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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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11 months ago

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

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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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11 months ago

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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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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11 months ago

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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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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11 months ago

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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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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11 months ago

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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I have 58 marks in MHT CET PCM. What is my percentile?

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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11 months ago

MHTCET

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Selva Priya

Beginner-Level 4

Actually percentile isn't calculated by marks it's based on your performance compared to other candidates.

Also, to know the exact percentile we need the official result data of your session but still we can calculate by checking previous year results and it would come around 70-80% and also it may vary based on other candidates results.

If you have further queries feel free to connect :)

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11 months ago

4.30 The following data were obtained during the first order thermal decomposition of SO₂Cl₂ at a constant volume. SO₂Cl_2(g)→ SO₂_(g) + Cl₂_(g)

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

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Contributor-Level 10

4.30 When t = 0, the total partial pressure is P₀ = 0.5 atm

When time t = t, the total partial pressure is P_t = P₀ + p

P₀-p = P_t-2p, but by the above equation, we know p = P_t-P₀

Hence, P₀-p = P_t-2 (P_t-P₀)

Thus, P₀-p = 2P₀ – P_t

We know that time

t= 2.303/K log R₀ / 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 P₀ / P₀ - P

t= 2.303/K log P₀ / 2P₀ - P_t

? equation 1

At time t = 100 s, P_t = 0.6 atm and P₀ = 0.5 atm,

Substituting in equation 1,

100 = 2.303/k log 0.5 /

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11 months ago

4.30 The following data were obtained during the first order thermal decomposition of SO₂Cl₂ at a constant volume. SO₂Cl_2(g)→ SO₂_(g) + Cl₂_(g)

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11 months ago

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

Calculate the rate constant.

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Lakehead University Thompson Rivers University

Contributor-Level 10

4.29 When t = 0, the total partial pressure is P₀ = 35.0 mm of Hg

When time t = t, the total partial pressure is P_t = P₀ + p

P₀-p = P_t-2p, but by the above equation, we know p = P_t-P₀ Hence, P₀-p = P_t-2 ( P_t-P₀)

Thus, P₀-p = 2P₀ – P_t

We know that time

t= 2.303/K log R₀ / 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 P₀ / P₀ - P

t= 2.303/K log P₀ / 2P₀ - P_t

? equation 1

At time t = 360 s, P_t = 54 mm of Hg and P₀ = 30 mm of Hg, Substituting in equation 1,

360 = 2

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11 months ago

Which one is better: Lakehead University or Thompson Rivers University?

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PRIYANKA

Contributor-Level 10

Both Lakehead University as well as Thompson Rivers University offer globally recognised degrees. The differences are mentioned in the below table.

Parameters	Lakehead University	Thompson Rivers University
World University Ranking	#1241	#2294
Ranking for Programs	#325	#60
Ranked in Canada	#36	#55
Acceptance Rate	Around 82-90%	Around 82-84%
Top Majors	Psychology, Civil Engineering, Computer Science, Mechanical Engineering, Computational Science	Arts, Law, Nursing, Science, Trades and Technology
Duration of Courses	UG: 3 - 4 years PG: 1 - 2 years	UG: 4 - 5 years PG: 16 months - 2 years
Annual Tuition Cost	UG: INR 19 L - 25 L PG: INR 4 L - 9.2 L	UG: INR 10 L - 15 L PG: INR 9.3L - 50.8L
Application Fees	INR 8.3K	INR 8.5K
Scholarships	UG: 3 PG: 2	6

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11 months ago

What are the specialised courses offered by the top distance BA colleges in Bangalore?

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Contributor-Level 10

There are many specialised courses offered by the top distance BA colleges in Bangalore. Some of the courses are mentioned below:

Top Specialisations	No. of Colleges
English	4
Languages	4
Economics	3
History	3
Political Science	3

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11 months ago

Are there any private distance BA colleges in Bangalore?

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Contributor-Level 10

Yes, there are many private distance BA colleges in Bangalore. Some of them are mentioned below along with their tuition fees:

Private Colleges	Tuition Fee
Wisdom School of Management, Bangalore Admission	INR 20,000
Bright International Educational Trust, Jayanagar Admission	INR 8,000

Disclaimer: This information is sourced from official website and may vary.

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11 months ago

Are there any affordable distance BA colleges in Bangalore?

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Contributor-Level 10

Yes, there are many affordable distance BA colleges in Bangalore. Some of them are mentioned below along with their tuition fees of less than INR 1 lakh:

Low-cost Colleges	Tuition Fee
Jamia Millia Islamia - Centre for Distance and Open Learning Fees	INR 7,000
Aaruhi Institute Fees	INR 45,000
Wisdom School of Management, Bangalore Fees	INR 20,000
Geetanjali Institute Fees	INR 45,000
Bright International Educational Trust, Jayanagar Fees	INR 8,000

Disclaimer: This information is sourced from official website and may vary.

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11 months ago

What is the admission process of distance BA colleges in Bangalore?

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Which is the best distance BA colleges in Bangalore?

Contributor-Level 10

Students can check the basic eligibility criteria below for Distance BA course in Bangalore:

Students must have passed Class 12 or equivalent in any stream from a recognised board of India.
Many popular Distance BA colleges require at least 50% marks in Class 12 for accepting admission and may vary.
There are no age limit criteria for Distance BA course.

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11 months ago

B.A Karnataka

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How many distance BA colleges are there in Bangalore?

Contributor-Level 10

As per popularity basis, listed below are the top distance BA colleges in Bangalore along with their tuition fees:

Top Colleges	Tuition Fee
Directorate of Correspondence Courses and Distance Education, Bangalore University	INR 14,000
Center For Distance Education and Virtual Learning, Jain Deemed to be University	INR 27,000 – INR 38,000
Sri Sai Correspondence College	-
Jamia Millia Islamia - Centre for Distance and Open Learning	INR 7,000
Mysore Correspondence College	-

Disclaimer: This information is sourced from official website and may vary.

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11 months ago

B.A Karnataka

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