| Given is the graph between \[{{(a-x)}^{-1}}\] and time. |
 |
| Hence, rate at the start of the reaction is: |
A) \[1.25\,mol\,\,{{L}^{-1}}{{\min }^{-1}}\]
B) \[0.125\,\,mol\,\,{{L}^{-1}}{{\min }^{-1}}\]
C) \[0.5\,\,mol\,\,{{L}^{-1}}{{\min }^{-1}}\]
D) \[1.25\,\,mol\,\,{{\min }^{-1}}\]
Correct Answer: B
Solution :
| Since, the graph of t vs \[{{(a-x)}^{-1}}\] is a straight line, it must be a second order reaction. |
| \[\therefore \] \[K=\frac{1}{t}\,\,\left[ \frac{1}{(a-x)}-\frac{1}{a} \right]\] |
| or \[\frac{1}{a-x}=Kt+\frac{1}{a}\] |
| On comparing, slope \[K=\tan \theta =0.5\,\,\text{mo}{{\text{l}}^{-1}}L{{\min }^{-1}}\] |
| \[OA=\frac{1}{a}=2\,\,L\text{mo}{{\text{l}}^{-1}}\] |
| or \[a=\text{0}\text{.5}\,\,\text{mol}\,\,{{L}^{-1}}\] |
| \[\text{Rate}=K\,{{(a)}^{2}}=0.5\times {{(0.5)}^{2}}\] |
| \[=0.125\,\,mol\,\,{{L}^{-1}}{{\min }^{-1}}.\] |
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