Verified Solution • 2025–26 Session moderate

QJEEM19Chd147: JEE Main 2019 Chemistry - Class 12 JEEM19Chd147 Solution: For a reaction consider the plot of n k versus 1/T given in the …

For a reaction consider the plot of ℓ\ell n k versus 1/T given in the figure. If the rate constant of this reaction at 400 K is 10–5 s–1, then the rate constant at 500 K is –

JEE Main 2019 (Online) 12th January Evening Slot Chemistry - Chemical Kinetics and Nuclear Chemistry Question 130 English

Multiple Choice Options

  • A 10 −- 4 s −- 1
  • B 4 ×\times 10 −- 4 s −- 1
  • C 10 −- 6 s −- 1
  • D 2 ×\times 10 −- 4 s −- 1

Step-by-Step Verified Solution

From Arhenius equation,

K = Ae −EaRT^{ - {{Ea} \over {RT}}}

taking log both sides,

lnk = lnA −- EaRT{{{Ea} \over {RT}}}

So in lnK vs 1T{1 \over T} graph

slope = −- EaR{{{Ea} \over R}}

Rate constant at 400 K is = 10 −- 5 s −- 1

Let rate constant at 500 K is = K

∴\therefore    ln(10 −- 5) = ln A −- EaR(400){{Ea} \over {R\left( {400} \right)}} . . . (1)

ln K = ln A −- EaR(500){{Ea} \over {R\left( {500} \right)}} . . . .(2)

Subtracting (2) from (1) we get,

ln K10−5{K \over {{{10}^{ - 5}}}} = EaR{{{Ea} \over R}} [1400−1500]\left[ {{1 \over {400}} - {1 \over {500}}} \right]

⇒\Rightarrow    2.303 log K10−5{K \over {{{10}^{ - 5}}}} = 4606 [1400−1500]\left[ {{1 \over {400}} - {1 \over {500}}} \right]

⇒\Rightarrow    K = 10 −- 4 s −- 1

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