**Question****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.**

**Solution :**

Average rate of reaction

= 6.67 × 10^{−6 M s−1}

**Question****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?**

**Solution :**

= 0.005 mol L^{−1 min−1}

= 5 × 10^{−3 }M min^{−1}

**Question****3. For a reaction, A + B → Product; the rate law is given by, r = k [A] ^{½ }[B]^{2}. What is the order of the reaction?**

**Solution :**

The order of the reaction r = k [A]^{½ }[B]^{2}

= 1 / 2 + 2

= 2.5

**Question****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?**

**Solution :**

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] = 3a mol L^{−1}

Now, the rate equation will be:

Rate = k (3a)^{2}

= 9(ka^{2)}

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

**Question****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?**

**Solution :**

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****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.**

**Solution :**

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

t_{½ }= 0.693 / k

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

k = 0.693 / t½

= 0.693 / 60

= 0.01155 min^{-1}

= 1.155 min^{-1}

Or

k = 1.925 x 10-2 s^{-1}

**Question****7. What will be the effect of temperature on rate constant?**

**Solution :**

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,

k = Ae – E_{a} / RT

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

**Solution :**

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} = 2k

Also, R = 8.314 J K^{−1 mol−1}

Now, substituting these values in the equation:

= 52897.78 J mol^{−1}

= 52.9 kJ mol^{−1}

**Question****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?**

**Solution :**

In the given case:

E_{a} = 209.5 kJ mol^{−1} = 209500 J mo^{l−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}

⇒Inx= −Ea / RT

⇒logx=−Ea / 2.303RT

⇒logx= −209500Jmol^{−1} / 2.303 × 8.314JK^{−1}mol^{−1}×581

=−18.8323

Now,x= Antilog (−18.8323)

=1.471×10^{−19}

**Question****10. 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) + 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]**

**Solution :**

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

Therefore, order of the reaction = 2

Dimension of k = Rate / [NO]^{2}

= 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

Dimension of

k = Rate / [H_{2}O_{2}][I ^{–} ]

= 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

Dimension of k = Rate / [CH3CHO]^{3/2}

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

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

= L½ mol-½ s-1

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

Therefore, order of the reaction = 1

Dimension of k = Rate / [C_{2}H_{5}Cl]

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

= s^{-1}

**Question****11. 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}.**

**Solution :**

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****12. 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}?**

**Solution :**

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

= 7.5 × 10^{−4 }mol L^{−1 }s^{−1}

**Question****13. 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 (P_{CH3OCH3})^{3/2}

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

**Solution :**

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

Unit of rate = bar min^{−1}

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

⇒ k =**Rate / [CH _{3}OCH_{3}]^{3/2}**

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

= bar^{-½} min ^{-1}

**Question****14. Mention the factors that affect the rate of a chemical reaction.**

**Solution :**

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

**15. 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?**

**Solution :**

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] = 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(1/2a)^{2}

= 1/4 Ka^{2}

= 1/4 R

Therefore, the rate of the reaction would be reduced to

**16. What change would happen in the rate constant of a reaction when there is a change in its temperature? How can this temperature effect on rate constant be represented quantitatively?**

**Solution :**

When a temperature of 10∘ rises for a chemical reaction then the rate constant increases and becomes near to double of its original value.

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

k=Ae^{−Ea/RT}

Where,

k = rate constant,

A = Frequency factor / Arrhenius factor,

R = gas constant

T = temperature

Ea = activation energy for the reaction.

**Question****17. In a pseudo first order hydrolysis of ester in water, the following results were obtained:**

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

[Ester]mol L^{−1} | 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.**

**Solution :**

(i) Average rate of reaction between the time interval, 30 to 60 seconds, d[ester] / dt

= (0.31-0.17) / (60-30)

= 0.14 / 30

= 4.67 × 10^{−3} mol L^{−1} s^{−1}

(ii) For a pseudo first order reaction,

k = 2.303/ t log [R]º / [R]

For t = 30 s, k_{1}

= 1.911 × 10^{−2 }s^{−1}

For t = 60 s, k1 = 2.303/ 30 log 0.55 / 0.31

= 1.957 × 10^{−2 }s^{−1}

For t = 90 s, k3 = 2.303/ 90 log 0.55 / 0.085

= 2.075 × 10 – 2s – 1

= 2.075 × 10^{−2 }s^{−1}

Then, average rate constant, k = k_{1} + k_{2}+ k_{3} / 3

= 1.911 × 10 ^{– 2 } + 1.957 × 10 ^{– 2} + 2.075 × 10^{ – 2 / 3}

= 1.981 x 10^{-2} s – 1

**Question****18. 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?**

**Solution :**

(i) The differential rate equation will be

**Question****19. 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?**

**Solution :**

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

Therefore,

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****20. 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.**

**Solution :**

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,

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

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

**Question****21. 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} |

**Solution :**

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****22. 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}**

**Solution :**

(i) Half life, t _{1/2} = 0.693 / k

= 0.693 / 200 s^{-1}

= 3.47×10 ^{-3 }s (approximately)

(ii) Half life, t _{1/2} = 0.693 / k

= 0.693 / 2 min-1

= 0.35 min (approximately)

(iii) Half life, t _{1/2} = 0.693 / k

= 0.693 **/ **4 years^{-1}

= 0.173 years (approximately)

**Question****23. 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.**

**Solution :**

Here, k = 0.693 / t_{1/2}

= 0.693 / 5730 years^{-1}

It is known that,

= 1845 years (approximately)

Hence, the age of the sample is 1845 years.

**Question****24. The experimental data for decomposition of N _{2}O_{5}**

**[2N2O5 → 4NO2 + O2] in gas phase at 318K are given below:**

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

10^{2} × [N_{2}O_{5}] mol L^{-1} | 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).**

**Solution :**

(ii) Time corresponding to the concentration, 1630×10^{2} / 2 mol L^{-1} = 81.5 mol L^{-1} is the half life. From the graph, the half life is obtained as 1450 s.

(iii)

t(s) | 10^{2} × [N_{2}O_{5}] mol L^{-1} | Log[N_{2}O_{5}] |

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, 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}]

(v) From the plot, Log[N_{2}O_{5}] v/s t, we obtain

– k /2.303

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

– k / 2.303. = -0.67 / 3200

Therefore, we obtain,

– k / 2.303 = – 0.67 / 3200

⇒ k = 4.82 x 10-4 s-1

(vi) Half-life is given by,

t_{½} = 0.693 / k

= 0.639 / 4.82×10^{-4} s

=1.438 x 103

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

**25. 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?**

**Solution :**

It is known that,

**Question****26. 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.**

**Solution :**

,

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

**Solution :**

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

= 2x 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****28. A first order reaction takes 40 min for 30% decomposition. Calculate t _{1/2}.**

**Solution :**

For a first order reaction,

t = 2.303/k Log [R] º / [R]

k = 2.303/40min Log 100 / 100-30

= 2.303/40min Log 10 / 7

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

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

t1/2 = 0.693/k

= 0.693 / 8.918 x 10-3 min

= 77.7 min (approximately)

= 77.7 min (approximately)

**Question****29. 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.**

**Solution :**

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

After time, t, total pressure, Pt = (P_{º} – p) + p + p

⇒ Pt = (P_{º }+ p)

⇒ p = P_{t} – P_{0}

therefore, P_{º} – p = P_{0} – Pt – P_{0}

= 2P_{0} − P_{t}

For a first order reaction,

k = 2.303/t Log P_{0} /P_{0} – p

= 2.303/t Log P_{0} / 2 P_{0} – P_{t}

When t = 360 s, k = 2.303 / 360s log 35.0 / 2×35.0 – 54.0

= 2.175 × 10^{−3 }s^{−1}

When t = 720 s, k = 2.303 / 720s log 35.0 / 2×35.0 – 63.0

= 2.235 × 10^{−3} s^{−1}

Hence, the average value of rate constant is

k = (2.175 × 10^{ – 3} + 2.235 × 10 ^{– 3} ) / 2 s ^{– 1}

= 2.21 × 10^{−3 }s^{−1}

**Question****30. 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^{−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.**

**Solution :**

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

After time, t, total pressure, P_{t} = (P_{º} – p) + p + p

⇒ P_{t} = (P_{º} + p)

⇒ p = Pt – Pº

therefore, Pº – p = Pº – Pt – Pº

= 2 P_{º }– P_{t}

For a first order reaction,

k = 2.303/t Log P_{º }/ P_{º} – p

= 2.303/t Log P_{º} / 2 P_{º} – Pt

When t= 100 s,

k = 2.303 / 100s log 0.5 / 2×0.5 – 0.6

= 2.231 × 10 – 3s – 1

When Pt= 0.65 atm,

P0+ p= 0.65

⇒ p= 0.65 – P0

= 0.65 – 0.5

= 0.15 atm

Therefore, when the total pressure is 0.65 atm, pressure of SOCl2 is_{PSOCL2} = 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(p_{SOCL2})

= (2.23 × 10 – 3s – 1) (0.35 atm)

= 7.8 × 10^{ – 4} atm s^{ – 1}

**Question****31. The rate constant for the decomposition of N _{2}O_{5} at various temperatures is given below:**

T/°C | 0 | 20 | 40 | 60 | 80 |

10^{5} X K /S^{-1} | 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.**

**Solution :**

From the given data, we obtain

T/°C | 0 | 20 | 40 | 60 | 80 |

T/K | 273 | 293 | 313 | 333 | 353 |

1/T / k^{-1} | 3.66×10^{−3} | 3.41×10^{−3} | 3.19×10^{−3} | 3.0×10^{−3} | 2.83 ×10^{−3} |

10^{5} X K /S^{-1} | 0.0787 | 1.70 | 25.7 | 178 | 2140 |

ln k | −7.147 | − 4.075 | −1.359 | −0.577 | 3.063 |

Slope of the line,

In k= – 2.8

Therefore, k = 6.08×10-2s-1

Again when T = 50 + 273K = 323K,

1/T = 3.1 x 10-3 K

In k = – 0.5

Therefore, k = 0.607 s-1

**Question****32. 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.**

**Solution :**

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****33. 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}.**

**Solution :**

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, k = 2.303/t Log [A]º / [A]

⇒2.0 × 110-2 s-1 = 2.303/100s Log 1.0 / [A]

⇒2.0 × 110-2 s-1 = 2.303/100s ( – Log [A] )

⇒ – Log [A] = – (2.0 x 10-2 x 100) / 2.303

⇒ [A] = antilog [- (2.0 x 10-2 x 100) / 2.303]

= 0.135 mol L^{−1 (approximately)}

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

**Question****34. 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?**

**Solution :**^{For a first order reaction,k = 2.303/t Log [R]º / [R]It is given that, t1/2 = 3.00 hoursTherefore, k = 0.693 / t1/2= 0.693 / 3 h-1= 0.231 h – 1Then, 0.231 h – 1 = 2.303 / 8h Log [R]º / [R]}

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

**Question****35. The decomposition of hydrocarbon follows the equation k = (4.5 × 10 ^{11} s^{−1}) e^{−28000} K/T Calculate E_{a}.**

**Solution :**

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

Ea / RT = 28000K / T

⇒ Ea = R x 28000K

= 8.314 J K^{−1} mol^{−1} × 28000 K

= 232792 J mol^{−1}

= 232.792 kJ mol^{−1}

**Question****36. 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?**

**Solution :**

Arrhenius equation is given by,

k= Ae -Ea/RT

⇒In k = In A – Ea/RT

⇒In k = Log A – Ea/RT

⇒ Log k = Log A – Ea/2.303RT (i)

The given equation is

Log k = 14.34 – 1.25 104 K/T (ii)

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

Ea/2.303RT = 1.25 104 K/T

= 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 – 5s^{ – 1}

= 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****37. 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}?**

**Solution :**

From Arrhenius equation, we obtain

log k2/k1 = Ea / 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,

= 297 K

= 24°C

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

**Question****38. 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}.**

**Solution :**

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

To calculate k at 318 K,

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

Again, from Arrhenius equation, we obtain

Therefore, k = Antilog (-1.9855)

= 1.034 x 10-2 s -1

**Question****39. 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.**

**Solution :**

From Arrhenius equation, we obtain

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

## NCERT Solutions For Class 12 Chemistry Chapter Wise.

- Chapter-1 The Solid States
- Chapter-2 Solutions
- Chapter-3 Electrochemistry
- Chapter-4 Chemical Kinetics
- Chapter-5 Surface Chemistry
- Chapter-6 General Priciples and Processes of Isolation of Elements
- Chapter-7 The P Block Elements
- Chapter-8 The D and F Elements
- Chapter-9 Coordination Compounds
- Chapter-10 Haloalkanes and Haloarenes
- Chapter-11 Alcohols, Phenols and Ethers
- Chapter-12 Aldehydes, ketones and Carboxylics Acids
- Chapter-13 Amines
- Chapter-14 Biomolecules
- Chapter-15 Polymers
- Chapter-16 Chemistry In Everyday Life

**Question****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?**

**Solution :**

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] = 3a mol L^{−1}

Now, the rate equation will be:

Rate = k (3a)^{2}

= 9(ka^{2)}

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

**Question****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?**

**Solution :**

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****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.**

**Solution :**

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

t_{½ }= 0.693 / k

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

k = 0.693 / t½

= 0.693 / 60

= 0.01155 min^{-1}

= 1.155 min^{-1}

Or

k = 1.925 x 10-2 s^{-1}

**Question****7. What will be the effect of temperature on rate constant?**

**Solution :**

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,

k = Ae – E_{a} / RT

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

**Solution :**

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} = 2k

Also, R = 8.314 J K^{−1 mol−1}

Now, substituting these values in the equation:

= 52897.78 J mol^{−1}

= 52.9 kJ mol^{−1}

**Question****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?**

**Solution :**

In the given case:

E_{a} = 209.5 kJ mol^{−1} = 209500 J mo^{l−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}

⇒Inx= −Ea / RT

⇒logx=−Ea / 2.303RT

⇒logx= −209500Jmol^{−1} / 2.303 × 8.314JK^{−1}mol^{−1}×581

=−18.8323

Now,x= Antilog (−18.8323)

=1.471×10^{−19}

**Question****10. 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) + 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]**

**Solution :**

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

Therefore, order of the reaction = 2

Dimension of k = Rate / [NO]^{2}

= 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

Dimension of

k = Rate / [H_{2}O_{2}][I ^{–} ]

= 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

Dimension of k = Rate / [CH3CHO]^{3/2}

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

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

= L½ mol-½ s-1

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

Therefore, order of the reaction = 1

Dimension of k = Rate / [C_{2}H_{5}Cl]

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

= s^{-1}

**Question****11. 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}.**

**Solution :**

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****12. 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}?**

**Solution :**

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

= 7.5 × 10^{−4 }mol L^{−1 }s^{−1}

**Question****13. 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 (P_{CH3OCH3})^{3/2}

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

**Solution :**

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

Unit of rate = bar min^{−1}

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

⇒ k =**Rate / [CH _{3}OCH_{3}]^{3/2}**

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

= bar^{-½} min ^{-1}

**Question****14. Mention the factors that affect the rate of a chemical reaction.**

**Solution :**

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

**15. 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?**

**Solution :**

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] = 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(1/2a)^{2}

= 1/4 Ka^{2}

= 1/4 R

Therefore, the rate of the reaction would be reduced to

**16. What change would happen in the rate constant of a reaction when there is a change in its temperature? How can this temperature effect on rate constant be represented quantitatively?**

**Solution :**

When a temperature of 10∘ rises for a chemical reaction then the rate constant increases and becomes near to double of its original value.

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

k=Ae^{−Ea/RT}

Where,

k = rate constant,

A = Frequency factor / Arrhenius factor,

R = gas constant

T = temperature

Ea = activation energy for the reaction.

**Question****17. In a pseudo first order hydrolysis of ester in water, the following results were obtained:**

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

[Ester]mol L^{−1} | 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.**

**Solution :**

(i) Average rate of reaction between the time interval, 30 to 60 seconds, d[ester] / dt

= (0.31-0.17) / (60-30)

= 0.14 / 30

= 4.67 × 10^{−3} mol L^{−1} s^{−1}

(ii) For a pseudo first order reaction,

k = 2.303/ t log [R]º / [R]

For t = 30 s, k_{1}

= 1.911 × 10^{−2 }s^{−1}

For t = 60 s, k1 = 2.303/ 30 log 0.55 / 0.31

= 1.957 × 10^{−2 }s^{−1}

For t = 90 s, k3 = 2.303/ 90 log 0.55 / 0.085

= 2.075 × 10 – 2s – 1

= 2.075 × 10^{−2 }s^{−1}

Then, average rate constant, k = k_{1} + k_{2}+ k_{3} / 3

= 1.911 × 10 ^{– 2 } + 1.957 × 10 ^{– 2} + 2.075 × 10^{ – 2 / 3}

= 1.981 x 10^{-2} s – 1

**Question****18. 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?**

**Solution :**

(i) The differential rate equation will be

**Question****19. 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?**

**Solution :**

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

Therefore,

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****20. 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.**

**Solution :**

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,

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

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

**Question****21. 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} |

**Solution :**

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****22. 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}**

**Solution :**

(i) Half life, t _{1/2} = 0.693 / k

= 0.693 / 200 s^{-1}

= 3.47×10 ^{-3 }s (approximately)

(ii) Half life, t _{1/2} = 0.693 / k

= 0.693 / 2 min-1

= 0.35 min (approximately)

(iii) Half life, t _{1/2} = 0.693 / k

= 0.693 **/ **4 years^{-1}

= 0.173 years (approximately)

**Question****23. 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.**

**Solution :**

Here, k = 0.693 / t_{1/2}

= 0.693 / 5730 years^{-1}

It is known that,

= 1845 years (approximately)

Hence, the age of the sample is 1845 years.

**Question****24. The experimental data for decomposition of N _{2}O_{5}**

**[2N2O5 → 4NO2 + O2] in gas phase at 318K are given below:**

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

10^{2} × [N_{2}O_{5}] mol L^{-1} | 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).**

**Solution :**

(ii) Time corresponding to the concentration, 1630×10^{2} / 2 mol L^{-1} = 81.5 mol L^{-1} is the half life. From the graph, the half life is obtained as 1450 s.

(iii)

t(s) | 10^{2} × [N_{2}O_{5}] mol L^{-1} | Log[N_{2}O_{5}] |

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, 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}]

(v) From the plot, Log[N_{2}O_{5}] v/s t, we obtain

– k /2.303

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

– k / 2.303. = -0.67 / 3200

Therefore, we obtain,

– k / 2.303 = – 0.67 / 3200

⇒ k = 4.82 x 10-4 s-1

(vi) Half-life is given by,

t_{½} = 0.693 / k

= 0.639 / 4.82×10^{-4} s

=1.438 x 103

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

**25. 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?**

**Solution :**

It is known that,

**Question****26. 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.**

**Solution :**

,

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

**Solution :**

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

= 2x 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****28. A first order reaction takes 40 min for 30% decomposition. Calculate t _{1/2}.**

**Solution :**

For a first order reaction,

t = 2.303/k Log [R] º / [R]

k = 2.303/40min Log 100 / 100-30

= 2.303/40min Log 10 / 7

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

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

t1/2 = 0.693/k

= 0.693 / 8.918 x 10-3 min

= 77.7 min (approximately)

= 77.7 min (approximately)

**Question****29. 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.**

**Solution :**

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

After time, t, total pressure, Pt = (P_{º} – p) + p + p

⇒ Pt = (P_{º }+ p)

⇒ p = P_{t} – P_{0}

therefore, P_{º} – p = P_{0} – Pt – P_{0}

= 2P_{0} − P_{t}

For a first order reaction,

k = 2.303/t Log P_{0} /P_{0} – p

= 2.303/t Log P_{0} / 2 P_{0} – P_{t}

When t = 360 s, k = 2.303 / 360s log 35.0 / 2×35.0 – 54.0

= 2.175 × 10^{−3 }s^{−1}

When t = 720 s, k = 2.303 / 720s log 35.0 / 2×35.0 – 63.0

= 2.235 × 10^{−3} s^{−1}

Hence, the average value of rate constant is

k = (2.175 × 10^{ – 3} + 2.235 × 10 ^{– 3} ) / 2 s ^{– 1}

= 2.21 × 10^{−3 }s^{−1}

**Question****30. 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^{−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.**

**Solution :**

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

After time, t, total pressure, P_{t} = (P_{º} – p) + p + p

⇒ P_{t} = (P_{º} + p)

⇒ p = Pt – Pº

therefore, Pº – p = Pº – Pt – Pº

= 2 P_{º }– P_{t}

For a first order reaction,

k = 2.303/t Log P_{º }/ P_{º} – p

= 2.303/t Log P_{º} / 2 P_{º} – Pt

When t= 100 s,

k = 2.303 / 100s log 0.5 / 2×0.5 – 0.6

= 2.231 × 10 – 3s – 1

When Pt= 0.65 atm,

P0+ p= 0.65

⇒ p= 0.65 – P0

= 0.65 – 0.5

= 0.15 atm

Therefore, when the total pressure is 0.65 atm, pressure of SOCl2 is_{PSOCL2} = 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(p_{SOCL2})

= (2.23 × 10 – 3s – 1) (0.35 atm)

= 7.8 × 10^{ – 4} atm s^{ – 1}

**Question****31. The rate constant for the decomposition of N _{2}O_{5} at various temperatures is given below:**

T/°C | 0 | 20 | 40 | 60 | 80 |

10^{5} X K /S^{-1} | 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.**

**Solution :**

From the given data, we obtain

T/°C | 0 | 20 | 40 | 60 | 80 |

T/K | 273 | 293 | 313 | 333 | 353 |

1/T / k^{-1} | 3.66×10^{−3} | 3.41×10^{−3} | 3.19×10^{−3} | 3.0×10^{−3} | 2.83 ×10^{−3} |

10^{5} X K /S^{-1} | 0.0787 | 1.70 | 25.7 | 178 | 2140 |

ln k | −7.147 | − 4.075 | −1.359 | −0.577 | 3.063 |

Slope of the line,

In k= – 2.8

Therefore, k = 6.08×10-2s-1

Again when T = 50 + 273K = 323K,

1/T = 3.1 x 10-3 K

In k = – 0.5

Therefore, k = 0.607 s-1

**Question****32. 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.**

**Solution :**

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****33. 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}.**

**Solution :**

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, k = 2.303/t Log [A]º / [A]

⇒2.0 × 110-2 s-1 = 2.303/100s Log 1.0 / [A]

⇒2.0 × 110-2 s-1 = 2.303/100s ( – Log [A] )

⇒ – Log [A] = – (2.0 x 10-2 x 100) / 2.303

⇒ [A] = antilog [- (2.0 x 10-2 x 100) / 2.303]

= 0.135 mol L^{−1 (approximately)}

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

**Question****34. 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?**

**Solution :**^{For a first order reaction,k = 2.303/t Log [R]º / [R]It is given that, t1/2 = 3.00 hoursTherefore, k = 0.693 / t1/2= 0.693 / 3 h-1= 0.231 h – 1Then, 0.231 h – 1 = 2.303 / 8h Log [R]º / [R]}

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

**Question****35. The decomposition of hydrocarbon follows the equation k = (4.5 × 10 ^{11} s^{−1}) e^{−28000} K/T Calculate E_{a}.**

**Solution :**

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

Ea / RT = 28000K / T

⇒ Ea = R x 28000K

= 8.314 J K^{−1} mol^{−1} × 28000 K

= 232792 J mol^{−1}

= 232.792 kJ mol^{−1}

**Question****36. 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?**

**Solution :**

Arrhenius equation is given by,

k= Ae -Ea/RT

⇒In k = In A – Ea/RT

⇒In k = Log A – Ea/RT

⇒ Log k = Log A – Ea/2.303RT (i)

The given equation is

Log k = 14.34 – 1.25 104 K/T (ii)

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

Ea/2.303RT = 1.25 104 K/T

= 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 – 5s^{ – 1}

= 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****37. 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}?**

**Solution :**

From Arrhenius equation, we obtain

log k2/k1 = Ea / 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,

= 297 K

= 24°C

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

**Question****38. 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}.**

**Solution :**

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

To calculate k at 318 K,

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

Again, from Arrhenius equation, we obtain

Therefore, k = Antilog (-1.9855)

= 1.034 x 10-2 s -1

**Question****39. 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.**

**Solution :**

From Arrhenius equation, we obtain

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