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JEE Main 201912 JanuaryMorning Shift
Exam
JEE Main
Year
2019
Date
12 January
Shift
Morning Shift
Subject
Chemistry
Chapter
Chemical Equilibrium

JEE Main 2019 · 12 January · Morning Shift · Chemistry · Chemical Equilibrium

Two solids dissociate as follows: A(s) ⇌ B(g) + C(g); K _ P _ 1 = x...

Two solids dissociate as follows:

A(s)⇌B(g)+C(g);KP1=xatm2A\left(s\right) ⇌ B\left(g\right) + C\left(g\right); {K}_{{P}_{1}} = x{atm}^{2}

D(s)⇌C(g)+E(g);KP2=yatm2D\left(s\right) ⇌ C\left(g\right) + E\left(g\right); {K}_{{P}_{2}} = y{atm}^{2}

The total pressure when both the solids dissociate simultaneously is:Two solids dissociate as follows: A(s) ⇌ B(g) + C(g); K _ P _ 1 = x atm ^ 2 D(s) ⇌ C(g) + E(g); K _ P _ 2 = y atm ^ 2

Options

  1. Ax+yatm\sqrt{x + y}atm
  2. Bx2+y2atm{x}^{2} + {y}^{2}atm
  3. C(x+y)atm\left(x + y\right)atm
  4. D2(x+y)atm2\left(\sqrt{x + y}\right)atm

Correct answer

D. 2(x+y)atm2\left(\sqrt{x + y}\right)atm

Step-by-step solution

Two solids dissociate as follows: A(s) ⇌ B(g) + C(g); K _ P _ 1 = x atm ^ 2 D(s) ⇌ C(g) + E(g); K _ P _ 2 = y atm ^ 2

kp1=P1(P1+P2){k}_{{p}_{1}} = {P}_{1}\left({P}_{1} + {P}_{2}\right)

kp2=P2(P1+P2){k}_{{p}_{2}} = {P}_{2} ( {P}_{1} + {P}_{2} )

kP1+kP2=(P1+P2)2{k}_{{P}_{1}} + {k}_{{P}_{2}} = {( {P}_{1} + {P}_{2} )}^{2}

x+y=(P1+P2)2x + y = {( {P}_{1} + {P}_{2} )}^{2}

P1+P2=x+y{P}_{1} + P{}_{2} = \sqrt{x + y}

2(P1+P2)=2x+y2 ( {P}_{1} + {P}_{2} ) = 2\sqrt{x + y}

PTotal=PB+PC+PE{P}_{Total} = {P}_{B} + {P}_{C} + {P}_{E}

=2(P1+P2)=2x+y= 2 ( {P}_{1} + {P}_{2} ) = 2\sqrt{x + y}

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