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Ncert Solutions Chemistry Class 12th

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Ncert Solutions Chemistry Class 12th Solutions

1. Define the following modes of expressing the concentration of a solution. Which of these modes are independent of temperature and why?

(i) $\frac{w}{w}$ (percentage mass by mass)

(ii) $\frac{V}{V}$ (volume by volume percentage)

(iii) $\frac{w}{V}$ (mass by volume percentage)

(iv) ppm. (parts per million)

(v) x, (mole fraction)

(vi) M (Molarity)

(vii) m (Molality)

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6 months ago

Ncert Solutions Chemistry Class 12th Chemical Kinetics

4.39 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.

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Payal Gupta

Contributor-Level 10

4.39 Given, k₂ = 4k₁, T₁ = 293K and T₂ = 313K

We know that from the Arrhenius equation, we obtain

On solving, we get,

E_a = 58263.33 J mol^-1 or 58.26 kJ mol^-1

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6 months ago

Ncert Solutions Chemistry Class 12th Chemical Kinetics

4.38 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.

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Payal Gupta

Contributor-Level 10

4.38 We know, time t = (2.303/k) * log ( [R]₀/ [R])

Where, k- rate constant

[R]₀-Initial concentration

[R]-Concentration at time 't'

At 298K, If 10% is completed, then 90% is remaining. t = (2.303/k) * log ( [R]₀/0.9 [R]₀)

t = (2.303/k) * log (1/0.9) t = 0.1054 / k

At temperature 308K, 25% is completed, 75% is remaining t' = (2.303/k') * log ( [R]₀/0.75 [R]₀)

t' = (2.303/k') * log (1/0.75) t' = 2.2877 / k'

But, t = t'

0.1054 / k = 2.2877 / k' / k = 2.7296

From Arrhenius equation, we obtain log k₂/k₁ = (E_a / 2.303 R) * (T₂ - T₁) / T₁T₂

Substituting the values,

E_a = 76640.09 J mol^-1 or 76.64 kJ mol^-1 We know, log k = log A –E_a/RT

Log k =

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6 months ago

Ncert Solutions Chemistry Class 12th Chemical Kinetics

4.37 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 * 104s^–1?

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Payal Gupta

Contributor-Level 10

4.37 From Arrhenius equation, we obtain

Also, k₁ = 4.5 * 10³ s ^-¹

T₁ = 273 + 10 = 283 K

k₂ = 1.5 * 10⁴ s ^-¹

E_a = 60 kJ mol ^-¹ = 6.0 * 10⁴ J mol ^-¹

Then,

→ 0.5229 = 3133.627 * (T₂-283)/ (283 * T₂)

→ 0.0472T₂ = T₂-283 T₂ = 297K or T₂ = 24⁰ C

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6 months ago

Ncert Solutions Chemistry Class 12th Chemical Kinetics

4.36 The rate constant for the first order decomposition of H₂O₂ is given by the following equation: log k = 14.34 – 1.25 * 104K/T. Calculate E_a for this reaction and at what temperature will its half-period be 256 minutes?

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Payal Gupta

Contributor-Level 10

4.36 We know, The Arrhenius equation is given by k = Ae^-Ea/RT Taking natural log on both sides,

Ln k = ln A- (E_a/RT)

Thus, log k = log A - (E_a/2.303RT). eqn 1

The given equation is log k = 14.34 – 1.25 * 10⁴K/T. eqn 2

Comparing 2 equations, E_a/2.303R = 1.25 * 10⁴K

E_a = 1.25 * 10⁴K * 2.303 * 8.314

E_a = 239339.3 J mol^-1 (approximately) E_a = 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 k = 4.51 * 10^-5s^–1

Substitute k = 4.51 * 10^-5s^–1 in eqn 2,

log 4.51 * 10^-5 s^–1 = 14.34 – 1.25 * 10⁴K/T

log (0.654-5) = 14.34– 1.25 * 10⁴K/T = 1.25 * 10⁴/ [ 14.34- log (0.654-5)] T = 668.9K or T =

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6 months ago

Ncert Solutions Chemistry Class 12th Chemical Kinetics

4.35 The decomposition of hydrocarbon follows the equation k = (4.5 * 10¹¹s^–1) e^-28000K/T. Calculate E_a

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Payal Gupta

Contributor-Level 10

4.35 The given equation is

k = (4.5 x 10¹¹ s^-1) e^-28000^K/T (i)

Comparing, Arrhenius equation

k = Ae ^-E^/RT (ii)

We get, E_a / RT = 28000K / T

⇒E_a = R x 28000K

= 8.314 J K^-1mol^-1 * 28000 K

= 232792 J mol^–1 or 232.792 kJ mol^–1

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6 months ago

Ncert Solutions Chemistry Class 12th Chemical Kinetics

4.34 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?

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Payal Gupta

Contributor-Level 10

4.34 t_1/2 = 3.00 hours

We know, t_1/2 = 0.693/k

? k = 0.693/3 k = 0.231 hrs^-1

We know, time

Where, k- rate constant

[R]_° -Initial concentration

[R]-Concentration at time 't'

Thus, substituting the values,

log ( [R]₀/ [R]) = 0.8

log ( [R]/ [R]₀) = -0.8

[R]/ [R]₀ = 0.158

Hence, 0.158 fraction of sucrose remains.

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6 months ago

Ncert Solutions Chemistry Class 12th Chemical Kinetics

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

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Payal Gupta

Contributor-Level 10

4.33 Given,

k = 2.0 * 10^–2s^-1

time t = 100s

Concentration [A₀] = 1.0 mol L^-1

We know,

On substituting the values, Log (1/ [A]) = 2.303/2

Log [A] = -2.303/2 [A] = 0.135 mol L^–1

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6 months ago

Ncert Solutions Chemistry Class 12th Chemical Kinetics

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

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Payal Gupta

Contributor-Level 10

4.32 Given,

k = 2.418 * 10^-5 s^-1

T = 546 K

E_a = 179.9 kJ mol^-1 = 179.9 * 10³J mol^-1

The Arrhenius equation is given by k = Ae^-Ea/RT Taking natural log on both sides,

Ln k = ln A- (E_a/RT) Substituting the values,

ln (2.418 * 10^-5 ) = ln A-179.9/ (8.314 * 546)

ln A = 12.5917

A = 3.9 * 10¹² s^-1 (approximately)

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6 months ago

Ncert Solutions Chemistry Class 12th Chemical Kinetics

4.31 The rate constant for the decomposition of N₂O₅ at various temperatures is given below:

Draw a graph between ln k and 1/T and calculate the values of A and Ea . Predict the rate constant at 30° and 50°C.

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Payal Gupta

Contributor-Level 10

4.31 To convert the temperature in °C to °K we add 273 K.

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 = (y₂-y₁)/ (x₂-x₁)

By substituting the values, slope = -12.301

? –E_a/R = -12.301

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

? _aE= 8.314 JK^-1mol^-1 * 12.301 K

? _aE= 102.27 kJ mol^-1

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

(? At T = 273K, ln k =-7.147)

On solving, we get ln A = 37.

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