A_{(g)} rightarrow 2B_{(g)}. Initially,2 mol | vedclass

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(A) For the reaction $A_{(g)} \rightarrow 2B_{(g)}$,let the initial concentration of $A$ be $[A]_0 = \frac{2 \ mol}{5 \ L} = 0.4 \ M$.Let $x$ be the a

Vedclass · Accepted Answer

$20$

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(A) For the reaction $A_{(g)} \rightarrow 2B_{(g)}$,let the initial concentration of $A$ be $[A]_0 = \frac{2 \ mol}{5 \ L} = 0.4 \ M$.Let $x$ be the amount of $A$ reacted at time $t = 20 \ min$.At $t = 20 \ min$,$[A]_t = 0.4 - x$ and $[B]_t = 2x$.Given that $[A]_t = \frac{[B]_t}{2}$,we have $0.4 - x = \frac{2x}{2} = x$.Thus,$2x = 0.4$,which means $x = 0.2 \ M$.At $t = 20 \ min$,$[A]_t = 0.4 - 0.2 = 0.2 \ M$.Since $[A]_t = \frac{[A]_0}{2}$,the time $t = 20 \ min$ is exactly the half-life $t_{1/2}$ of the reaction.Therefore,the half-life time is $20 \ min$.

$A_{(g)} \rightarrow 2B_{(g)}$. Initially,$2 \ mol$ of $A$ are taken in a $5 \ L$ vessel. After $20 \ min$,$[A]_t = \frac{[B]_t}{2}$. Find the half-life time of $A$ in the first-order reaction in $\min$.

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