A) \[e\]
B) \[-e\]
C) \[\text{2bc}-\text{1}\]
D) \[\text{2bc}+\text{1}\]
Correct Answer: C
Solution :
When \[[F{{L}^{2}}{{L}^{-2}}]\] then \[[F{{L}^{-1}}{{T}^{2}}]\] \[[{{F}^{2}}L{{T}^{-2}}]\] \[2{{O}_{3}}\to 3{{O}_{2}}\] \[{{O}_{3}}{{O}_{2}}+O......(fast)\] Now, \[O+{{O}_{3}}\xrightarrow{{}}2{{O}_{2}}\,\,\,......(Slow)\] \[r=k{{[{{O}_{3}}]}^{2}}\] \[r=k{{[{{O}_{3}}]}^{2}}{{[{{O}_{2}}]}^{-1}}\] \[r=k[{{O}_{3}}][{{O}_{2}}]\] \[Mg(s)+Z{{n}^{2+}}(aq)(0.1M)\to M{{g}^{2+}}(aq)\]
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