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JEE Main & Advanced AIEEE Solved Paper-2006

  • question_answer
    The following mechanism has been proposed for the reaction of\[NO\]with\[B{{r}_{2}}\]to form\[NOBr\]. \[NO(g)+B{{r}_{2}}(g)NOB{{r}_{2}}(g)\] \[NOB{{r}_{2}}(g)+NO(g)\xrightarrow{{}}2NOBr(g)\] If the second step is the rate determining step, the order of the reaction with respect to\[NO(g)\]is     AIEEE  Solved  Paper-2006

    A) 1       

    B)                        0             

    C)        3             

    D)        2

    Correct Answer: D

    Solution :

    Rate\[=k[NOB{{r}_{2}}][NO]\]                   ???..(i) But\[NOB{{r}_{2}}\]is in equilibrium. \[{{K}_{eq}}=\frac{[NOB{{r}_{2}}]}{[No][B{{r}_{2}}]}\] \[[NOB{{r}_{2}}]={{K}_{eq}}[NO][B{{r}_{2}}]\]        ...(ii) Putting the value of\[[NOB{{r}_{2}}]\]in Eq. (i) Rate\[=k.{{K}_{eq}}[NO][B{{r}_{2}}][NO]\] Hence, rate\[=k.{{K}_{eq}}{{[NO]}^{2}}[B{{r}_{2}}]\] Rate\[=k'{{[NO]}^{2}}[B{{r}_{2}}]\] where  \[k'=k.{{K}_{eq}}\] Alternative Solution \[NO(g)+B{{r}_{2}}(g)NOB{{r}_{2}}(g)\] \[NOB{{r}_{2}}(g)+NO(g)\xrightarrow{\,}NOBr(g)\] Rate\[=k[NOB{{r}_{2}}].[NO]=k.{{K}_{eq}}[NO][B{{r}_{2}}][NO]\] \[\left[ where,{{K}_{eq}}=\frac{[NOB{{r}_{2}}]}{[NO][B{{r}_{2}}]} \right]\] \[=k'{{[NO]}^{2}}[B{{r}_{2}}]\]


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