Ammonium carbonate decomposes as NH2COONH4 s - 2NH3 g - CO2 g For the reaction- KP-2-9 -10-5 atm3- Calculate the total pressure at equilibrium if 1 mole of NH2COONH4is taken-

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Standard XIII

Chemistry

Derivation of Kp and Kc

Ammonium carb...

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, $K_{P} = 2.9 \times 10^{- 5} a t m^{3}$ . Calculate the total pressure at equilibrium if $1 m o l e o f N H_{2} C O O N H_{4}$ is taken.

0.0766 atm

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0.0582 atm

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0.0388 atm

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0.0194 atm

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Solution

The correct option is B
0.0582 atm

Correct option is b).
$N H_{2} C O O N H_{4} (s) ⇋ 2 N H_{3} (g) + C O_{2} (g)$
At equilibrium, if partial pressure of $C O_{2} = p,$ and that of $N H_{3} = 2 p .$
$K_{p} = (p_{N H 3})^{2} \times (p_{c o 2}) = (2 p)^{2} \times p = 4 p^{3} = 2.9 \times 10^{- 5}$
Solving for p, $p = 1.935 x 10^{- 2} .$
Hence, total pressure $= 3 p = 0.0581 a t m$

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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, $K_{P} = 2.9 \times 10^{- 5} a t m^{3}$ . Calculate the total pressure at equilibrium if $1 m o l e o f N H_{2} C O O N H_{4}$ is taken.

Q. 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 \times 10}^{- 5} {a t m}^{3}

. If we start with1 mole of the compound , the total pressure at equilibrium would be?

Q. For the decomposition,

N H_{2} C O O N H_{4} (s) ⇌ 2 N H_{3} (g) + C O_{2} (g)

;

K_{p} = 2.9 \times 10^{- 5} a t m^{3}

The total pressure of gases at equilibrium when 1.0 mol of

N H_{2} C O O N H_{4} (s)

was taken to start with, would be :

Q. For the decomposition of the compound represented as

{N H}_{2} C O O {N H}_{4} (s) ⇌ 2 {N H}_{3} (g) + {C O}_{2} (g)

the

K_{p} = 2.9 \times 10^{- 5} {a t m}^{3}

IF the reaction is started with

1

mol of the compound, the total pressure at equilibrium would be:

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Ammonium carbonate decomposes as NH2COONH4 (s) ⇋ 2NH3 (g) + CO2 (g) For the reaction, KP=2.9×10−5 atm3. Calculate the total pressure at equilibrium if 1 mole of NH2COONH4is taken.

The correct option is B 0.0582 atm Correct option is b). NH2COONH4 (s) ⇋ 2NH3(g) + CO2 (g) At equilibrium, if partial pressure of CO2=p, and that of NH3 =2p. Kp = (pNH3)2 × (pco2) =(2p)2 × p = 4p3 =2.9× 10−5 Solving for p, p = 1.935x 10−2. Hence, total pressure = 3p = 0.0581 atm

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, $K_{P} = 2.9 \times 10^{- 5} a t m^{3}$ . Calculate the total pressure at equilibrium if $1 m o l e o f N H_{2} C O O N H_{4}$ is taken.