For the reaction A arrow B the following graph was obtained. The...

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

[We can assume in a hypothetical situation that the graph is correct, assuming its first order reaction, in reality the question is wrong.]
For the reaction A arrow B the following graph was obtained. The time required (in seconds) for the concentration of A

For the reaction A arrow B the following graph was obtained. The time required (in seconds) for the concentration of A

Correct answer

Step-by-step solution

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}<br />& \mathrm{K}=\frac{1}{\mathrm{t}} \ln \frac{\mathrm{~A}_0}{\mathrm{~A}_{\mathrm{t}}} \\<br>& \mathrm{~K}=\frac{1}{10} \ln \frac{40}{20}<br />\end{aligned}

Time required to reduce to

2.5 \mathrm{~g} / \mathrm{L}

\begin{aligned}<br />& \mathrm{K}=\frac{1}{\mathrm{t}} \ln \frac{50}{2.5} \\<br>& \frac{1}{10} \ln 2=\frac{1}{\mathrm{t}} \ln 20 \\<br>& \mathrm{t}=\frac{1.3010 \times 10}{0.3010}=43.3 \mathrm{sec}<br />\end{aligned}

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