Class 12 - Chemistry - Chemical Kinetics

Question 4.1.

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

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

(ii) H_{2}O_{2} (aq) + 3I^{–} (aq) + 2H^{+} → 2H_{2}O (l) + I_{3}^{−} 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

K= (Rate)/ [NO]^{ 2}

Dimension of = (mol L^{-1} s^{-1})/ (mol L^{-1})^{2}

= (mol L^{-1} s^{-1})/ (mol^{2} L^{-2})

= L mol^{-1} s^{-1}

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

Therefore, order of the reaction = 2

k= (Rate)/ [H_{2}O_{2}] [I^{−}]

Dimension of = (mol L^{-1} s^{-1})/ ((mol L^{-1}) (mol L^{-1}))

= L mol^{-1} s^{-1}

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

Therefore, order of reaction = (3/2)

k= (Rate)/ [CH_{3}CHO]^{ (3/2)}

Dimension of = (mol L^{-1} s^{-1})/ (mol L^{-1})^{ (3/2)}

= (mol L^{-1} s^{-1})/ (mol ^{(3/2)} L ^{(3/2)})

=L^{ (1/2)} mol^{ (1/2)} s^{-1}

(iv) Given rate = k [C_{2}H_{5}Cl] Therefore, order of the reaction = 1

k= (Rate)/ [C_{2}H_{5}Cl]

Dimension of = (mol L^{-1} s^{-1})/ (mol L^{-1})

= s^{-1}

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 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 = (1/2) x (0.04) mol L^{-1} =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:

2NH_{3} (g) → N_{2} (g) + 3H_{2} (g)

Rate of zero order reaction is equal to rate constant

I.e. Rate = 2.5×10^{-4}molL^{-1}sec^{-1}.

According to rate law, - d [NH_{3}]/2dt = d [N_{2}]/ (dt)

2.5 × 10^{-4}mol L^{-1}sec^{-1} = d [N_{2}]/ (dt)

i.e. the rate of production of N_{2} is 2.5 × 10^{-4}mol L^{-1} sec^{-1}.

According to rate law,

- d [NH_{3}]/2dt = d [H_{2}]/ (3dt)

d [H_{2}]/(dt) =(-3) x d[NH_{3}]/[2dt]

i.e. rate of formation of H_{2} is 3 times rate of reaction

= 3 × 2.5 × 10^{-4}mol L^{-1}sec^{-1}

= 7.5 × 10^{-4}mol L^{-1}sec^{-1}

Rate of formation of N_{2} and H_{2} is 2.5 × 10^{-4} mol L^{-1}sec^{-1} and

7.5 × 10^{-4} mol L^{-1}sec^{-1} respectively

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.,

Rate = k (pCH_{3}OCH_{3})^{ (3/2)}

If the pressure is measured in bar and time 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}

Rate = k (pCH_{3}OCH_{3})^{ (3/2)}

- k= (Rate)/ (pCH
_{3}OCH_{3})^{ (3/2)}

Therefore, unit of rate constants (k) = (bar min^{-1})/ (bar)^{ (3/2)}

=bar^{-(1/2)} min^{-1}

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

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:

Let the 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] = 2a, then the rate of the reaction would be

R’ = k (2a)^{ 2}

= 4ka^{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., [A] = (1/2) a then the rate of the reaction would be

R’’ =k (1/2) (a)^{ 2}

= (1/4) ka

= (1/4) R

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,

K= A e^{-(Ea/RT)}

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 intervals 30 to 60 seconds.

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

Answer:

- Average rate of reaction over interval

= [change in concentration]/ [time taken] i.e.

= [0.31 -0.17]/ [60-30]

= (0.14)/ (30)

=0.00467 mol L^{-1} sec^{-1}

= 4.67 x 10^{-3} mol L^{-1} s^{-1}

- the pseudo first-order rate constant can be calculated by

K = (2.303/t) log (C_{i}/C_{t})

Where K = Rate constant

t =s time taken

C_{i} = initial concentration

C_{t} = Concentration at time t.

K_{1} = (2.303/30) log (0.55/0.31) [at t=30 sec]

= 1.911 x 10^{-2} s^{-1}

K_{2} = (2.303/60) log (0.55/0.17) [at t=60 sec]

= 1.957 x 10^{-2} s^{-1}

K_{3} = (2.303/90) log (0.55/0.31) [at t=90 sec]

=2.075 x 10^{-2} s^{-1}

Then, average rate constant k = (k_{1} +k_{2} +k_{3})/ (3) = 1.911 x 10^{-2} s^{-1}

= (1.911 x 10^{-2}) + (1.957 x 10^{-2}) + (2.075 x 10^{-2})

⇒ K = 1.98 × 10^{-2} sec^{-1}

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

- (d[R])/ (dt) =k ([A]^{ 1}[B]^{ 2}

(i) If conc. of B is increased to 3 times then:

- (d[R])/ (dt) =k ([A] ([3B])^{ 2}

- (d[R])/ (dt) =k ([A]^{ 9}[B]^{ 2}

=9 x k ([A]^{ 1}[B]^{ 2}

Therefore rate increases 9 times

(iii) When conc. of both A and B are doubled

- (d[R])/ (dt) =k [A] [B]^{ 2}

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

=8 k [A] [B]^{ 2}

Therefore, the rate of reaction increases by 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 |
0.20 |
0.20 |
0.40 |

B/mol L |
0.30 |
0.10 |
0.05 |

r |
5.07 x 10 |
5.07 x 10 |
1.43 x 10 |

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,

r_{0} = k [A]^{x} [B]^{y}

5.07×10^{−5} = k [0.20]^{ x} [0.30]^{ y} (1)

5.07×10^{−5} = k [0.20]^{ x} [0.10]^{ y} (2)

1.43×10^{−4} = k [0.40]^{ x} [0.05]^{ y} (3)

Dividing (i) and (ii)

(5.07×10^{−5}) / (5.07×10^{−5})

= ([0.20]^{ x} [0.30]^{ y})/ ([0.20]^{ x} [0.10]^{ y})

Or, 1= ([0.30]^{y}/[0.10]^{y})

Or, 1^{y} = 3^{y}

Using formula, 1=x^{0}

3^{0}=3^{y}

Or, y=0

Dividing (ii) and (iii)

(1.43×10^{−4}) / (5.07×10^{−5}) = ([0.40]^{ x} [0.05]^{ y})/ ([0.20]^{ x} [0.10]^{ y})

(1.43×10^{−4}) / (5.07×10^{−5}) = ([0.40]^{ x})/ ([0.20]^{ x}) (since y=0, [0.05]^{ x}= [0.30]^{ y}=1)

Or, 2.821= (2)^{ x}

Taking log both sides, log (2.821) =log (2)^{ x}

- log 2.821 = x log 2
- x=(log 2.821)/(log 2)
- =1.496
- 5 approx.

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 |
[B]/mol L |
Initial rate of formation of [D]/mol L |

I |
0.1 |
0.1 |
6.0 x 10 |

II |
0.3 |
0.2 |
7.2 x 10 |

III |
0.3 |
0.4 |
2.88 x 10 |

IV |
0.4 |
0.1 |
2.40 x 10 |

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,

Rate = k [A]^{x} [B]^{y}

According to the question,

6.0 x 10^{-3} = k [0.1]^{ x} [0.1]^{ y} (1)

7.2 x 10^{-2} = k [0.3]^{ x} [0.2]^{ y} (2)

= k [0.3]^{ x} [0.4]^{ y} (3)

2.40 x 10^{-2} =k [0.4]^{ x} [0.1]^{ y} (4)

Dividing equations (4) by (1),

(2.40 x 10^{-2}) / (6.0 x 10^{-3}) = (k [0.4]^{ x} [0.1]^{ y)}/ (k [0.1]^{ x} [0.1]^{ y})

4 = [0.4]^{x}/[0.1]^{x}

4 = (0.4/0.1)^{x}

(4)^{1} = 4^{x}

=> x=1

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

(2.88 x 10^{-1})/ (7.2 x 10^{-2}) = (k [0.3]^{ x} [0.4]^{ y)}/ (k [0.3]^{ x} [0.2]^{ y})

4= (0.4/0.2)^{y}

- 4 = 2y
- (2)
^{2}= 2^{y}

Or y=2

Therefore, the rate law is

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

k = (Rate)/ [A] [B]^{ 2}

From experiment I, we obtain

k= (6.0 x 10^{-3} molL^{-1}min^{-1})/ (0.1molL^{-1}) (0.1mol L^{-1})^{2}

= 6.0 L^{2} mol^{−2} min^{−1}

From experiment II, we obtain

k= (7.2 x 10^{-2} molL^{-1}min^{-1})/ (0.3molL^{-1}) (0.2mol L^{-1})^{2}

=6.0 L^{2} mol^{−2} min^{−1}

= 6.0 L2 mol−2 min−1

From experiment III, we obtain

k= (2.88 x 10^{-1} molL^{-1}min^{-1})/ (0.3molL^{-1}) (0.4mol L^{-1})^{2}

= 6.0 L^{2} mol^{−2} min^{−1}

From experiment IV, we obtain

k= (2.40 x 10^{-2} molL^{-1}min^{-1})/ (0.4molL^{-1}) (0.1mol L^{-1})^{2}

= 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 |
[B]/mol L |
Initial rate of formation of [D]/mol L |

I |
0.1 |
0.1 |
2.0 x 10 |

II |
- |
0.2 |
4.0 x 10 |

III |
0.4 |
0.4 |
- |

IV |
- |
0.2 |
2.0 x 10 |

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}

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, t_{ (½)} = (0.693 / k)

= (0.693 / 200) s^{-1}

= 3.47 ×10^{-3} s (approximately)

(ii) Half-life, t_{ (½)} = (0.693 / k)

= (0.693 / 2) min^{-1}

= 0.35 min (approximately)

(iii) Half-life, t_{ (½)} = (0.693 / k)

= (0.693 / 4) years^{-1}

= 0.173 years (approximately)

Question 4.14.

The half-life for radioactive decay of 14C is 5730 years.

An archaeological artifact containing wood had only 80% of the 14C found in a living tree. Estimate the age of the sample?

Answer:

k= (0.693)/ (t_{ (1/2)})

= (0.693)/ (5730) years^{-1}

Also, t = (2.303)/ (k) log[R]_{o}/[R]

= (2.303)/ ((0.693/5730)) log (100/80)

=1845 years (approx.)

Hence, the age of the sample is 1845 years.

Question 4.15.

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

[2N_{2}O_{5} → 4NO_{2} + O_{2}] in gas phase at 318K are given below:

t/s |
0 |
400 |
800 |
1200 |
1600 |
2000 |
2400 |
2800 |
3200 |

10 |
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:

(i)

(ii)

t/s |
0 |
400 |
800 |
1200 |
1600 |
2000 |
2400 |
2800 |
3200 |

10 |
1.63 |
1.36 |
1.14 |
0.93 |
0.78 |
0.64 |
0.53 |
0.43 |
0.35 |

log [N |
-1.79 |
-1.87 |
-1.94 |
-2.03 |
-2.11 |
-2.19 |
-2.28 |
-2.37 |
-2.46 |

- Time corresponding to the concentration (1.603 x 10
^{2})/ (2) molL^{-1}, is the half-life. - From the graph, the half-life is obtained as 1450 s.
- The given reaction is of the first order as the plot, log [N
_{2}O_{5}] v/s t, is a straight line.

Therefore, the rate law of the reaction is

Rate =k [N_{2}O_{5}]

- From the plot, log [N
_{2}O_{5}]

Slope = (-2.46)-(-1.79)/(3200-0)

=-(0.67/3200)

Again, slope of the line of the plot log [N_{2}O_{5}] v/s t is given by

(-k/2.303)

Therefore,

(-k/2.303) = - (0.67/3200)

=> k= 4.82 x 10^{-4} s^{-1}

^{ }

- Half-life is given by,

t_{ (1/2)} = (0.693)/(k)

= (0.693)/ (4.82 x 10^{-4}) s

=1.438 x 10^{3} sec

= 1438 sec

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,

t= (2.303/k) log[R]_{ 0}/[R]

= (2.303)/ (60 s^{-1}) log ((1)/ (1/16))

= (2.303)/ (60 s^{-1}) log (16)

=4.6 x 10-2 sec (approx.)

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

Question 4.17.

During nuclear explosion, one of the products is 90Sr with half-life of 28.1 years.

If 1μg of 90Sr 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:

Initial concentration, [R]_{ 0} = 1μg

Final concentration= [R]

Half-life t_{ (1/2)} = 28.1 years

We know, t_{ (1/2)} = (0.693/k) Where, k = rate constant

⇒ k = (0.693/ t_{ (1/2)})

Or k = (0.693/28.1 years)

k=0.0246 years^{-1}

Also, t = (2.303/k) log[R]_{o}/[R]

If t = 10yrs, then, using the formula, we get,

t = (2.303/k) log[R]_{o}/[R]

10 = (2.303)/ (0.0246) log (1/[R])

10 = (2.303)/ (0.0246) log (- [R])

log [R] = - (10 x 0.693)/(2.303 x 28.1)

[R] = antilog (-0.1071)

=0.7814 μg

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

Again,

t = (2.303/k) log[R]_{o}/[R]

60 = (2.303)/ (0.0246) log (1/[R])

60 = (2.303)/ (0.0246) log (1/ [R])

log [R] = - (60 x 0.693)/(2.303 x 28.1)

[R] = antilog (-0.6425)

=0.2278 μg

Therefore, 0.2278 μg of 90Sr 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

t_{1}= (2.303/k) log [(100)/ (100-99)]

= (2.303/k) log (100)

= 2 x (2.303/k)

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

T_{2}= (2.303/k) log [(100)/ (100-90)]

= (2.303/k) log (10)

= (2.303/k)

Therefore, t_{1} = 2t_{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,

t= (2.303/k) log [R]_{ 0}/[R]

k= (2.303/40 min) log [(100)/ (100-30)]

= (2.303/40 min) log (10/7)

=8.918 x 10^{-3} min^{-1}

Therefore t_{ (1/2)} of the decomposition reaction is

t_{ (1/2)} =(0.693/k)

= (0.6913)/ (8.918 x 10^{-3}) min

=77.7 approx.

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.

(CH_{3})_{2}CHN=NCH (CH_{3})_{2(g)} à N_{2 (g)} + C_{6}H_{14 (g)}

At t=0 P_{0} 0 0

At t= t (P_{0} –p) 0 0

When time t = t, the total partial pressure is P_{t }= P_{0} + p

(P_{0}-p) = (P_{t}-2p), but by the above equation, we know p = (P_{t}-P_{0})

Hence, (P_{0}-p) = Pt-2(Pt-P_{0})

Thus, (P_{0}-p) = 2P_{0} – P_{t}

We know that time t= (2.303/k) log[R]_{ 0}/[R]

Where, k=rate constant

[R]_{0} =Initial concentration of reactant

[R]=Concentration of reactant at time‘t’

Here concentration can be replaced by the corresponding partial pressures.

Hence, the equation becomes, t= (2.303/k) log [(P_{0})/ (P_{0} – p)] (equation 1)

At time t = 360 s, P_{t} = 54 mm of Hg and P_{0} = 30 mm of Hg,

Substituting in equation 1,

⇒ 360 = (2.303/k) log [(30)/ (2x30) – 54)]

⇒ k = 2.175 × 10^{-3} s^{-1}

At time t = 720 s, P_{t} = 63 mm of Hg and P_{0} = 30 mm of Hg,

Substituting in equation 1,

720 = (2.303/k) log [(30)/ (2x30) – 63)]

Thus, k= (2.235 x 10^{-3} s^{-1 }+ 2.175 x 10^{-3} s^{-1})/ (2)

Therefore k = 2.21 x 10^{-3} s^{-1}

Question 4.21.

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

SO_{2}Cl_{2} (g) -->SO_{2} (g) + Cl_{2 (g)}

Experiment |
Time/s |
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:

When t = 0, the total partial pressure is P_{0} = 0.5 atm

SO_{2}Cl_{2} (g) --> SO_{2} (g) + Cl_{2 (g)}

At t=0 P_{0} 0 0

At t= t (P_{0} –p) 0 0

When time t = t, the total partial pressure is P_{t} = (P_{0} + p)

(P_{0}-p) = (P_{t}-2p), but by the above equation, we know p = (P_{t}-P_{0})

Hence, (P_{0}-p) = (P_{t})-(2(P_{t}-P_{0}))

Thus, (P_{0}-p) = (2P_{0} – P_{t})

We know that time, t= (2.303)/ (k) log[R]_{ 0}/[R]

Where, k=rate constant

[R]_{°} =Initial concentration of reactant

[R]-Concentration of reactant at time‘t’

Here concentration can be replaced by the corresponding partial pressures.

Hence, the equation becomes, t= (2.303)/ (k) log (P_{o})/ (P_{0}-p)

⇒t= (2.303)/ (k) log (P_{o})/ (2P_{0}-P_{t}) (equation 1)

At time t = 100 s, P_{t} = 0.6 atm and P_{0} = 0.5 atm,

Substituting in equation 1,

100 = (2.303)/ (k) log (0.5)/ ((2 x 0.5) -0.6)

Thus, k = 2.231 × 10^{-3} s^{-1}

The rate of reaction R = (k × P P_{S02Cl2})

When total pressure P_{t} = 0.65 atm and P_{0} = 0.5 atm, then

P_{S02Cl2} = (2P_{0}-P_{t})

Thus, substituting the values, P_{S02Cl2} = (2(0.5)-0.6) = 0.35 atm

R =k × P_{S02Cl2 }= (2.231 × 10^{-3} s^{–1} × 0.35)

Rate of the reaction R = 7.8 × 10^{-4}atm s^{–1}

Question 4.22.

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

T/ |
0 |
20 |
40 |
60 |
80 |

10 |
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 30° and 50°C.

Answer:

T/ |
0 |
20 |
40 |
60 |
80 |

10 |
0.0787 |
1.70 |
25.7 |
178 |
2140 |

T (in Kelvin) |
(0+273) =273 |
(20 + 273)=293 |
(40 + 273)=313 |
(60 + 273) =333 |
(80 + 273)=353 |

(1/T) K |
3.66 x 10 |
3.41 x 10 |
3.19 x 10 |
3.0 x 10 |
2.83 x 10 |

ln K |
-7.147 |
-4.075 |
-1.359 |
-0.577 |
3.063 |

The graph is given as:

The Arrhenius equation is given by k = Ae^{-(Ea/RT)}

Where, k= Rate constant

A= Constant

E_{a}=Activation Energy

R= Gas constant

T=Temperature

Taking natural log on both sides,

ln k = ln A-(E_{a}/RT) (equation 1)

By plotting a graph, ln K Vs (1/T), we get y-intercept as ln A and Slope is – (E_{a}/R).

Slope = (y_{2}-y_{1})/(x_{2}-x_{1})

By substituting the values, slope = -12.301

⇒ – (E_{a}/R) = -12.301

But, R = 8.314 JK^{-1}mol^{-1}

⇒ E_{a} = (8.314 JK^{-1}mol^{-1} × 12.301 K)

⇒ E_{a} = 102.27 kJ mol^{-1}

Substituting the values in equation 1 for data at T = 273K

⇒ ln K = ln A – (E_{a}/RT)

=>-7.147 = ln A – (102.27 x 10^{3})/ (8.314 x 273)

=By taking (T=273 K, ln k =-7.147)

On solving, we get ln A = 37.911

Therefore, A = 2.91×10^{6}

When T = 300C, hence T = (30 + 273) = 303K

⇒ ln k = -2.8

⇒ k = 6.08x10^{-2}s^{-1}.

When T = 500C, hence T = (50 + 273) = 323K

Substituting in equation 1,

A = 2.91 × 10^{6}, E_{a} = 102.27 kJ mol^{-1}

Ln K = ln (2.91 × 10^{6}) – (102.27)/ (8.314 × 323)

Thus, ln k = -0.5 and k = 0.607s^{-1}.

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 × 103J mol^{-1}

According to the Arrhenius equation k = Ae^{-(E}_{a}^{/RT)}

Taking log on both sides,

log k =ln A – (E_{a})/(RT)

log k =ln A – (E_{a})/(2.303 RT)

log A = log k + (E_{a})/(2.303 RT)

=log (2.418 x 10^{-5} s^{-1}) + (179.9 x 103) Jmol^{-1}/ (2.303x 8.314 Jk^{-1}mol^{-1} x 546k)

= (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}

Time t = 100s

Concentration [A_{0}] = 1.0 mol L^{-1}

We know, t= (2.303)/ (k) log [A_{0}]/ [A]

On substituting the values, 100 = (2.303)/ (0.02) log [1]/ [A]

log (1/ [A]) = (2.303/2)

Log [A] = - (2.303/2)

[A] =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:

t_{ (1/2)} = 3.00 hours

We know, t_{ (1/2)} = (0.693/k)

Therefore, k = (0.693/3)

k = 0.231hrs^{-1}

We know, time t= (2.303)/ (k) log [R_{0}]/[R]

Where, k= rate constant

[R]_{°} =Initial concentration

[R]=Concentration at time‘t’

Thus, substituting the values, 8= (2.303)/ (0.231) log ([R]_{ 0}/[R])

log ([R]_{0}/[R]) = 0.8

log ([R]/[R]_{0}) = -0.8

[R]/[R]_{ 0} = 0.158

Hence, 0.158 fraction of sucrose remains.

Question 4.26.

The decomposition of hydrocarbon follows the equation

k = (4.5 × 10^{11}s^{–1}) e^{ (-28000K/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,

K = Ae^{-(Ea/RT)} (ii)

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

(E_{a}/RT) = (28000K)/ (T)

Ea = (R x 28000 K)

= (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,

k= Ae ^{– (Ea/RT)}

⇒In k = In A – (E_{a}/RT)

⇒In k = Log A – (E_{a}/RT)

⇒ Log k = Log A – (E_{a}/2.303RT) (i)

The given equation is

Log k = (14.34 - 1.25 10^{4 }(K/T)) (ii)

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

(E_{a}/2.303RT) = 1.25 10^{4} (K/T)

⇒ E_{a} = (1.25 × 10^{4}K × 2.303 × R)

= (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,

k = (0.693 / t_{ (1/2)})

= (0.693 / 256)

= 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))

=>log (4.51 × 10^{-5}) = (14.34 - 1.25 × 10^{4}(K/T))

=> log (0.654 – 05) = (14.34 - 1.25 × 10^{4}(K/T))

=> (1.25 x 10^{4} K)/ (T) = 18.686

T= (1.25 x 10^{4} K)/ (18.686)

= 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

log (k_{2}/k_{1}) = (E_{a})/ (2.303 R) (T_{2}-T_{1})/(T_{1}T_{2})

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, log (1.5 x10^{4})/ (4.5 x 10^{4}) = (6 x 10^{4})/ (2.303 x 8.314) (T_{2} -283)/ (283 – T_{2})

0.5229 = 3133.627 – (T_{2} -283)/ (283 – T_{2})

0 .0472T_{2} = (T_{2}-283)

T_{2} = 297K or T_{2} = 24^{0} C

Question 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 4 × 10^{10 }s^{–1} Calculate k at 318K and E_{a}?

Answer:

For a first order reaction,

t = (2.303) / (k log a) / (a – x)

At 298 K,

t = (2.303) / (k log 100) / (90)

= (0.1054 / k)

At 308 K,

t' = (2.303) / (k' log 100) / (75)

= (2.2877 / k')

According to the question,

t = t'

⇒ (0.1054 / k) = (2.2877 / k')

⇒ (k' / k) = 2.7296

From Arrhenius equation, we obtain

log (k_{2}/k_{1}) = (E_{a})/ (2.303 R) (T_{2}-T_{1})/(T_{1}T_{2})

log (2.7296) = (E_{a})/ (2.303 x 8.314) (308-298)/ (298 x 308)

- E
_{a}= (2.303 x 314 x298 x 308 x log(2.7296))/(308-298) - =76640.096 Jmol
^{-1} - =76.64 kJmol
^{-1}

To calculate k at 318 K,

It is given that, A = 4 x 1010 s^{-1}, T = 318K

Again, from Arrhenius equation, we obtain

log k = log A – (E_{a})/ (2.303 RT)

=log (4 x 1010) – (76.64 x 10^{3})/ (2.303 x 8.314 x 318)

= (0.6021+ 10) – (12.5876)

=-1.9855

Therefore, k = Antilog (-1.9855)

= 1.034 x 10^{-2} s ^{-1}

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

log (k_{2}/k_{1}) = (E_{a})/ (2.303 R) (T_{2}-T_{1})/(T_{1}T_{2})

It is given that k_{2} = 4k_{1}

T_{1} = 293 K

T_{2} = 313 K

Therefore, log (4k_{1}/k_{1}) = (E_{a})/ (2.303 x 8.314) (313-293)/ (293 x 313)

- 6021 = (20 x E
_{a})/( 2.303 x 8.314 x293 x 313) - E
_{a}= (0.6021 x 2.303 x 8.314 x293 x 313)/(20) - = ( 52863.33 ) J mol
^{-1} - = 52.86 kJ mol
^{-1}

Hence, the required energy of activation is 52.86 kJ mol^{- 1}

.