PrepRung
Sign inSign up free

JEE Advanced 2018 Paper 2 · Chemistry · Chemical Kinetics · easy

Reaction kinetics (take R=8.314 J K^-1 mol^-1).

One or more options correct

Consider the following statements about reaction kinetics (take R=8.314 J K−1 mol−1R=8.314\ \mathrm{J\,K^{-1}\,mol^{-1}}). The correct statement(s) is/are
  1. A. For a first-order reaction the time required for 99%99\% completion is about 6.646.64 times the half-life
  2. B. For a zero-order reaction with [A]0=0.80 M[\mathrm{A}]_0=0.80\ \mathrm{M} and k=0.020 M s−1k=0.020\ \mathrm{M\,s^{-1}}, the time for 75%75\% conversion is 40 s40\ \mathrm{s}
  3. C. For a second-order reaction of the type 2A→products\mathrm{2A\rightarrow products}, doubling the initial concentration doubles the half-life
  4. D. For a reaction with Ea=100 kJ mol−1E_a=100\ \mathrm{kJ\,mol^{-1}} the rate constant roughly doubles when the temperature is raised from 298 K298\ \mathrm{K} to 308 K308\ \mathrm{K}
Show answer and solution

Answer: A

(A) For first order, t=1kln⁡1001t=\dfrac{1}{k}\ln\dfrac{100}{1} and t1/2=ln⁡2kt_{1/2}=\dfrac{\ln2}{k}, so t99/t1/2=log⁡2100=6.64t_{99}/t_{1/2}=\log_2100=6.64: correct. (B) For zero order, [A]0−[A]=kt[\mathrm{A}]_0-[\mathrm{A}]=kt; 75% conversion means 0.60 M0.60\ \mathrm{M} consumed, so t=0.60/0.020=30 st=0.60/0.020=30\ \mathrm{s}, not 40 s: wrong. (C) For second order, t1/2=1/(k[A]0)t_{1/2}=1/(k[\mathrm{A}]_0) (up to a numerical factor), so doubling [A]0[\mathrm{A}]_0 halves the half-life; it does not double it: wrong. (D) From the Arrhenius equation, ln⁡(k2/k1)=EaR(1298−1308)=1000008.314×1.0896×10−4=1.31\ln(k_2/k_1)=\dfrac{E_a}{R}\left(\dfrac{1}{298}-\dfrac{1}{308}\right)=\dfrac{100000}{8.314}\times1.0896\times10^{-4}=1.31, so k2/k1=3.7k_2/k_1=3.7; the rule of doubling per 10 K holds only near Ea≈50 kJ mol−1E_a\approx50\ \mathrm{kJ\,mol^{-1}} (ratio 1.93): wrong. Answer: A only. Each false option fails on a specific quantitative or conceptual slip: the arithmetic of zero-order kinetics, the dependence of half-life on concentration, and the magnitude of the temperature effect at high activation energy.
Report a mistake

Original question written in the JEE Advanced 2018 Paper 2 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake