For the reaction $\mathrm{A} \rightarrow \mathrm{B}$ the following graph was obtained. The time required (in seconds) for the concentration of A to reduce to $2.5 \mathrm{~g} \mathrm{~L}^{-1}$ (if the initial concentration of A was $50 \mathrm{~g} \mathrm{~L}^{-1}$) is _______ (Nearest integer)
Given : $\log 2=0.3010$

Question

Aicharya · Accepted Answer

As it is difficult to predict order using data provided in graph.
For specific time interval $0-5 \mathrm{sec}, 5-10 \mathrm{sec}$ and $10-15 \mathrm{sec}$. order comes to be zero, but graph is not a straight line.
Assuming $1^{\text {st }}$ order kinetics
$\begin{aligned}
& \mathrm{K}=\frac{1}{\mathrm{t}} \ln \frac{\mathrm{~A}_0}{\mathrm{~A}_{\mathrm{t}}} \\
& \mathrm{~K}=\frac{1}{10} \ln \frac{40}{20}
\end{aligned}$
Time required to reduce to $2.5 \mathrm{~g} / \mathrm{L}$
$\begin{aligned}
& \mathrm{K}=\frac{1}{\mathrm{t}} \ln \frac{50}{2.5} \\
& \frac{1}{10} \ln 2=\frac{1}{\mathrm{t}} \ln 20 \\
& \mathrm{t}=\frac{1.3010 \times 10}{0.3010}=43.3 \mathrm{sec}
\end{aligned}$

For the reaction $\mathrm{A} \rightarrow \mathrm{B}$ the following graph was obtained. The time required (in…

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