Class 12th

13. $I f s i n^{- 1} x = y, t h e n$

$\begin{array}{l} (A) 0 \leq y \leq π (B) - \frac{π}{2} \leq y \leq \frac{π}{2} \\ (C) 0 < y < π (D) - \frac{π}{2} < y < \frac{π}{2} \end{array}$

0 Follower 3 Views

Contributor-Level 10

Given, Sin⁻¹x=y.

(E) We know that the principal value branch of Sin⁻¹ is

$[\frac{- π}{2}, \frac{π}{2}]$ Hence, $\frac{- π}{2}$ ≤ y ≤ $\frac{π}{2}$

Option B is correct.

0 0

New answer posted

11 months ago

12. $c o s^{- 1} \frac{1}{2} + 2 s i n^{- 1} \frac{1}{2}$

0 Follower 2 Views

Contributor-Level 10

cos⁻¹ $\frac{1}{2}$ + 2Sin⁻¹ $\frac{1}{2}$ = cos⁻¹ $(c o s \frac{π}{3})$ + 2*Sin⁻¹ $(s i n \frac{π}{6})$

= $\frac{π}{3} + 2 * \frac{π}{6}$

= $\frac{π}{3} + \frac{π}{3}$

= $\frac{2 π}{3}$

0 0

New answer posted

11 months ago

11. $t a n^{- 1} (1) + c o s^{- 1} - \frac{1}{2} + s i n^{- 1} - \frac{1}{2}$

0 Follower 10 Views

Contributor-Level 10

tan⁻¹ (1) + cos⁻¹ $(\frac{1}{2})$ + sin ⁻¹ $(\frac{- 1}{2})$

0 0

New answer posted

11 months ago

5. $c o s^{- 1} (- \frac{1}{2})$

0 Follower 4 Views

Contributor-Level 10

Let cos ^-1 $(\frac{- 1}{2})$ =y Then cos y = $\frac{- 1}{2}$ = − cos
$\frac{π}{3}$ $\frac{}{}$ cos $(π - \frac{x}{3})$

= cos $\frac{3 π - π}{3}$

= cos $\frac{2 π}{3}$

(E) We know that the range of principal value

branch of cos⁻¹ is [0, $π$ ] and cos $\frac{2 x}{3}$ = $\frac{- 1}{2}$

Principal value of cos−1 $(\frac{- 1}{2})$ is $\frac{2 x}{3}$

0 0

New answer posted

11 months ago

1. $s i n^{- 1} (- \frac{1}{2})$

0 Follower 2 Views

Contributor-Level 10

Let sin⁻¹ $(\frac{- 1}{2})$ =y. Then, sin y=- $\frac{- 1}{2}$

We know that the range of principal value branch of sin⁻¹ is $- \frac{π}{2},$ $- \frac{π}{2}$

and sin⁻¹ $(\frac{- 1}{2})$ =−sin⁻¹ $\frac{- 1}{2}$ (sin (-x) = -sin x)

= $(- \frac{π}{6})$ = sin y (as sin
$\frac{π}{6}$
= $\frac{1}{2}$ )

Principal value of sin⁻¹ $(- \frac{1}{2})$ is $(- \frac{π}{6})$

0 0

New answer posted

11 months ago

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.

0 Follower 9 Views

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

0 0

New answer posted

11 months ago

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.

0 Follower 8 Views

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 =

...more

0 0

New answer posted

11 months ago

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?

0 Follower 36 Views

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

0 0

New answer posted

11 months ago

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?

0 Follower 89 Views

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 =

...more

0 0

New answer posted

11 months ago

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

0 Follower 23 Views