For the two reactions I : A →B; II: C→D following graph is obtained. Which of the following is true:

(1) True A → BInitial 0.5 0At eq. 0.25 0.25When [A] = [B] = 0.25 M[B]t=k1t=tan⁡30∘t=13t0.25=13t, ∴t=0.253.[D]t=kIIt=tan⁡60∘t=3t(t=0.253⋅) =3×0.25×3 =0.75M.(2) False if [C] = [A] then at that time [B] < [D](3) Falset100%=ak( for zero order )kI=1/3, kII=3t100%t100%=aIaII⋅kIIkI=0.51×31/3=32(4) [A]t=[A]0−kIt or [A]t=0.5−13t[C]t=[C]0−kIIt or [C]t=1−3t if [.4]=[C]t i.e., 0.5−13t=1−3t or 3−13t=0.5t=34min.

For the two reactions I : A →B; II: C→D following graph is obtained. Which of the following is true:

(1) True A → BInitial 0.5 0At eq. 0.25 0.25When [A] = [B] = 0.25 M[B]t=k1t=tan⁡30∘t=13t0.25=13t, ∴t=0.253.[D]t=kIIt=tan⁡60∘t=3t(t=0.253⋅) =3×0.25×3 =0.75M.(2) False if [C] = [A] then at that time [B] < [D](3) Falset100%=ak( for zero order )kI=1/3, kII=3t100%t100%=aIaII⋅kIIkI=0.51×31/3=32(4) [A]t=[A]0−kIt or [A]t=0.5−13t[C]t=[C]0−kIIt or [C]t=1−3t if [.4]=[C]t i.e., 0.5−13t=1−3t or 3−13t=0.5t=34min.

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

For the two reactions I : A $\to$ B; II: C $\to$ D following graph is obtained.
Which of the following is true:

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$[A] = [C] at t = \frac{\sqrt{3}}{2} min$

${(t_{100 %})}_{Reaction I} = {(t_{100 %})}_{Reaction II}$

If [B] = [A] then at that time [D] = 0.75

It [C] : [A] then at that time [B] > [D]

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

(1) True

A $\to$ B

Initial 0.5 0

At eq. 0.25 0.25

When [A] = [B] = 0.25 M

$\begin{array}{l} [B]_{t} = k_{1} t = \tan 30^{\circ} t = \frac{1}{\sqrt{3}} t \\ 0.25 = \frac{1}{\sqrt{3}} t, ∴ t = 0.25 \sqrt{3} . \\ [D]_{t} = k_{II} t = \tan 60^{\circ} t = \sqrt{3} t (t = 0.25 \sqrt{3} \cdot) \\ = \sqrt{3} \times 0.25 \times \sqrt{3} \\ = 0.75 M . \end{array}$

(2) False

if [C] = [A] then at that time [B] < [D]

(3) False

$\begin{array}{l} t_{100 %} = \frac{a}{k} (for zero order) \\ k_{I} = 1 / \sqrt{3}, k_{II} = \sqrt{3} \\ \frac{(t_{100 %})}{(t_{100 %})} = \frac{a_{I}}{a_{II}} \cdot \frac{k_{II}}{k_{I}} = \frac{0.5}{1} \times \frac{\sqrt{3}}{1 / \sqrt{3}} = \frac{3}{2} \end{array}$

(4) $[A]_{t} = [A]_{0} - k_{I} t or [A]_{t} = 0.5 - \frac{1}{\sqrt{3}} t$

$[C]_{t} = [C]_{0} - k_{II} t or [C]_{t} = 1 - \sqrt{3} t if [. 4] = [C]_{t}$

$\begin{array}{l} i.e., 0.5 - \frac{1}{\sqrt{3}} t = 1 - \sqrt{3} t \\ or (\sqrt{3} - \frac{1}{\sqrt{3}}) t = 0.5 \\ t = \frac{\sqrt{3}}{4} min . \end{array}$