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  • Which of the following is a correct expression for the integrated rate law of first order reaction: A→Products? Here, a = [A]0 = Initial concentration of the reactant A<br /

Which of the following is a correct expression for the integrated rate law of first order reaction: A→Products?
Here,
a = [A]0 = Initial concentration of the reactant A
(a−x) = [A] = Concentration of the reactant A at time t

Option: 1

[\mathrm{A}]=[\mathrm{A}]_0-\mathrm{kt}


Option: 2

\mathrm{[\mathrm{A}]=[\mathrm{A}]_0 \mathrm{e}^{-k t}}


Option: 3

[\mathrm{A}]_0=[\mathrm{A}] \mathrm{e}^{-\mathrm{kt}}


Option: 4

[\mathrm{A}]=[\mathrm{A}]_0+\mathrm{kt}


Answers (1)

best_answer

For a first order reaction:
\begin{matrix} &A \rightarrow Products \\ t=0 & a \\ t=t & a-x \end{matrix}
a is intial concentration i.e. \mathrm{[A]_0 } and (a-x) is the concentration at time t i.e. \mathrm{[A] }
\mathrm{ -\frac{d[A]}{d t}=k[A] }

\mathrm{ -\frac{d(a-x)}{d t}=k(a-x)}

a is a constant value
\mathrm{ \frac{d x}{d t}=k(a-x) \\ }

\mathrm{ \frac{d x}{a-x}=k \cdot d t \\ }

\mathrm{ -\int_0^x \frac{d x}{a-x}=k \int_0^t d t }

\mathrm{\ln \frac{a}{a-x}=k t \\ }

\mathrm{ 2.303 \log \frac{a}{a-x}=k t }

This is the integrated rate law of first order reaction
\mathrm{ \ln \frac{[A]_0}{[A]}=k t \\ }

\mathrm{ \Rightarrow[A]=[A]_0 e^{-k t} }

Posted by

Divya Prakash Singh

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