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:

2.303

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

0.02303

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

0.2303

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

0.002303

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

Open in App

Solution

The correct option is B 0.02303
The expression for the rate constant is
$k = \frac{2.303}{t} l o g \frac{a}{a - x}$

$k = \frac{2.303}{100 s} l o g \frac{0.1}{0.01}$

$k = \frac{2.303}{100 s} l o g 10$

$k = \frac{2.303}{100 s} \times 1$

$k = 0.02303 / s$
Hence, the rate constant is 0.02303 /s

Suggest Corrections

Similar questions

Q. In a first order reaction the concentration of reactant decreases from 800

m o l / d m^{3}

50 m o l / d m^{3}

2 \times 10^{4} s e c

. The rate constant of reaction in

s e c^{- 1}

is:

In a first-order reaction the concentration of the reactant is decreased from $1.0 M to 0 .25 M$ in 20 min.The rate constant of the reaction would be:

In a first order reaction the concentration of reactant decreases from $800 m o l / d m^{3}$ to $50 m o l / d m^{3}$ in $2 \times 10^{4} s$ . The rate constant of reaction in $s^{- 1}$ is:

l n

a first order reaction the concentration of reactant decreases from 800

m o l / d m^{3}

to 50

m o l / d m^{3}

2 \times 10^{4}

sec. The rate constant of reaction in

{sec}^{- 1}

is:

Q. In a first-order reaction, the concentration of the reactant is reduced to one-fourth of its initial value in

50

seconds. Calculate the rate constant of the reaction.

Join BYJU'S Learning Program

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:

The correct option is B 0.02303The expression for the rate constant isk=2.303tlogaa−xk=2.303100slog0.10.01k=2.303100slog10k=2.303100s×1k=0.02303/sHence, the rate constant is 0.02303 /s

The correct option is B 0.02303
The expression for the rate constant is
$k = \frac{2.303}{t} l o g \frac{a}{a - x}$

$k = \frac{2.303}{100 s} l o g \frac{0.1}{0.01}$

$k = \frac{2.303}{100 s} l o g 10$

$k = \frac{2.303}{100 s} \times 1$

$k = 0.02303 / s$
Hence, the rate constant is 0.02303 /s