The reaction $2 \mathrm{~A} \rightarrow 2 \mathrm{~B}+\mathrm{C}$ has a rate constant of $1.2 \times 10^{-2}…
The reaction $2 \mathrm{~A} \rightarrow 2 \mathrm{~B}+\mathrm{C}$ has a rate constant of $1.2 \times 10^{-2} \mathrm{~s}^{-1}$. Which of the following is correct?
Plot of $[\mathrm{A}]$ vs $" \frac{1}{t}$ " will be straight line
Plot of $\frac{1}{[\mathrm{~A}]}$ vs $t^2$ will be a straight line
Plot of $\ln [\mathrm{A}]$ vs $t$ will be a straight line
Plot of [A] vs $t^2$ will be a straight line
Solution
$\mathrm{K}=1.2 \times 10^{-2} \mathrm{~s}^{-1}$; This is a 1 st order reaction. Hence, $[\mathrm{A}]=[\mathrm{A}]_0 \mathrm{e}^{-\mathrm{kt}} ; \ln [\mathrm{A}]=\ln [\mathrm{A}]_0-\mathrm{kt}$ $\therefore \quad \ln [\mathrm{A}]$ vs $\mathrm{t}$ is a straight line with negative slope $(\mathrm{k})$.