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JEE Main2014ChemistryChemical EquilibriumActual

For the decomposition of the compound, represented as NH 2 COONH 4 s ⇌ 2 NH 3 g + CO 2 g the K p = 2 .9 × 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 × 1 0 - 2 atm
  2. B38 .8 × 1 0 - 2 atm
  3. C5 .82 × 1 0 - 2 atm
  4. D1 .94 × 10 - 2   atm

Correct answer

C. 5 .82 × 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 K p   and   K c NH 2 COONH 4 s ⇌ 2 NH 3 g + CO 2 g At equilibrium ⇒ 2 p and p will be partial pressures of ammonia and carbon dioxide gas respectively. K P   =   P NH 3 2 P CO 2   =   2P 2 P   =   4P 3 4 P 3 = 2 · 9 × 1 0 - 5 atm 3 P 3 = 0 · 7 2 5 × 1 0 - 5 atm 3 = 7 · 2 5 × 1 0 - 6 atm 3 P = 7 · 2 5 × 1 0 - 6

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