JEE Main 2017 · Chemistry · Chemical Kinetics · medium

A first order reaction is 75% complete in 70 min at 300 K. Its activation energy is 50 kJ mol^-1 (R=8.314 J K^-1mol^-1, e^1.25≈3.5). The time…

Single correct option

A first order reaction is

75\%

complete in

70

min at

300\ \mathrm{K}

. Its activation energy is

50\ \mathrm{kJ\,mol^{-1}}

(

R=8.314\ \mathrm{J\,K^{-1}mol^{-1}}

e^{1.25}\approx3.5

). The time required for

87.5\%

completion at

320\ \mathrm{K}

is nearly

Show answer and solution

Answer: (C) $30$ min

75\%

completion is two half-lives, so

t_{1/2}(300)=35

min. Arrhenius:

\ln\dfrac{k_{320}}{k_{300}}=\dfrac{50000}{8.314}\Big(\dfrac1{300}-\dfrac1{320}\Big)=1.25

, so

k_{320}=3.5\,k_{300}

and

t_{1/2}(320)=35/3.5=10

min.

87.5\%

completion needs three half-lives:

3\times10=30

min. (

105

min is the answer at

300

10

min is a single half-life.)

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