After how many seconds will the concentration of a reactant in a reaction is halved in a first order reaction, if the rate constant is $6.93 \times 10^{- 3} s e c^{- 1}$ :

10^{4}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

100

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

10^{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B 100
For $1^{s t}$ order reaction,

$t_{^{1} /_{2}} = \frac{l n 2}{K}$

$= \frac{l n 2}{6.93 \times 10^{- 3}}$

$t_{^{1} /_{2}} = 100 s e c$
Hence, the correct option is $B$

Suggest Corrections

Similar questions

Q. After how many seconds will the concentration of the reactant in a first order reaction be halved, if the rate constant is 1.155x10

^{- 3} s e c^{- 1}

Q. In a first order reaction, the initial concentrated of the reactant was

M / 10.

After

8

minutes

20

seconds the concentrated becomes

M / 100.

. What is the rate constant?

Q. In a first order reaction, molar concentration of a reactant decreases from 0.1 to 0.01 in 100 seconds. The rate constant of the reaction is:

Q. A first order reaction has a rate constant

1.5 \times 10^{- 3} s e c^{- 1}

. How long will 5.0 g of this reactant take to reduce to 1.25 g.

In a first order reaction, the concentration of the reactant decreases from 800 mol $d m^{- 3} i n 2 \times 10^{4} s .$ The rate constant of the reaction $i n (s^{- 1})$ is

(IIT-JEE, 2003)

Join BYJU'S Learning Program

After how many seconds will the concentration of a reactant in a reaction is halved in a first order reaction, if the rate constant is 6.93×10−3sec−1:

The correct option is B 100For 1st order reaction,t1/2=ln 2K=ln 26.93×10−3t1/2=100 secHence, the correct option is B

After how many seconds will the concentration of a reactant in a reaction is halved in a first order reaction, if the rate constant is $6.93 \times 10^{- 3} s e c^{- 1}$ :

The correct option is B 100
For $1^{s t}$ order reaction,

$t_{^{1} /_{2}} = \frac{l n 2}{K}$

$= \frac{l n 2}{6.93 \times 10^{- 3}}$

$t_{^{1} /_{2}} = 100 s e c$
Hence, the correct option is $B$