Hence, rate at the start of the reaction is
A) \[1.25\,\,mo{{l}^{-1}}{{\min }^{-1}}\]
B) \[0.5\,\,mol\,{{L}^{-1}}{{\min }^{-1}}\]
C) \[0.125\,\,mol\,{{L}^{-1}}\,{{\min }^{-1}}\]
D) \[1.25\,mol\,{{L}^{-1}}\,{{\min }^{-1}}\]
Correct Answer: C
Solution :
Since, the graph of \[t\,vs\,\,{{(a-x)}^{-1}}\] or \[\frac{1}{a-x}\] is a straight line, it must be a second order reaction and for a second order reaction, \[K=\frac{1}{t}\,\left[ \frac{1}{(a-x)}-\frac{1}{a} \right]\] \[\frac{1}{a-x}={{K}_{t}}+\frac{1}{a}\] On comparing, slope \[=K=\tan \,\theta =0.5\,mo{{l}^{-1}}\,{{\sin }^{-1}}\] \[OA=\frac{1}{a}=2L\,mol\] or \[a=0.5\,\,mol\,{{L}^{-1}}\] Rate \[=\,K{{(a)}^{2}}=0.5\times {{(0.5)}^{2}}\] \[=0.125\,mol\,{{L}^{-1}}\,{{\min }^{-1}}\]
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