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JEE Main201910 Jan 2019Evening ShiftChemistryChemical EquilibriumActual

5 .1 g NH 4 SH is introduced in 3 .0 L evacuated flask at 327 o C . 30 % of the solid NH 4 SH is decomposed to NH 3 and H 2 S as gases. The K P of the reaction at 327 o C is R = 0 .082 L atm mol - 1 K - 1 , Molar mass of S = 32 g mol - 1 , Molar mass of N = 14 g mol - 1

Options

  1. A0 .242   atm 2
  2. B0 .242 × 10 - 4   atm 2
  3. C1 × 10 - 4   atm 2
  4. D4 .9 × 10 - 3   atm 2

Correct answer

A. 0 .242   atm 2

Step-by-step solution

NH 4 SH   s ⇌ NH 3   g + H 2 S   g Number of moles = 5 .1 51 = 0 .1   mol NH 4 SH NH 3 H 2 S Initial concentration 0 . 1   mol 0 0 Equilibrium concentration 0 . 1 1 - α 0 . 1 α 0 . 1 α α = 30 % = 0 .3 = Degree of dissociation So, number of moles at equilibrium, NH 3 = H 2 S = 0 . 1 × 0 . 3 = 0 . 03 NH 4 SH is not considered as it is a solid. K c = NH 3 H 2 S K c = 0 . 03 3 0 . 03 3 = 10 - 4 K p = 10 - 4   0 . 082 × 600 2 K p = 0 .242   atm 2

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