JEE Main · 2019 · Shift-ImediumCK-109

For a reaction, consider the plot of k versus 1/T given in the figure. If the rate constant of this reaction at 400 K…

Chemical Kinetics · Class 12 · JEE Main Previous Year Question

Question

For a reaction, consider the plot of ln⁡k\ln k versus 1/T1/T given in the figure. If the rate constant of this reaction at 400 K is 10−5 s−110^{-5}\,\mathrm{s^{-1}}, then the rate constant at 500 K is: image

Options
  1. a

    10−4 s−110^{-4}\,\mathrm{s^{-1}}

    ✓
  2. b

    10−6 s−110^{-6}\,\mathrm{s^{-1}}

  3. c

    2×10−4 s−12 \times 10^{-4}\,\mathrm{s^{-1}}

  4. d

    4×10−4 s−14 \times 10^{-4}\,\mathrm{s^{-1}}

Correct Answera

10−4 s−110^{-4}\,\mathrm{s^{-1}}

Detailed Solution

From the graph, the slope of ln⁡k\ln k vs 1/T1/T can be read. The slope =−Ea/R= -E_a/R.

Reading the graph: when 1/T1/T decreases from 1/4001/400 to 1/5001/500 (i.e., TT increases from 400 to 500 K), ln⁡k\ln k increases by approximately 2.303 (one log unit).

This means kk increases by a factor of 10: k500=10×k400=10×10−5=10−4 s−1k_{500} = 10 \times k_{400} = 10 \times 10^{-5} = 10^{-4}\,\mathrm{s^{-1}}

Answer: Option (1) — 10−4 s−110^{-4}\,\mathrm{s^{-1}}

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