constant for the reaction.

#### Answer:

Let
the order of the reaction with respect to A be *x* and with respect to B be *y*.

Therefore, rate of the reaction is given by,

According to the question,

Dividing equation (iv) by (i), we obtain

Dividing equation (iii) by (ii), we obtain

Therefore, the rate law is

Rate
= *k *[A]
[B]^{2}

From experiment I, we obtain

=
6.0 L^{2} mol^{−2} min^{−1}

From experiment II, we obtain

=
6.0 L^{2} mol^{−2} min^{−1}

From experiment III, we obtain

=
6.0 L^{2} mol^{−2} min^{−1}

From experiment IV, we obtain

=
6.0 L^{2} mol^{−2} min^{−1}

Therefore,
rate constant, *k* = 6.0 L^{2} mol^{−2} min^{−1}

#### Question 4.12:

The reaction between A and B is first order with respect to A and zero order with respect to B. Fill in the blanks in the following table:

Experiment

A/ mol L

^{−1}B/ mol L

^{−1}Initial rate/mol L

^{−1}min^{−1}I

0.1

0.1

2.0 × 10

^{−2}II

—

0.2

4.0 × 10

^{−2}III

0.4

0.4

—

IV

—

0.2

2.0 × 10

^{−2}

#### Answer:

The given reaction is of the first order with respect to A and of zero order with respect to B.

Therefore, the rate of the reaction is given by,

Rate
= *k *[A]^{1} [B]^{0}

⇒
Rate = *k *[A]

From experiment I, we obtain

2.0
× 10^{−2} mol L^{−1} min^{−1} = k (0.1 mol L^{−1})

⇒ *k* = 0.2 min^{−1}

From experiment II, we obtain

4.0
× 10^{−2} mol L^{−1} min^{−1} = 0.2 min^{−1} [A]

⇒
[A] = 0.2 mol L^{−1}

From experiment III, we obtain

Rate
= 0.2 min^{−1} × 0.4 mol L^{−1}

= 0.08 mol L^{−1} min^{−1}

From experiment IV, we obtain

2.0
× 10^{−2} mol L^{−1} min^{−1} = 0.2 min^{−1} [A]

⇒
[A] = 0.1 mol L^{−1}

#### Page No 119:

#### Question 4.13:

Calculate the half-life of a first order reaction from their rate constants given below:

**(i)** 200 s^{−1} **(ii)** 2 min^{−1} **(iii)** 4 years^{−1}

#### Answer:

**(i)** Half life,

= 3.47

×10 ^{-3} s (approximately)

**(ii)** Half life,

= 0.35 min (approximately)

**(iii)** Half life,

= 0.173 years (approximately)

#### Question 4.14:

The
half-life for radioactive decay of ^{14}C
is 5730 years. An archaeological artifact containing wood had only
80% of the ^{14}C
found in a living tree. Estimate the age of the sample.

#### Answer:

Here,

It is known that,

= 1845 years (approximately)

Hence, the age of the sample is 1845 years.

#### Question 4.15:

The
experimental data for decomposition of N_{2}O_{5}

in gas phase at 318K are given below:

*t*(s)0 400 800 1200 1600 2000 2400 2800 3200 1.63 1.36 1.14 0.93 0.78 0.64 0.53 0.43 0.35

**(i) **Plot
[N_{2}O_{5}]
against *t*.

**(ii) ** Find the half-life period for the
reaction.

**(iii)** Draw
a graph between log [N_{2}O_{5}]
and *t.*

**(iv) ** What is the rate law?

**(v) ** Calculate the rate constant.

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

#### Answer:

**(ii) **Time
corresponding to the concentration, _{ }is the half life. From the graph, the
half life is obtained as 1450 s.

**(iii) **

**t(s)**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

**(iv) **The given
reaction is of the first order as the plot, v/s *t*, is
a straight line. Therefore, the rate law of the reaction is

**(v) **From
the plot, v/s *t*, we
obtain

Again,
slope of the line of the plot v/s *t* is
given by

.

Therefore, we obtain,

**(vi) **Half-life
is given by,

This value, 1438 s, is very close to the value that was obtained from the graph.

#### Question 4.16:

The
rate constant for a first order reaction is 60 s^{−1}.
How much time will it take to reduce the initial concentration of the
reactant to its 1/16^{th} value?

#### Answer:

It is known that,

Hence,
the required time is 4.6 × 10^{−2} s.

#### Question 4.17:

During
nuclear explosion, one of the products is ^{90}Sr
with half-life of 28.1 years. If 1μg of ^{90}Sr
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.

#### Answer:

Here,

It is known that,

Therefore,
0.7814 μg
of ^{90}Sr
will remain after 10 years.

Again,

Therefore,
0.2278 μg
of ^{90}Sr
will remain after 60 years.

#### Question 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.

#### Answer:

For a first order reaction, the time required for 99% completion is

For a first order reaction, the time required for 90% completion is

Therefore, *t*_{1} = 2*t*_{2}

Hence, the time required for 99% completion of a first order reaction is twice the time required for the completion of 90% of the reaction.

#### Question 4.19:

A
first order reaction takes 40 min for 30% decomposition. Calculate *t*_{1/2}.

#### Answer:

For a first order reaction,

Therefore, *t*_{1/2} of the decomposition reaction is

= 77.7 min (approximately)

#### Question 4.20:

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

*t*(sec)P(mm of Hg)

0

35.0

360

54.0

720

63.0

Calculate the rate constant.

#### Answer:

The decomposition of azoisopropane to hexane and nitrogen at 543 K is represented by the following equation.

After time, *t*, total pressure,

= 2P_{0} − P_{t}

For a first order reaction,

When *t* = 360 s,

= 2.175 × 10^{−3} s^{−1}

When *t* = 720 s,

= 2.235 × 10^{−3} s^{−1}

Hence, the average value of rate constant is

= 2.21 × 10^{−3} s^{−1}

**Note:** There is
a slight variation in this answer and the one given in the NCERT
textbook.

#### Question 4.21:

The
following data were obtained during the first order thermal
decomposition of SO_{2}Cl_{2} at a constant volume.

**Experiment****Time/s**^{−1}**Total pressure/atm**1

0

0.5

2

100

0.6

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

#### Answer:

The
thermal decomposition of SO_{2}Cl_{2} at a constant volume is represented by the following equation.

After
time, *t*,
total pressure,

Therefore,

=
2 P_{0} −
P_{t}

For a first order reaction,

When *t* = 100 s,

= 2.231 × 10^{−3} s^{−1}

When
P_{t} = 0.65 atm,

P_{0} + *p* = 0.65

⇒ *p* = 0.65 − P_{0}

= 0.65 − 0.5

= 0.15 atm

Therefore,
when the total pressure is 0.65 atm, pressure of SOCl_{2} is

=
P_{0} −
p

= 0.5 − 0.15

= 0.35 atm

Therefore, the rate of equation, when total pressure is 0.65 atm, is given by,

Rate
= *k*()

=
(2.23 × 10^{−3} s^{−1})
(0.35 atm)

=
7.8 × 10^{−4} atm s^{−1}

#### Page No 120:

#### Question 4.22:

The
rate constant for the decomposition of N_{2}O_{5} at
various temperatures is given below:

*T*/°C0 20 40 60 80 0.0787 1.70 25.7 178 2140

Draw
a graph between ln *k *and 1/*T *and
calculate the values of *A *and *E*_{a}.

Predict the rate constant at 30º and 50ºC.

#### Answer:

From the given data, we obtain

*T*/°C0

20

40

60

80

*T*/K273

293

313

333

353

3.66×10

^{−3}3.41×10

^{−3}3.19×10

^{−3}3.0×10

^{−3}2.83 ×10

^{−3}0.0787

1.70

25.7

178

2140

ln

*k*−7.147

− 4.075

−1.359

−0.577

3.063

Slope of the line,

According to Arrhenius equation,

Again,

When ,

Then,

Again, when ,

Then, at ,

#### Question 4.23:

The
rate constant for the decomposition of hydrocarbons is 2.418 ×
10^{−5 }s^{−1} at 546 K. If the energy of activation is 179.9 kJ/mol, what will be
the value of pre-exponential factor.

#### Answer:

*k* = 2.418 × 10^{−5} s^{−1}

*T* = 546 K

*E*_{a} = 179.9 kJ mol^{−1} = 179.9 × 10^{3} J mol^{−1}

According to the Arrhenius equation,

= (0.3835 − 5) + 17.2082

= 12.5917

Therefore, A = antilog (12.5917)

= 3.9 ×
10^{12} s^{−1} (approximately)

#### Question 4.24:

Consider
a certain reaction A → Products with *k *= 2.0 ×
10^{−2 }s^{−1}.
Calculate the concentration of *A *remaining
after 100 s if the initial concentration of *A *is 1.0 mol
L^{−1}.

#### Answer:

*k* = 2.0 × 10^{−2} s^{−1}

*T* = 100 s

[A]_{o} = 1.0 moL^{−1}

Since
the unit of *k* is s^{−1},
the given reaction is a first order reaction.

Therefore,

=
0.135 mol L^{−1} (approximately)

Hence,
the remaining concentration of A is 0.135 mol L^{−1}.

#### Question 4.25:

Sucrose
decomposes in acid solution into glucose and fructose according to
the first order rate law, with *t*_{1/2 }= 3.00
hours. What fraction of sample of sucrose remains after 8 hours?

#### Answer:

For a first order reaction,

It
is given that, *t*_{1/2} = 3.00 hours

Therefore,

=
0.231 h^{−1}

Then,
0.231 h^{−1}

Hence, the fraction of sample of sucrose that remains after 8 hours is 0.158.

#### Question 4.26:

The decomposition of hydrocarbon follows the equation

*k *= (4.5 ×
10^{11 }s^{−1})
e^{−28000 }^{K}^{/}^{T}

Calculate *E*_{a}.

#### Answer:

The given equation is

*k *= (4.5 ×
10^{11 }s^{−1})
e^{−28000 }^{K}^{/}^{T} (i)

Arrhenius equation is given by,

(ii)

From equation (i) and (ii), we obtain

= 8.314 J
K^{−1} mol^{−1} × 28000 K

= 232792 J
mol^{−1}

= 232.792 kJ
mol^{−1}

#### Question 4.27:

The
rate constant for the first order decomposition of H_{2}O_{2} is given by the following equation:

log *k *=
14.34 − 1.25 × 10^{4 }K/*T*

Calculate *E*_{a} for this reaction and at what temperature will its half-period be 256
minutes?

#### Answer:

Arrhenius equation is given by,

The given equation is

From equation (i) and (ii), we obtain

= 1.25 ×
10^{4} K × 2.303 × 8.314 J K^{−1} mol^{−1}

= 239339.3 J
mol^{−}1
(approximately)

= 239.34 kJ
mol^{−1}

Also,
when *t*_{1/2} = 256 minutes,

=
2.707 × 10^{−3} min^{−1}

=
4.51 × 10^{−5} s^{−1}

It
is also given that, log *k* = 14.34 − 1.25 × 10^{4} K/*T*

= 668.95 K

= 669 K (approximately)

#### Question 4.28:

The decomposition of A into product has value of *k *as 4.5 ×
10^{3} s^{−1} at 10°C and energy of
activation 60 kJ mol^{−1}. At what temperature would *k *be 1.5 × 10^{4} s^{−1}?

#### Answer:

From Arrhenius equation, we obtain

Also, *k*_{1} = 4.5 × 10^{3} s^{−1}

*T*_{1} = 273 + 10 = 283 K

*k*_{2} = 1.5 × 10^{4} s^{−1}

*E*_{a} = 60 kJ mol^{−1} = 6.0 ×
10^{4} J mol^{−1}

Then,

= 297 K

= 24°C

Hence, *k* would be 1.5 × 10^{4} s^{−1} at 24°C.

**Note:** There is
a slight variation in this answer and the one given in the NCERT
textbook.

#### Question 4.29:

The time required for 10% completion of a first order reaction at 298 K is

equal
to that required for its 25% completion at 308 K. If the value of *A *is

4 ×
10^{10 }s^{−1}.
Calculate *k *at 318 K and *E*_{a}.

#### Answer:

For a first order reaction,

At 298 K,

At 308 K,

According to the question,

From Arrhenius equation, we obtain

To
calculate *k *at 318 K,

It is given that,

Again, from Arrhenius equation, we obtain

#### Question 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.

#### Answer:

From Arrhenius equation, we obtain

Hence,
the required energy of activation is 52.86 kJmol^{−1}.

**NCERT Solutions for Class 12 Chemistry Chapters**

**Part I**

- Chapter 1 – The Solid State
- Chapter 2 – Solutions
- Chapter 3 – Electrochemistry
- Chapter 4 – Chemical Kinetics
- Chapter 5 – Surface Chemistry
- Chapter 6 – General Principles and Processes of Isolation of Elements
- Chapter 7 – The p-Block Elements
- Chapter 8 – The d-and f-Block Elements
- Chapter 9 – Coordination Compounds

**Part II**

- Chapter 1 – Haloalkanes and Haloarenes
- Chapter 2 – Alcohols, Phenols and Ethers
- Chapter 3 – Aldehydes Ketones and Carboxylic Acids
- Chapter 4 – Amines
- Chapter 5 – Biomolecules
- Chapter 6 – Polymers
- Chapter 7 – Chemistry in Everyday Life

**NCERT Solutions for Class 12:**