Q15 / 90 · JEE Main 2014 (19 Apr Online) · Chemistry

For the decomposition of the compound, represented as

NH 2 COONH 4 s ⇌ 2 NH 3 g + CO 2 g

the Kp=2.9×10-5atm3. If the reaction is started with 1 mole of the compound, the total pressure at equilibrium would be :

  • A
    7.66×10-2atm
  • B
    38.8×10-2atm
  • Correct answer
    C
    5.82×10-2atm
  • D
    1.94×10-2 atm

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 3 = 1 · 9 4 × 1 0 - 2 atm

Σ P = P Total equilibrium = 2 P + P = 3 P

= 3 × 1 · 9 4 × 1 0 - 2 atm

= 5 · 8 2 × 1 0 - 2 atm

Hence, the correct answer is option: 'C'.

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