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NCERT Class XII Chemistry
Chapter - Chemical Kinetics
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Question : 23 of 39
Marks: +1, -0
The rate constant for the decomposition of a hydrocarbon is 2.418×10−5 s−12.418 \times 10^{-5}\,\mathrm{s}^{-1} at 546 K. If the energy of activation is 179.9 kJ/mol, what will be the value of pre-exponential factor.
Solution:  
Here, k=2.418×10−5 s−1,k=2.418 \times 10^{-5}\,\mathrm{s}^{-1},
Ea=179.9 kJ mol−1,E_{a}=179.9\,\mathrm{kJ}\,\mathrm{mol}^{-1},
T=546 K,A=?T=546\,\mathrm{K}, A=?
According to Arrhenius equation,
log⁡A=log⁡k+Ea2.303RT\log A = \log k + \frac{E_a}{2.303 R T}
=log⁡(2.418×10−5)+179.92.303×8.314×10−3×546=\log (2.418 \times 10^{-5}) + \frac{179.9}{2.303 \times 8.314 \times 10^{-3} \times 546}
=(−5+0.3834)+17.2081=(-5+0.3834)+17.2081
=12.5924 s−1=12.5924\,\mathrm{s}^{-1}
or, A=A= Antilog (12.5924) s−1(12.5924)\,\mathrm{s}^{-1}
=3.912×1012 s−1=3.912 \times 10^{12}\,\mathrm{s}^{-1}
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