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JEE Main 201419 AprilOnline
Exam
JEE Main
Year
2014
Date
19 April
Mode
Online
Subject
Chemistry
Chapter
Chemical Equilibrium

JEE Main 2014 · 19 April · Online · Chemistry · Chemical Equilibrium

For the decomposition of the compound, represented as NH _ 2 COONH...

For the decomposition of the compound, represented as

NH2COONH4(s)⇌2NH3(g)+CO2(g){\text{NH}}_{2} {\text{COONH}}_{4} \left(\text{s}\right) ⇌ 2 {\text{NH}}_{3} \left(\text{g}\right) + {\text{CO}}_{2} \left(\text{g}\right)

the Kp=2.9×10−5atm3{K}_{p} = 2 .9 \times {10}^{- 5}{atm}^{3}. If the reaction is started with 1 mole of the compound, the total pressure at equilibrium would be :

Options

  1. A7.66×10−2atm7 .66 \times 1{0}^{- 2} atm
  2. B38.8×10−2atm38 .8 \times 1{0}^{- 2} atm
  3. C5.82×10−2atm5 .82 \times 1{0}^{- 2} atm
  4. D1.94×10−2atm1 .94 \times {10}^{- 2}atm

Correct answer

C. 5.82×10−2atm5 .82 \times 1{0}^{- 2} atm

Step-by-step solution

In case of heterogeneous equilibrium. The achive mass of solid is unit so not consider in the expression of KpandKc{\text{K}}_{\text{p}} \text{and}{\text{K}}_{\text{c}}

NH2COONH4(s)⇌2NH3(g)+CO2(g){\text{NH}}_{2} {\text{COONH}}_{4} \left(\text{s}\right) ⇌ 2 {\text{NH}}_{3} \left(\text{g}\right) + {\text{CO}}_{2} \left(\text{g}\right)

At equilibrium ⇒⇒ 2 p and p will be partial pressures of ammonia and carbon dioxide gas respectively.

KP=[PNH3]2[PCO2]=[2P]2[P]=4P3{\text{K}}_{\text{P}} \text{=}{\left[{\text{P}}_{{\text{NH}}_{\text{3}}}\right]}^{\text{2}} \left[{\text{P}}_{{\text{CO}}_{\text{2}}}\right]\text{=}{\left[\text{2P}\right]}^{\text{2}} \left[\text{P}\right]\text{=}{\text{4P}}^{\text{3}}

4P3=2⋅9×10−5atm34 {\text{P}}^{3} = 2 \cdot 9 \times 1 {0}^{- 5} {\text{atm}}^{3}

P3=0⋅725×10−5atm3=7⋅25×10−6atm3{\text{P}}^{3} = 0 \cdot 7 2 5 \times 1 {0}^{- 5} {\text{atm}}^{3} = 7 \cdot 2 5 \times 1 {0}^{- 6} {\text{atm}}^{3}

P=7⋅25×10−63=1⋅94×10−2atm\text{P} = 7 \cdot 2 5 \times 1 {0}^{- 6} 3 = 1 \cdot 9 4 \times 1 {0}^{- 2} \text{atm}

ΣP=(PTotal)equilibrium=2P+P=3PΣ \text{P} = {\left({\text{P}}_{\text{Total}}\right)}_{\text{equilibrium}} = 2 \text{P} + \text{P} = 3 \text{P}

=3×1⋅94×10−2atm3 \times 1 \cdot 9 4 \times 1 {0}^{- 2} \text{atm}

=5⋅82×10−2atm5 \cdot 8 2 \times 1 {0}^{- 2} \text{atm}
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