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**NCERT Class 12 ****Chemistry** Chapter 4 **Chemical Kinetics**

**Chemistry**Chapter 4

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**Chemical Kinetics**

**Chemical Kinetics****Chapter: 4**

Part – I |

**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}

Ans: Rate = k[NO]^{2}

**Order of Reaction:**

Order with respect to NOis 2

Overall order of the reaction is 2.

The dimensions of the rate constants [k] is mol^{−1}Ls^{−1}

**(ii) H**_{2}**O**_{2}** (aq) + 3I**^{–}** (aq) + 2H**^{+}** → 2H**_{2}**O (l) + 3 I Rate = k[H**_{2}**O**_{2}** ][I-] **

Ans: **Order of Reaction:**

Rate = k[H_{2}O_{2}][I^{−}].

Order with respect to H_{2}O_{2} is 1

Order with respect to I^{−} is 1.

Overall order of the reaction is 2.

The dimensions of the rate constants [k] is mol^{−1}Ls^{−1}

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

Ans: **Order of Reaction:**

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

The reaction order with respect to CH_{3}CHO is 3/2.

The overall order of the reaction is 3/2.

The dimensions of the rate constants [k] = mol^{−1/2}L^{1/2}s^{−1}

**(iv) C**_{2}**H**_{5}**Cl (g) → C2H**_{4}** (g) + HCl (g) Rate = k [C2H**_{5}**Cl]**

Ans: **Order of Reaction:**

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

The reaction order with respect to C_{2}H_{5}Cl is 1.

The overall order of the reaction is 1.

The dimensions of the rate constants [k] = s^{−1}

**2. For the reaction:**** **

2A + B → A_{2}B

the rate = k[A][B]^{2} with k = 2.0 × 10^{–6} mol^{–2} L2 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}.

Ans: 2A + B → A2B

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

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

= 8.0 × 10^{-9} mol L^{-1} s^{-1}

Rate After Reduction of [A]

**New concentration:** [A] = 0.06 mol L^{−1}[A] = 0.06,

[B] = 0.2 mol L^{−1}

**New concentration of B:** 0.2−0.02

= 0.18 mol L^{−1}

R = (2.0 × 10−6 mol−2L2s^{−1}) × (0.06mol L^{−1}) × (0.18mol L^{−1})^{2}

= 0.0324 mol ^{2} L^{−2}

Ratenew = 3.89 × 10−9mol L^{−}1s^{−1}

**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}**?**

Ans:

For the zero order reaction rate: K

= 2.5 × x10^{−4} mol/L/s

N2 = d[N_{2}] / dt

= 2.5 × 10^{–}4 mol/L/s

Rate of production of

d[H2]/dt = 3 × k

= 3 × 2.5 ×1 0^{−4}

= 7.5 × x10^{−4} molL/s

**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 (P_{CH3}O_{CH3 })^{3/2}

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

Ans: Unit of rate of reaction = bar min^{−1}

Units of Rate (K) = Rate/ (PCH_{3}OCH_{3})^{3/2}

= Units of Rate Constant (K): bar min^{−1}/ (bar)^{3/2}

= bar^{-1/2} min^{−1}

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

Ans: **The various factors which affecting the rate of a chemical reaction are:**

(i) Concentration of reactants.

(ii) Temperature of reaction.

(iii) Nature of reactants.

(iv) Surface area.

(v) Exposure to radiation.

(vi) Presence of catalyst.

**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.**

Ans: **The initial rate of the reaction is:**

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

When construction of A is double

I.e. [A] = 2a

Rate of reaction becomes 4 times.

**(ii) reduced to half? **

Ans: If the concentration of the reactant is reduced to half, the rate of reaction becomes one fourth.

I.e [A] = ½(a)

Rate = k(a/2)^{2}

= 1/4ka^{2}

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

Ans: The rate constant of a reaction increases with increase of temperature and becomes nearly double for every 10° rise of temperature. This is also called temperature coefficient. It is the ratio of rate constants of the reaction at two temperatures differing by ten degrees. According to Arrhenius equation,

k = Ae ^{-EART}

where Ea represents the activation energy of the reaction and A represents the frequency factor.

**8. In a pseudo first order reaction in water, the following results were obtained:**

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

[A]/ mol L^{–1} | 0.55 | 0.31 | 0.17 | 0.085 |

Calculate the average rate of reaction between the time interval 30 to 60 seconds.

Ans: (b) The average rate of reaction between the time interval 30 to 60 is

Average rate =

= 0.14/30

= – 0.00467

= -4.67 ×10^{-3}Ms^{-1}

Minus sign shows that rate of reaction is decreasing with time as conc. Of ester is decreasing with time.

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

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

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

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

Ans: Rate = k[A][3B]² = 9k [A] [B]^{2}

Thus, the rate of the reaction will increase by a factor of 9 when the concentration of B is increased three times

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

Ans: Rate = k[2A] [2B]² = 8k [A] [B]

Thus, the rate of the reaction will increase by a factor of 8 when the concentrations of both A and B are doubled.

**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?

Ans: r_{0} = k[A]m[B]n

Given,

[A] and [B] are the concentrations of reactants A and B,

m and n are the orders of the reaction with respect to A and B, respectively,

k is the rate constant.

r_{0} is the initial rate of reaction,

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} |

Dividing equation (i) by equation (ii)

= 1

[3]_{y} = [3]^{0}

= 1

y = 0

Now dividing equation (ii) by Equation (iii)

[½]^{x} = ½.82

2x = 2.82

X log 2 = log 2.82

X = 1.4957 ≃1.5

Order with respect to A: 1.5

Order with respect to B: 0.

**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 × 0^{–2} |

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

Ans: **The rate law expression:** k[A]^{x}[B]^{y}

**Here:**

k is the rate constant,

m and n are the reaction orders with respect to A and B,

[A] and [B] are the concentrations of the reactants.

(Rate_{1}) = 6.0 × 10^{−3 }= k(0.1)^{x }(0.1)^{y}……………(i)

(Rate_{2}) = 7.2 × 10^{−2 }= k(0.3)^{x }(0.2)^{y}……………(ii)

(Rate_{3}) = 2.88 × 10^{−1 }= k(0.3)^{x }(0.4)^{y}………….(iii)

(Rate_{4}) = 2.40 × 10^{−2 }= k(0.4)^{x }(0.1)^{y}…………(iv)

Divide equation (ii) by equation (i)

**Or**

¼ = (0.1)^{x}/(0.4)^{y}

= (¼)^{x}

∴ x = 1

From experiments II and III, we get

**Or**

¼ = (0.2)^{y }/ (0.4)^{y}

= (½)y

∴ y = 2

The rate law expression is, k[A][B]^{2}

Substitute values in equation (i)

= 6.0 mol^{-2} min^{-1}

K = 6.0mol^{-2} L^{2} min^{-1}.

**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} |

Ans: Rate law for the reaction = k[A]^{1}[B]^{0 }= k[A]

From the table for Experiment I, we know

Rate = 2.0 × 10^{−2 }mol L^{−1 }min^{−1}

[A] = 0.1mol L^{−1}

[B] = 0.1mol L^{−1}

For I nd Experiment

Rate = k[A]

2.0 × 10^{−2 }= k × 0.1

k = (2.0 × 10^{−2})/0.1

= 0.2Min^{-1}

For II nd Experiment

Rate = 4.0 × 10^{−2 }mol L^{−1 }min^{−1}

[B] = 0.2mol L^{−1}

4.0 × 10^{−2} = 0.2 × [A]

[A] = 4.0 ×10^{−2}/0.2

= 0.2mol L^{−1}

For III nd Experiment

[A] = 0.4mol L^{−1}

[B] = 0.4mol L^{−1}

Rate = k[A]

= 0.2 × 0.4 mol L^{−1} min ^{−1}

= 8.0 × 10^{−2} mol L^{−1} min^{−1}

For IV nd Experiment

Rate = k[A]

Rate = 2.0 × 10^{−2 }mol L^{−1 }min^{−1}

[B] = 0.2mol L^{−1}

= 2.0 × 10^{−2 }mol L^{−1} min^{−1}

= 0.2 × [A]

[A] = (2.0 × 10^{−2})/0.2

= 0.1mol L^{−1}

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 | 0.2 | 4.0 × 10^{–2} |

III | 0.4 | 0.4 | 8.0 × 10^{–2} |

IV | 0.1 | 0.2 | 2.0 × 10^{–2} |

**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}

Ans: **The half-life t**_{1/2}** for a first-order reaction is given by the formula:**

t_{1/2 }= 0.693/k

**Where:**

t_{1/2} is the half-life,

k is the rate constant.

(a) k = 200s^{−1}

t_{1/2 }= 0.693/200

= 3.465 × 10^{-3} second

(b) k = 2min^{−1}

t_{1/2 }= 0.693/2min^{−1}

= 0.3465 min.

(c) k = 4 years^{−1}

t_{1/2 }= 0.693/4years^{−1}

= 0.17325 years

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

Ans: N_{t} = N_{0}⋅e ^{−λt}

k = 1.209 × 10^{−4}years^{−1},

NO = 100 (original amount),

N = 80(remaining amount).

λ is the decay constant,

t is the time elapsed.

λ = 0.693 /t_{1/2}

= 0.693/5730

= 1.21 × 10^{-4} year^{-1}

All radioactive nuclear are first order process,

Therefore, decay constant,

= 1845 year

So, the estimated age of the sample is approximately 1844 years

**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 | 2400 | 2800 | 3200 | 3200 |

102×[N2O5 ]/mol^{-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.**

Ans:

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

Ans: Initial concentration ([N₂O₅]₀) = 1.63 × 10² mol⁻¹

Half of this value = 1.63/2 = 0.815 × 10² mol⁻¹

Time for half of initial concentration (i.e half life period) from plot is 1440 s

Hence, T_{1/2} =1440s

**(iii) Draw a graph between log[N**_{2}**O**_{5}**] and t.**

Ans:

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

Ans: As plot of log (N_{2}O_{5}) vs. time is a straight line, hence it is a reaction of first order i.e., rate law is

Thus, the rate law is: Rate = k[N_{2}O_{5}]

Here k is the rate constant.

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

Ans: = -(k/2303)

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

= -0.67 / 3200

I.e., k / 2303 = 0.67 / 3200

K = (0.67/3200) × 2.303

= 4.82 × 10^{-4} mol L^{-1} s^{-1}

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

Ans: t_{1/2} = 0.693/k

Given

k = 4.84 × 10^{−4}s^{−1}

t_{1/2} = (0.693)/(4.84 × 10^{−4})

t_{1/2} = (0.693)/0.000484s

= 1432 seconds.

**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 1/16th value?**

Ans: **For the first-order reaction we have:**

**Here:**

The rate constant k = 60 s^{−1}.

t_{15/16 }= ?

Rate constant of first order reaction is given by

Where 15/16th the reaction is over then

= 0.046 S

= 4.6 × 10^{-2}sec.

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

Ans: λ = 0.693 / t1/2

= 0.693/28.1

= 0.0247 year-1

(i) After 10 years, let xμ be the concentration of^{ 90}Sr.

λ = (2.303/t) log (N_{o}/N)

t =10 years

N_{o }= 1 microgram

= 1 × 10-6g,

N = ?

(ii) After 60 years, let yμ be the concentration of^{ 90}Sr.

N = (1 × 10^{-6})/4.400

= 0.227ug.

**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.**** **

Ans:

When reaction is 99% complete,

t_{99%} = 4.602/k…………..(i)

When reaction is 90% complete,

(2.303/k)log10 = 2.303/k ………….(ii)

Divide equation (1) by (2), we get

t_{99%}/_{ }t_{90% }= (4.606/h) × (k/2.303)

= 2

The time required for 99% completion of a first-order reaction is indeed twice the time required for the completion of 90% of the reaction.

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

Ans:

**Where:**

t = 40 min,

x = 30% decomposition

[A_{0}−x] = 100−30 = 70%.

A = 100.

k = 8.19 × 10^{−3 }min^{−1}

t_{1/2} = 0.693/(8.919 × 10^{−3})

= 77.6 minutes.

Therefore the half-life t_{1/2} is indeed 77.6 minutes.

**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.

Ans: (CH_{3})2CHN = N(CH_{3})2 → N_{2 }+ C_{6}H1_{4}

**Given:**

Initial pressure P0 = 35.0 mm of Hg

Pressure at t = 360 sect = 3,

Pt = 54.0mm of Hg

Pressure at t = 720 sect

Pt = 63.0mm of Hg

**Where:**

P0 is the initial pressure, 35.0mm of Hg,

Pt is the pressure at time t, 54.0mm of Hg.

= 2.2 × 10^{-3}s^{-1}

K = 2.303 / 720s

= 2.2 ×10^{-3}s^{-1}.

**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^{–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.

Ans: SO_{2}Cl_{2}(g) → SO_{2}(g)Cl_{2}(g)

Att = 0 | P_{0} | 0 | 0 |

Att = t | P_{o}-P | P | P |

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

= Pt = (P_{o}+P)

= p= P_{t}-P_{o}

∴ P_{o} – p = P_{o} – P_{t} -P_{o}

= 2P_{o} – P_{t}

Initial pressure, P_{0 }= 0.6 atm

Total pressure at t = 100 sect is 0.6 atm

Total pressure we need to evaluate is 0.65 atm

t = 100 sect

= 2.23 × 10-3s-1

For total pressure of 0.65atm

Rate of reaction = kp(SO_{2}CI_{2})

From the reaction stoichiometry

p(SO_{2}CI_{2}) = 2P_{i} -P_{t}

= 1 atm -0.65atm.

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

= 2.23 × 10^{-3}s^{-1} × 0.35atm

= 7.8 × 10^{-4} s^{-1} atm.

**22. 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 }× 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 Ea . Predict the rate constant at 30° and 50°C.

Ans: Convert the Celsius temperatures to Kelvin by adding 273.15:

T0 = 0°C = 273.15 K

T20 = 20°C = 293.15 K

T40 = 40°C = 313.15 K

T60 = 60°C = 333.15 K

T80 = 80°C = 353.15 K

**Calculation of 1/T and Ink:**

1/T (reciprocal of temperature in Kelvin)

In k (natural logarithm of the rate constant)

Temperature (°C) | Temperature (K) | 1/T (K^{−1}) | k (×10^{−}3 s^{−1}) | lnk |

0 | 273.15 | 0.00366 | 0.0787 | -2.553 |

20 | 293.15 | 0.00341 | 1.70 | 0.526 |

40 | 313.15 | 0.00319 | 25.7 | 3.440 |

60 | 333.15 | 0.00300 | 178 | 5.182 |

80 | 353.15 | 0.00283 | 2140 | 7.329 |

Slope = -2.4 / 0.00047

= Ea/ 2.303R

Therefor Activation energy

(Ea) = (2.4 × 2.303 × 8.314J mol^{-1}) / (0.00047)

= 97875J mol

= 97.875kJmol^{-1}

Ea = 97.875kJmol^{-1}

Now,

**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.**

Ans: According to Arrhenius equation

Log k = log A-(Ea/2.303Rt)

K = 2.418 × 10-5s^{-1}

Ea = 179.98 kJ mol^{-1}

= 179900 J mol^{-1}

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

T = 546K

= 12.5916

A = Antilog 12.5916

= 3.9 × 10^{12}s^{-1}

A = 3.9 × 10^{12}s^{-1}.

**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}**.**

Ans: According to Arrhenius equation

k = 2.1418 × 10^{-5}s^{-1}

Ea = 179.9kJmol^{-1}

= 179900 J^{-1} mol^{-1}

R = 8.314 JK^{-1}

T = 546K

= -4.6184 + 17.21

= 12.5916

A = Antilog 12.5916

= 3.9 × 10^{12}s^{-1}

A = 3.9 × 10^{12}s^{-1}.

**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?**

Ans: K = 2.0 × 10^{-2}s^{-1}

t_{1/2 }= 100 sec, a = 1.0 molL^{-1}

Or 0.8684 = log (1/ (1-x))

Antilog 0.8684 = 1/(1-x)

Or 7.386 (1-x) =1

Or 1-x = 1/7.386

= 0.135

∴ Concentration of A remaining after 100S = 0.135M.

The relationship between the rate constant k and the half-life t_{1/2} for a first-order reaction is

K = 0.693 / t_{ 1/2}

Given

t_{ 1/2 }= 3.00 hours

t = 8 hours.

For the first order reaction,

K = 0.693. 3.00

= 0.231hours^{-1}

Now

Calculation of first-order decay equation to find the fraction remaining after 8 hours:

a/(a-x) = 6.345

Or Fraction left = (a-x)/a

= 1/6.345 = 0.157M.

**26. The decomposition of hydrocarbon follows the equation**

k = (4.5 × 1011s –1) e ^{-28000K}/T

Calculate E_{a}.

Ans: According to Arrhenius equation

k = A^{eRT−Ea}

**Here:**

k is the rate constant,

A is the pre-exponential factor (here, 4.5 × 1011s^{−1}),

Ea is the activation energy,

R is the universal gas constant (8.314 J/mol\cdotpK 8.314),

T is the temperature in Kelvin.

On comparing both equations,

-(Ea/RT) = -28000K/T

Ea = (28000k) × R

= 28000 × 8.314 J/mol^{-1}

= 232792J/mol^{-1}

Thus, the activation energy (Ea) is approximately 232.8 kJ/mol.

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

Ans: The Arrhenius equation helps in determining the activation energy for a reaction.

K = Ae^{– Ea/Rt}

Calculation of activation energy Ea According to Arrhrnius equation,

On comparing both equation,

Ea/2.303Rt = 1.25 × 10^{4} K × 2.303 × 8.314 jK^{-1} Mol^{-1}

= 23.93 × 10^{4} J mol^{-1}

= 239.3 kJ mol^{-1}.

Calculation of required temperature if t_{1/2} = 256 min

**For I st order reaction: **

t_{1/2} = 0.693/k

= 0.693/ (256 min)

= 4.51 × 10^{-5} s^{-1}

According to Arrhenius theory

Log4.51 × 10^{−5} = 14.341.25 × 104K/T

-4.35 = 14.341.25 × 104K/T

T = 669 K

Therefore, the temperature at which the half-life period is 256 minutes is 669 K.

**28. The decomposition of A into product has value of k as 4.5 × 103 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}**?**

Ans:

k_{1} = 4.5 × 10^{3}s^{−1}

k_{2} = 1.5 × 10^{4}s^{−1}

T_{1 }= 10^{∘}C = 283

E_{a} = 60kJ/mol = 60000J/mol.

1-(283/T2) = 0.04776

Or

T_{2} = 283/1-0.04776

= 283/0.95224

T_{2} = 297.19 K

= (297.19 -273.0)

= 24.19^{∘}C

**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 × 1010s **^{–1}**. Calculate k at 318K and Ea.**

Ans: Calculation of Activation energy (Ea)

**Here:**

k_{1} is the rate constant at T_{1} = 298

k_{2}^{ }is the rate constant at T_{2} = 308

A is the pre-exponential factor

Ea is the activation energy

R is the gas constant (8.314 J/mol·K)

For 10% completion of the reaction, we have (i)

For 25% completion of the reaction, we have(ii)

Dividing Rq (ii) by (i)

**According to Arrhenius theory: **

Calculation of rate constant(K)

Logk = 10.6021-12.5870

= 1.9849

K = Antilog (-1.9849)

= Antilog (2^{–}.0151)

= 1.035 × 10^{-2}s^{-1}

Ea = 76.640kJ mpl^{-1}

k = 1.035 V10^{-2}s^{1}.

**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.**

Ans:** According to Arrhenius equation:**

= 5.2863 × 10^{4} J mol^{-1}

= 52.863 kJ mol^{-1}.