Given: 2A rightarrow B. The rate constant of | vedclass

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(B) For the reaction $2A \rightarrow B$,the rate of reaction is given by $Rate = -\frac{1}{2} \frac{d[A]}{dt} = \frac{d[B]}{dt} = K[A]$.Given $K = 10^

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Half-life period $t_{0.5}$ is equal to $70 \ min$

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(B) For the reaction $2A ightarrow B$,the rate of reaction is given by $Rate = -\frac{1}{2} \frac{d[A]}{dt} = \frac{d[B]}{dt} = K[A]$.Given $K = 10^{-2} \ min^{-1}$,this is a first-order reaction with respect to $A$.$(1)$ The rate of formation of $B$ is $\frac{d[B]}{dt} = K[A]$. Since $K$ is the rate constant of the reaction,$K_B$ (rate of formation of $B$) depends on the concentration of $A$,so option $A$ is incorrect.$(2)$ Half-life $t_{0.5} = \frac{\ln 2}{K} = \frac{0.7}{10^{-2}} = 70 \ min$. Thus,option $B$ is correct.$(3)$ For a first-order reaction,$t_{0.75} = 2 	imes t_{0.5} = 2 	imes 70 = 140 \ min$. Thus,option $C$ is incorrect.

Given: $2A \rightarrow B$. The rate constant of the reaction is $K = 10^{-2} \ min^{-1}$. Which statement is correct? $(\ln 2 = 0.7)$

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