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. |
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