Given above is the concentration vs time plot | vedclass

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(B) From the graph,at $t = 0 \ min$,$[A] = 0.05 \ M$ and $[B] = 0 \ M$.At $t = 10 \ min$,$[A] = 0.04 \ M$ and $[B] = 0.03 \ M$.The change in concentra

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$3$

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(B) From the graph,at $t = 0 \ min$,$[A] = 0.05 \ M$ and $[B] = 0 \ M$.At $t = 10 \ min$,$[A] = 0.04 \ M$ and $[B] = 0.03 \ M$.The change in concentration of $A$ is $\Delta[A] = 0.05 - 0.04 = 0.01 \ M$.The change in concentration of $B$ is $\Delta[B] = 0.03 - 0 = 0.03 \ M$.According to the stoichiometry of the reaction $A ightarrow nB$,the rate of disappearance of $A$ and appearance of $B$ are related as: $\Delta[B] = n 	imes \Delta[A]$.Substituting the values: $0.03 = n 	imes 0.01$.Therefore,$n = \frac{0.03}{0.01} = 3$.

Given above is the concentration vs time plot for a dissociation reaction: $A \rightarrow nB$. Based on the data of the initial phase of the reaction (initial $10 \ min$),the value of $n$ is . . . . . . .

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