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Chemical Kinetics question

2013 · Q90
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Chemical Kinetics question

2013 · Q90

NEETChemistryChemical KineticsMCQ+4 / −1
What is the activation energy for a reaction if its rate doubles when the temperature is raised from 20oC to 35oC?
(R = 8.314 J mol−-−1 K−-−1)
  1. A
    34.7 kJ mol−-−1
  2. B
    15.1 kJ mol−-−1
  3. C
    342 kJ mol−-−1
  4. D
    269 kJ mol−-−1
View written solutionFree

Correct answer: A

log⁡k2k1=Ea2.303R(1T1−1T2)\log {{{k_2}} \over {{k_1}}} = {{{E_a}} \over {2.303R}}\left( {{1 \over {{T_1}}} - {1 \over {{T_2}}}} \right)logk1​k2​​=2.303REa​​(T1​1​−T2​1​)

Initial temperature, T1 = 20 + 273 = 293 K

Final temperature, T2 = 35 + 273 = 308 K

R = 8.314 JK–1 mol–1



As rate becomes double on raising temperature

∴\therefore∴ r2 = 2r1

As rate constant, k ∞\infty ∞ r

k2 = 2k1

∴\therefore∴ log⁡2=Ea2.303×8.314(1293−1308)\log 2 = {{{E_a}} \over {2.303 \times 8.314}}\left( {{1 \over {293}} - {1 \over {308}}} \right)log2=2.303×8.314Ea​​(2931​−3081​)

⇒\Rightarrow⇒ 0.301=Ea19.147×15293×3080.301 = {{{E_a}} \over {19.147}} \times {{15} \over {293 \times 308}}0.301=19.147Ea​​×293×30815​

⇒\Rightarrow⇒ Eaaa = 34673 J mol–1 = 34.7 kJ mol–1

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