For the reaction A → B , the rate law expression is − dt d [ A ] = k [ A ] 1/2 . If initial concentration of [ A ] is [ A ] 0 , then

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For the reaction $A \to B$ , the rate law expression is $- \frac{d [ A ]}{dt} = k [A]^{1/2}$ . If initial concentration of $[A]$ is $[A]_{0}$ , then

Answer:

A, B, D

Solution:

$A \to B$

$\frac{- dA}{dt} = k [A]^{1/2}$

$\int \frac{d A}{( A ) ^{1/2}} = - k \int d t$

$2 (A)^{1/2} = - kt + C$

Condition

$A = A_{0}$ at $t = 0$

$c = 2 [A]_{0}^{1/2}$

$2 [A]^{1/2} = - kt + 2 [A_{0}]^{1/2}$

$k = \frac{2}{t} [A_{0}]^{1/2} - [A]^{1/2}]$

$t_{1/2} = \frac{2 [ A _{0} ] ^{1/2}}{k}$

Stream:JEESubject:ChemistryTopic:Chemical KineticsSubtopic:First Order Reactions

⏱ 2mℹ️ Source: QB

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