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JEE Main & Advanced Sample Paper JEE Main Sample Paper-33

  • question_answer
    Graph between log k and \[\frac{1}{T}\] (k is rate constant in \[{{s}^{-1}}\] and T is the temperature in K) is a straight line. As shown in figure if \[OX=5\]and slope of the line = \[-\frac{1}{2.303}\]then \[{{E}_{a}}\] is:

    A)  \[2.303\times 2cal\]   

    B)    \[\frac{2}{2.303}cal\]

    C)  \[2cal\]           

    D)                     None of these

    Correct Answer: C

    Solution :

    We know,     \[\log k=\log A-\frac{{{E}_{a}}}{2.303RT}\] compare this by \[y=mx+C\]             \[m=-\,\frac{{{E}_{a}}}{2.303R}\] slope of this             Given   \[-\frac{{{E}_{a}}}{2.303R}\,=-\frac{1}{2.303}\]             \[{{E}_{a}}=R=2\,cal\]


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