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NEET 2013
Exam
NEET
Year
2013
Subject
Chemistry
Chapter
Chemical Kinetics

NEET 2013 · Chemistry · Chemical Kinetics

What is the activation energy for a reaction if its rate doubles...

What is the activation energy for a reaction if its rate doubles when the temperature is raised from 20∘C20^{\circ} \mathrm{C} to 35^{\circ} \mathrm{C} ?\left(R=8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\right)

Options

  1. A342 kJ mol−1342 \mathrm{~kJ} \mathrm{~mol}^{-1}
  2. B269 kJ mol−1269 \mathrm{~kJ} \mathrm{~mol}^{-1}
  3. C34.7 kJ mol−134.7 \mathrm{~kJ} \mathrm{~mol}^{-1}
  4. D15.1 kJ mol−115.1 \mathrm{~kJ} \mathrm{~mol}^{-1}

Correct answer

C. 34.7 kJ mol−134.7 \mathrm{~kJ} \mathrm{~mol}^{-1}

Step-by-step solution

Given, initial temperature, T1=20+273=293 KT_1=20+273=293 \mathrm{~K} Final temperature T2=35+273=308 KR=8.314 J mol−1 K−1\begin{aligned} T_2 & =35+273 \\ & =308 \mathrm{~K} \\ R & =8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1} \end{aligned} Since, rate becomes double on raising temperature, ∴r2=2r1 or  r2r1=2\therefore r_2=2 r_1 \text{\text{ or } } \frac{r_2}{r_1}=2 As rate constant, k∝r∴k2k1=2k \propto r\therefore \frac{k_2}{k_1}=2 From Arrnhenius equation, we know that log⁡k2k1=−E22.303R[T1−T2T1T2]log⁡2=−E22.303×8.314[293−308293×308]0.3010=−E22.303×8.314[−15293×308]\begin{aligned} & \log \frac{k_2}{k_1}=-\frac{E_2}{2.303 R}\left[\frac{T_1-T_2}{T_1 T_2}\right] \\ & \log 2=-\frac{E_2}{2.303 \times 8.314}\left[\frac{293-308}{293 \times 308}\right] \\ & 0.3010=-\frac{E_2}{2.303 \times 8.314}\left[\frac{-15}{293 \times 308}\right] \end{aligned}∴E2=0.3010×2.303×8.314×293×30815=34673.48 J mol−1=34.7 kJ mol−1\begin{aligned} \therefore E_2 & =\frac{0.3010 \times 2.303 \times 8.314 \times 293 \times 308}{15} \\ & =34673.48 \mathrm{~J} \mathrm{~mol}^{-1}=34.7 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned}$
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