Chemical Kinetics

  • Question
    CBSEENCH12010497

    Following data are obtained for the reaction :
    N2O5 → 2NO2 + ½O2

    t/s 0 300 600
    [N2O5]/mol L–1 1.6 × 10–2 0.8 × 10–2 0.4 × 10–2

    (a) Show that it follows first order reaction.
    (b) Calculate the half-life.
    (Given log 2 = 0.3010 log 4 = 0.6021)

    Solution

    At space 300 space straight s
straight k space equals space fraction numerator 2.303 over denominator straight t end fraction space log space fraction numerator left square bracket straight A right square bracket subscript 0 over denominator left square bracket straight A right square bracket end fraction
equals space fraction numerator 2.303 over denominator 300 end fraction space log space fraction numerator 1.6 space straight x space 10 to the power of negative 2 end exponent over denominator 0.8 space straight x space 10 to the power of negative 2 end exponent end fraction
equals fraction numerator 2.303 over denominator 300 end fraction space log space 2 space equals 2.31 space straight x 10 to the power of negative 3 end exponent space straight s to the power of negative 1 end exponent
At space 600 space straight s comma
straight k equals space straight k space equals space fraction numerator 2.303 over denominator straight t end fraction space log space fraction numerator left square bracket straight A right square bracket subscript 0 over denominator left square bracket straight A right square bracket end fraction
equals space fraction numerator 2.303 over denominator 600 end fraction space log space fraction numerator 1.6 space straight x space 10 to the power of negative 2 end exponent over denominator 0.4 space straight x space 10 to the power of negative 2 end exponent end fraction
straight k equals 2.31 space straight x 10 to the power of negative 3 end exponent space straight s to the power of negative 1 end exponent
    In equal time interval, k is constant when using first order equation, therefore, it follows first order kinetics.

    b) Half order reaction,
    straight t subscript 1 divided by 2 end subscript space equals space fraction numerator 0.693 over denominator straight k end fraction
space equals space fraction numerator 0.693 over denominator 2.31 space straight x space 10 to the power of negative 3 end exponent end fraction space equals space 300 space straight s

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