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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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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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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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11 months ago

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.

...more

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11 months ago

Class 12th Chemical Kinetics

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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Payal Gupta

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 /

...more

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11 months ago

Class 12th Chemical Kinetics

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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New answer posted

11 months ago

Class 12th Chemical Kinetics

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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Payal Gupta

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

...more

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11 months ago

Class 12th Relations and Functions

74. Number of binary operations on the set {a, b} are:

(A) 10

(B) 16

(C) 20

(D) 8

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Vishal Baghel

Contributor-Level 10

A binary operation * on {a, b} is a function from {a, b} * {a, b} → {a, b}

i.e., * is a function from { (a, a), (a, b), (b, a), (b, b)} → {a, b}.

Hence, the total number of binary operations on the set {a, b} is 2⁴ i.e., 16.

The correct answer is B.

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11 months ago

Class 12th Relations and Functions

73. Let R → R be the Signum Function defined as $f (x) = {\begin{matrix} 1 & x > 0 \\ 0 & x = 0 \\ - 1 & x < 0 \end{matrix}$ and R → R be the Greatest Function given by g(x) = [x] where [x] is greatest integer less than or equal to x

Then, does fog and gof coincide in (0, 1)?

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Vishal Baghel

Contributor-Level 10

It is given that,

$f : R \to R$ is defined as $f (x) = {\begin{matrix} 1 & x > 0 \\ 0 & x = 0 \\ - 1 & x < 0 \end{matrix}$

Also, $g : R \to R$ is defined as $g (x) = [x]$ , where [x] is the greatest integer less than or equal to x.

Now, let $x \in (0, 1)$

Then, we have:

$[x] = 1$ if $x = 1$ and $[x] = 0$ if $0 < x < 1$

$\begin{array}{l} ∴ f o g (x) = f (g (x)) = f ([x]) = {\begin{matrix} f (1) & i f, x = 1 \\ f (0) & i f, x \in (0, 1) \end{matrix} = {(1, " i f, x = 1 "), (0, : i f, x \in (0, 1) ") :} \\ g o f (x) = g (f (x)) \\ = g (1) [x > 0] \\ = [1] = 1 \end{array}$

Thus, when $x \in (0, 1)$ , we have $f o g (x) = 0 a n d, g o f (x) = 1 .$

Hence, fog and gof do not coincide in (0, 1).

Therefore, option (B) is correct.

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11 months ago

Class 12th Relations and Functions

72. Let A = {1, 2, 3}. Then number of equivalence relations containing (1, 2) is:

(A) 1

(B) 2

(C) 3

(D) 4

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Vishal Baghel

Contributor-Level 10

2, Therefore, option (B) is correct.

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