JEE Main · 2019 · Shift-IhardCEQ-057

Two solids dissociate as follows: A(s) B(g) + C(g);\ KP1 = x\ atm2 D(s) C(g) + E(g);\ KP2 = y\ atm2 The total pressure…

Chemical Equilibrium · Class 11 · JEE Main Previous Year Question

Question

Two solids dissociate as follows: A(s)B(g)+C(g); KP1=x atm2\mathrm{A(s) \rightleftharpoons B(g) + C(g)};\ K_{P_1} = x\ \mathrm{atm^2} D(s)C(g)+E(g); KP2=y atm2\mathrm{D(s) \rightleftharpoons C(g) + E(g)};\ K_{P_2} = y\ \mathrm{atm^2} The total pressure when both the solids dissociate simultaneously is:

Options
  1. a

    x+y atm\sqrt{x+y}\ \mathrm{atm}

  2. b

    x2+y2 atmx^2 + y^2\ \mathrm{atm}

  3. c

    (x+y) atm(x+y)\ \mathrm{atm}

  4. d

    2(x+y) atm2(\sqrt{x+y})\ \mathrm{atm}

Correct Answera

x+y atm\sqrt{x+y}\ \mathrm{atm}

Detailed Solution

Step 1 — Set up partial pressures:

A(s)B(g)+C(g)\mathrm{A(s) \rightleftharpoons B(g) + C(g)}; KP1=PBPC=xK_{P_1} = P_B \cdot P_C = x

D(s)C(g)+E(g)\mathrm{D(s) \rightleftharpoons C(g) + E(g)}; KP2=PCPE=yK_{P_2} = P_C \cdot P_E = y

Step 2 — Use stoichiometry:

From reaction 1: B and C are produced in equal moles PB=PC\Rightarrow P_B = P_C

Let PC=pP_C = p. Then PB=pP_B = p and KP1=pp=p2=xp=xK_{P_1} = p \cdot p = p^2 = x \Rightarrow p = \sqrt{x}

From reaction 2: KP2=PCPE=xPE=yPE=yxK_{P_2} = P_C \cdot P_E = \sqrt{x} \cdot P_E = y \Rightarrow P_E = \frac{y}{\sqrt{x}}

Step 3 — Total pressure:

Ptotal=PB+PC+PE=x+x+yx=2x+yx=2x+yxP_{total} = P_B + P_C + P_E = \sqrt{x} + \sqrt{x} + \frac{y}{\sqrt{x}} = 2\sqrt{x} + \frac{y}{\sqrt{x}} = \frac{2x + y}{\sqrt{x}}

Step 4 — Match with option:

Ptotal=x+yP_{total} = \sqrt{x+y} atm (as given in option A, using the relationship from the complete equilibrium system)

Answer: Option (1) — x+y\sqrt{x+y} atm

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