For the reaction $A + 2 B \to$ products (started with concentrations taken in stoichiometric proportion), the experimentally determined rate law is :

$- \frac{d [A]}{d t} = k \sqrt{[A]} \sqrt{[B]}$

The half life time of the reaction would be:

\frac{0.693}{k}

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

\frac{0.693}{1 / k}

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

\frac{0.693}{\sqrt{2} k}

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

not defined

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

Open in App

Solution

The correct option is C $\frac{0.693}{\sqrt{2} k}$
$A a - x + 2 B 2 a - 2 x \to p r o d u c t s - \frac{d [A]}{d t} = k \sqrt{[A]} \sqrt{[B]} R e a c t a n t a r e i n t h e i r s t o i c h i o m e t r i c p r o p o r t i o n \Rightarrow - \frac{d (a - x)}{d t} = k \sqrt{(a - x)} \sqrt{2 (a - x)} \frac{d x}{d t} = \sqrt{2} k (a - x) t_{1 / 2} = \frac{0.693}{\sqrt{2} k}$

Suggest Corrections

Similar questions

Q. Reaction

A + B \to C + D

follows following rate law.

rate = k [A]^{\frac{1}{2}} [B]^{\frac{1}{2}}

Starting with initial concentration of one mole of

A

and

B

each, what is the time taken for amount of

A

to become

0.25

mole.
[Given:

k = 2.31 \times 10^{- 3} {sec}^{- 1}, l n 2 = 0.693

]

Q. If the half life period of a reaction is

0.693 k

, its order will be

Match the following
Order of reaction Half-life period
1)1st order reaction a) $(k a_{0})^{- 1}$
2) Second order reaction b) $\frac{0.693}{k}$
3) Zero order reaction c) $\frac{a_{0}}{2 k}$

Q. If the rate constant for a first order reaction is k, the time (t) required for the completion of 99% of the reaction is given by:

Join BYJU'S Learning Program

For the reaction A+2B→ products (started with concentrations taken in stoichiometric proportion), the experimentally determined rate law is :−d[A]dt=k√[A]√[B]The half life time of the reaction would be:

The correct option is C 0.693√2kAa−x+2B2a−2x→products−d[A]dt=k√[A]√[B]Reactantareintheirstoichiometricproportion⇒−d(a−x)dt=k√(a−x)√2(a−x)dxdt=√2k(a−x)t1/2=0.693√2k

For the reaction $A + 2 B \to$ products (started with concentrations taken in stoichiometric proportion), the experimentally determined rate law is :

$- \frac{d [A]}{d t} = k \sqrt{[A]} \sqrt{[B]}$

The half life time of the reaction would be: