A) First
B) Second
C) Third
D) Zero
Correct Answer: C
Solution :
\[r=k{{[NO]}^{\alpha }}{{[C{{l}_{2}}]}^{\beta }}=k{{a}^{\alpha }}{{b}^{\beta }}\] ...(i) or, \[2r={{k}^{\alpha }}(2b){{|}^{\beta }}={{2}^{\beta }}.k{{a}^{\alpha }}{{b}^{\beta }}\] ...(ii) Dividing (ii) by (i) \[2={{2}^{\beta }}\]or, \[\beta =1\] Further, \[8r=k{{(2a)}^{\alpha }}{{(2b)}^{\beta }}\] or, \[8r={{2}^{\alpha +\beta }}k{{a}^{\alpha }}{{b}^{\beta }}\] ...(iii) Dividing (iii) by (i) \[8={{2}^{\alpha +\beta }}\] Or, \[{{2}^{3}}={{2}^{\alpha +\beta }}\] or, \[\alpha +\beta =3\]or, \[\alpha =2\] Overall order = 3.
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