For a given reaction R arrow P, t_ 1/2 is related to [A]_0 as given in table. Given: 2=0.30 Which of the following is true? A. The order of the reaction is 1 / 2 . B. If [A]_0 is 1 M , then t_ 1/2 is 200 10 ~min C. The

Correct answer: B. A, B and D Only (t_1)/(2) (C_0)^ 1- aligned & (t_1)/(t_2)=((C_1)/(C_2))^ 1- & (200)/(100)=((0.100)/(0.025))^ 1- & 2=(4)^ 1- & (1- )=(1)/(2) & =(1)/(2) & For =(1)/(2) & (-d A)/(d t)=k(A)^ (1)/(2) & ∫_ C_0 ^C d A (A)^ (1)/(2) =-∫_0^t k d t & 2 ~A^ (1)/(2) =-kt & C - C_0 = -kt 2 & C - C_0 - kt 2 & For C_0=0.1 t_ (1)/(2) =200 ~min & (c_0)/(2) = c_0 -(k t)/(2) & kt 2 = c_0 - c_0 2 & kt 2 = c_0 ( 2 -1 2 ) & t_ (1)/(2) = 2 c_0 k ( 2 -1 2 ) & 200= 2 0.1 k ( 2 -1 2 ) aligned k= 0.1 100 ( 2 -1 2 ) For C_0=1 M t_ (1)/(2) = 2 × 100( 2 ) 0.1 ( 2 -1) × ( 2 -1) 2 200 10 ~min . B is correct C is incorrect For C_0=1.6 M aligned & t_ (1)/(2) = 2 1.6 ( 2 )( 2 -1) × 100 0.1 ( 2 -1)( 2 ) & t_ (1)/(2) =400 × 2 ~min & t_ (1)/(2) =800 ~min aligned

JEE Main 202528 JanuaryMorning Shift

Exam: JEE Main
Year: 2025
Date: 28 January
Shift: Morning Shift
Subject: Chemistry
Chapter: Chemical Kinetics

JEE Main 2025 · 28 January · Morning Shift · Chemistry · Chemical Kinetics

For a given reaction R arrow P, t_ 1/2 is related to [A]_0 as given...

For a given reaction

\mathrm{R} \rightarrow \mathrm{P}, \mathrm{t}_{1/2}

is related to

[\mathrm{A}]_0

as given in table. Given:

\log 2=0.30

Which of the following is true? A. The order of the reaction is

1 / 2

. B. If

[\mathrm{A}]_0

is 1 M , then

\mathrm{t}_{1/2}

200 \sqrt{10} \mathrm{~min}

C. The order of the reaction changes to 1 if the concentration of reactant changes from 0.100 M to 0.500 M . D.

\mathrm{t}_{1/2}

is 800 min for

[\mathrm{A}]_0=1.6 \mathrm{M}

Choose the correct answer from the options given below: Options

Options

AA and C Only
BA, B and D Only
CC and D Only
DA and B Only

Correct answer

B. A, B and D Only

Step-by-step solution

\frac{t_1}{2} \propto\left(C_0\right)^{1-\eta}

\begin{aligned} & \frac{t_1}{t_2}=\left(\frac{C_1}{C_2}\right)^{1-\eta} \\ & \Rightarrow \frac{200}{100}=\left(\frac{0.100}{0.025}\right)^{1-\eta} \\ & \Rightarrow 2=(4)^{1-\eta} \\ & (1-\eta)=\frac{1}{2} \\ & \eta=\frac{1}{2} \\ & \text{ For } \eta=\frac{1}{2} \\ & \frac{-d A}{d t}=k(A)^{\frac{1}{2}} \\ & \int_{C_0}^C \frac{d A}{(A)^{\frac{1}{2}}}=-\int_0^t k d t \\ & \Rightarrow 2 \mathrm{~A}^{\frac{1}{2}}=-\mathrm{kt} \\ & \Rightarrow \sqrt{\mathrm{C}}-\sqrt{\mathrm{C}_0}=\frac{-\mathrm{kt}}{2} \\ & \sqrt{\mathrm{C}}-\sqrt{\mathrm{C}_0}-\frac{\mathrm{kt}}{2} \\ & \text{ For } \mathrm{C}_0=0.1 \Rightarrow \mathrm{t}_{\frac{1}{2}}=200 \mathrm{~min} \\ & \sqrt{\frac{c_0}{2}}=\sqrt{c_0}-\frac{k t}{2} \\ & \frac{\mathrm{kt}}{2}=\sqrt{\mathrm{c}_0}-\sqrt{\frac{\mathrm{c}_0}{2}} \\ & \frac{\mathrm{kt}}{2}=\sqrt{\mathrm{c}_0}\left(\frac{\sqrt{2}-1}{\sqrt{2}}\right) \\ & t_{\frac{1}{2}}=\frac{2 \sqrt{c_0}}{k}\left(\frac{\sqrt{2}-1}{\sqrt{2}}\right) \\ & 200=\frac{2 \sqrt{0.1}}{k}\left(\frac{\sqrt{2}-1}{\sqrt{2}}\right) \end{aligned}

\mathrm{k}=\frac{\sqrt{0.1}}{100}\left(\frac{\sqrt{2}-1}{\sqrt{2}}\right)

For

\mathrm{C}_0=1 \mathrm{M}t_{\frac{1}{2}}=\frac{2 \times 100(\sqrt{2})}{\sqrt{0.1}(\sqrt{2}-1)} \times \frac{(\sqrt{2}-1)}{\sqrt{2}}\Rightarrow 200 \sqrt{10} \mathrm{~min}

.

\Rightarrow B

is correct C is incorrect For

\mathrm{C}_0=1.6 \mathrm{M}

\begin{aligned} & \mathrm{t}_{\frac{1}{2}}=\frac{2 \sqrt{1.6}(\sqrt{2})(\sqrt{2}-1) \times 100}{\sqrt{0.1}(\sqrt{2}-1)(\sqrt{2})} \\ & \mathrm{t}_{\frac{1}{2}}=400 \times 2 \mathrm{~min} \\ & \mathrm{t}_{\frac{1}{2}}=800 \mathrm{~min}\end{aligned}

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