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JEE Advanced 2018Paper 2
Exam
JEE Advanced
Year
2018
Paper
Paper 2
Subject
Chemistry
Chapter
Chemical Kinetics

JEE Advanced 2018 · Paper 2 · Chemistry · Chemical Kinetics

For a first order reaction A ( g ) → 2B ( g ) + C ( g ) at constant...

For a first order reaction A(g)→2B(g)+C(g)A ( g ) \to 2B ( g ) + C ( g ) at constant volume and 300 K, the total pressure at the beginning (t=0)( t = 0 ) and at time are P0andP0{P}_{0}and{P}_{0}, respectively. Initially, only A is present with concentration [A]0{\left[A\right]}_{0}, and t13{t}_{\dfrac{1}{3}} is the time required for the partial pressure of A to reach 1/3rd1 / {3}^{rd} of its initial value. The correct option(s) is (are)
(Assume that all these gases behave as ideal gases)For a first order reaction A ( g ) → 2B ( g ) + C ( g ) at constant volume and 300 K, the total pressure at theFor a first order reaction A ( g ) → 2B ( g ) + C ( g ) at constant volume and 300 K, the total pressure at theFor a first order reaction A ( g ) → 2B ( g ) + C ( g ) at constant volume and 300 K, the total pressure at theFor a first order reaction A ( g ) → 2B ( g ) + C ( g ) at constant volume and 300 K, the total pressure at the

Options

  1. AWhat is the initial product in the formation of phenol-formaldehyde polymers?
  2. BWhat is the initial product in the formation of phenol-formaldehyde polymers?
  3. CWhat is the initial product in the formation of phenol-formaldehyde polymers?
  4. DWhat is the initial product in the formation of phenol-formaldehyde polymers?

Correct answer

A. What is the initial product in the formation of phenol-formaldehyde polymers?

Step-by-step solution

A→2B+CA \to 2B + C
{t=0P0−−t=tP0−P2PP\begin{cases}t = 0 & {P}_{0} & - - \\ t = t & {P}_{0} - P & 2PP\end{cases}
P0+2P=Pt{P}_{0} + 2P = {P}_{t}
K=1tlnP0P0−P=1tlnP0P0−(Pt−P0)2K = \dfrac{1}{t}ln \dfrac{{P}_{0}}{{P}_{0} - P} = \dfrac{1}{t}ln \dfrac{{P}_{0}}{{P}_{0} - \dfrac{\left({P}_{t} - {P}_{0}\right)}{2}}
K=1tln2P03P0−Pt⇒−Kt+ln⁡2P0=ln⁡(3P0−Pt)K = \dfrac{1}{t}ln \dfrac{2{P}_{0}}{3{P}_{0} - {P}_{t}} ⇒ - Kt + \ln 2{P}_{0} = \ln \left(3{P}_{0} - {P}_{t}\right)
And t13=1KlnP0P03=1Kln⁡3={t}_{\dfrac{1}{3}} = \dfrac{1}{K}ln \dfrac{{P}_{0}}{\dfrac{{P}_{0}}{3}} = \dfrac{1}{K}\ln 3 = contant
Rate constant does not depends on concentration
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