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JEE Advanced 2009Paper 2
Exam
JEE Advanced
Year
2009
Paper
Paper 2
Subject
Chemistry
Chapter
Chemical Kinetics

JEE Advanced 2009 · Paper 2 · Chemistry · Chemical Kinetics

For a first order reaction, A arrow P , the temperature (T)...

For a first order reaction, A→PA \rightarrow P, the temperature (T)(T) dependent rate constant (k)(k) was found to follow the equation, log⁡k=−(2000)/T+6.0\log k=-(2000) / T+6.0 The pre-exponential factor Aandtheactivationenergy(Ea)A and the activation energy \left(E_a\right), respectively, are

Options

  1. A1.0×106 s−11.0 \times 10^6 \mathrm{~s}^{-1} and 9.2 kJ mol−19.2 \mathrm{~kJ} \mathrm{~mol}^{-1}
  2. B6.0 s−16.0 \mathrm{~s}^{-1} and 16.6 kJ mol−116.6 \mathrm{~kJ} \mathrm{~mol}^{-1}
  3. C1.0×106 s−11.0 \times 10^6 \mathrm{~s}^{-1} and 16.6 kJ mol−116.6 \mathrm{~kJ} \mathrm{~mol}^{-1}
  4. D1.0×106 s−11.0 \times 10^6 \mathrm{~s}^{-1} and 38.3 kJ mol−138.3 \mathrm{~kJ} \mathrm{~mol}^{-1}

Correct answer

D. 1.0×106 s−11.0 \times 10^6 \mathrm{~s}^{-1} and 38.3 kJ mol−138.3 \mathrm{~kJ} \mathrm{~mol}^{-1}

Step-by-step solution

Comparing the slope and intercept of the given equation with the following Arrhenius equation : log⁡k=−Ea2303RT+log⁡A\log k=-\frac{E_a}{2303 R T}+\log A Hence, log⁡A=6\log A=6 i.e. A=106 s−1A=10^6 \mathrm{~s}^{-1}. Comparing slope gives Ea=38.3 kJ/molE_a=38.3 \mathrm{~kJ} / \mathrm{mol}.
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