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JEE Advanced 2011Paper 2
Exam
JEE Advanced
Year
2011
Paper
Paper 2
Subject
Chemistry
Chapter
Chemical Kinetics

JEE Advanced 2011 · Paper 2 · Chemistry · Chemical Kinetics

For the first order reaction 2 ~N_2 O_5(~g) 4 NO_2(g)+O_2(g)

For the first order reaction 2 N2O5( g)⟶4NO2(g)+O2(g)2 \mathrm{~N}_2 \mathrm{O}_5(\mathrm{~g}) \longrightarrow 4 \mathrm{NO}_2(g)+\mathrm{O}_2(g)For the first order reaction 2 ~N_2 O_5(~g) 4 NO_2(g)+O_2(g)For the first order reaction 2 ~N_2 O_5(~g) 4 NO_2(g)+O_2(g)

Options

  1. Athe concentration of the reactant decreases exponentially with time.
  2. Bthe half-life of the reaction decreases with increasing temperature
  3. Cthe half-life of the reaction depends on the initial concentration of the reactant.
  4. Dthe reaction proceeds of 99.6%99.6 \% completion in eight half-life duration.

Correct answer

A. the concentration of the reactant decreases exponentially with time.

Step-by-step solution

(a) For a first order reaction, the concentration of reactant remaining after time tt is given by [A]=[A]0e−kt[A]=[A]_0 e^{-k t} Therefore, concentration of reactant decreases exponentially with time. (b) Rise in temperature increases rate constant (k)(k) and therefore decreases half-life \left(t_{1/2}\right)ast1/2=as t_{1/2}=\frac{\operatorname{In} 2}{k}For the first order reaction 2 ~N_2 O_5(~g) 4 NO_2(g)+O_2(g)(d)Forafirstorderreaction,if100molesofreactantistakeninitially,after\\ (d) For a first order reaction, if 100 moles of reactant is taken initially, afternhalf−lives,reactantremainingisgivenbyhalf-lives, reactant remaining is given by Percentage A=100(12)n=100(12)8=0.3906⇒ A reacted =100−0.3906=99.6%\begin{aligned}& \text{ Percentage }A=100\left(\frac{1}{2}\right)^n=100\left(\frac{1}{2}\right)^8=0.3906 \\ & \Rightarrow \text{ A reacted }=100-0.3906=99.6 \% \end{aligned}$(c) Half-life of first order reaction is independent of initial concentration.
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