JEE Main & Advanced Sample Paper JEE Main Sample Paper-29

  • question_answer
    For the reaction, \[2NO+B{{r}_{2}}\to 2NOBr\], the following mechanism has been given             
    a. \[NO+B{{r}_{2}}\overset{fast}{leftrightarrows}NOB{{r}_{2}}\]
    b. \[NOB{{r}_{2}}+NO\overset{slow}{leftrightarrows}2NOBr\]
    Hence rate law is

    A) \[r=k{{[NO]}^{2}}[B{{r}_{2}}]\] 

    B) \[r=k[NO][B{{r}_{2}}]\]

    C) \[r=k[NOB{{r}_{2}}][NO]\]          

    D) \[r=k[NO]{{[B{{r}_{2}}]}^{2}}\]

    Correct Answer: A

    Solution :

    \[9{{x}^{2}}\,-16{{y}^{2}}=144\]                 \[{{x}^{2}}+{{y}^{2}}=9\]                 \[y=mx+c\]                 \[y=mx+c\]


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