Ammonium carbonate decomposes as N H 2 C O O N H 4 s ⇌ 2 N H 3 g + C O 2 ( g ) For the reaction, K p = 2.9 × 10 - 5 a t m 3 . If we start with 1 mole of the compound, the total pressure at equilibrium would be

Question

Ammonium carbonate decomposes as N H 2 C O O N H 4 s ⇌ 2 N H 3 g + C O 2 ( g ) For the reaction, K p = 2.9 × 10 - 5 a t m 3 . If we start with 1 mole of the compound, the total pressure at equilibrium would be, 0.766 atm, 0.0582 atm, 0.0388 atm, 0.0194 atm

MARKS App · Accepted Answer

N H 2 C O O N H 4 s ⇌ 2 N H 3 g + C O 2 ( g ) At equilibrium if partial pressure of C O 2 is p then that of N H 3 is 2 p So, K p = p NH 3 2 × p CO 2 K p = 2 p 2 × p = 4 p 3 2.9 × 10 - 5 = 4 p 3 p = 1.935 × 10 - 2 hence, total pressure p = 3 p = 5.81 × 10 - 2 = 0.0581 atm.

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