Class 12th

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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a year 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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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 /

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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

a year 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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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

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a year ago

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

(A) 10

(B) 16

(C) 20

(D) 8

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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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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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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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a year ago

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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2, Therefore, option (B) is correct.

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71. Let A = {1, 2, 3}. Then number of relations containing (1, 2) and (1, 3) which are reflexive and symmetric but not transitive is:

(A) 1

(B) 2

(C) 3

(D) 4

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Contributor-Level 10

It is clear that 1 is reflexive and symmetric but not transitive.

Therefore, option (A) is correct.

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a year ago

70. Let A = {–1, 0, 1, 2}, B = {–4, –2, 0, 2} and f; g: A→B be the functions defined by f(x) = x² − x, x ∈ A and g(x) = 2x - 1/2 - 1. x∈ A. Are f and g equal? Justify your answer.

(Hint: One may note that two functions f: A → B and g: A → B such that f(a) = g(a) & mn For E; a ∈A, are called equal functions).

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It is given that A = {–1, 0, 1, 2}, B = {–4, –2, 0, 2}

Also, it is given that $f, g : A \to B$ are defined by $f (x) = x^{2} - x, x \in A$ and $g (x) = 2 x - \frac{1}{2} - 1, x \in A$ .

It is observed that:

$\begin{array}{l} f (- 1) = (1^{2}) - (- 1) = 1 + 1 = 2 \\ g (- 1) = 2 (- 1) - \frac{1}{2} - 1 = 2 (\frac{3}{2}) - 1 = 3 - 1 = 2 \\ \Rightarrow f (- 1) = g (- 1) \\ f (0) = {(0)}^{\land} 2 - 0 = 0 \\ g (0) = 2 (0) - \frac{1}{2} - 1 = 2 (\frac{1}{2}) - 1 = 1 - 1 = 0 \\ \Rightarrow f (0) = g (0) \\ f (1) = {(1)}^{\land} 2 - 1 = 1 - 1 = 0 \\ g (1) = 2 a - \frac{1}{2} - 1 = 2 (\frac{1}{2}) - 1 = 1 - 1 = 0 \\ \Rightarrow f (1) = g (1) \\ f (2) = {(2)}^{\land} 2 - 2 = 4 - 2 = 2 \\ g (2) = 2 (2) - \frac{1}{2} - 1 = 2 (\frac{3}{2}) - 1 = 3 - 1 = 2 \\ \Rightarrow f (2) = g (2) \\ ∴ f (a) = g (a) \forall a \in A \end{array}$

Hence, the functions f and g are equal.

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69. Define binary operation * on the set {0, 1, 2, 3, 4, 5} as $a * b = {\begin{matrix} a + b & i f, a + b < 6 \\ a + b - 6 & f, a + b \geq 6 \end{matrix}$

Show that zero is the identity for this operation and each element a ≠ 0 of the set is invertible with 6 - abeing the inverse of a

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