For the second-order reaction below,the initial concentration of A is 1.00 M.If the concentration of A is reduced to 0.43 M after 75 seconds,what is the rate constant? $2 \mathrm{~A} \rightarrow \mathrm{B} \quad\quad\quad\quad\quad \text { rate }=k[\mathrm{~A}]^{2}$

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Quizplus · Accepted Answer

For a second-order reaction, the rate law is given by $\frac{1}{[\mathrm{A}]} = \frac{1}{[\mathrm{A}]_0} + kt$. Plugging in the values: $\frac{1}{0.43} = \frac{1}{1.00} + k(75)$, solving for $k$ gives approximately $1.8 \times 10^{-2}$ L/mol?s.

For the Second-Order Reaction Below,the Initial Concentration of a Is 2 A→B rate =k[ A]22 \mathrm{~A} \rightarrow \mathrm{B} \quad\quad\quad\quad\quad \text { rate }=k[\mathrm{~A}]^{2}2 A→B rate =k[ A]2

For the Second-Order Reaction Below,the Initial Concentration of a Is $2 \mathrm{~A} \rightarrow \mathrm{B} \quad\quad\quad\quad\quad \text { rate }=k[\mathrm{~A}]^{2}$