question_answer

In the reaction : \[A+2B\to product\]\[Rate=k\,[A]\,\,{{[B]}^{0}}\] The initial concentration of both A and B are a M. The graph of concentration of B vs time is: \[({{t}_{1}}={{t}_{1/2\,}}for\,A)\]

A)

B)

C)

D)

Correct Answer: B

Solution :

\[{{[A]}_{t}}=a{{e}^{-kt}}\]

\[a-x=a{{e}^{-kt}}\]

\[\Rightarrow x=a\,\,(1-{{e}^{-\,kt}})\]

\[at\,\,t={{t}_{1}}\]                   \[\,a-x=\frac{a}{2}\]

\[x\]\[=\frac{a}{2}\]

\[[{{B}_{t}}]=a-2x\]\[=a-2a\,\,(1-{{e}^{-\,kt}})\]\[=a\,(2{{e}^{-\,kt}}-1)\]

\[at\,t=0\]                      \[[B]=a\]

\[at\,\,t={{t}_{1/2}}\]  of  \[a,[{{B}_{t}}]=a-2x\]

\[=a-a\] = 0

Only graph [b] matches this.