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. 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 m o l

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

Q. For the decompostion 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 compounds, the total pressure at equilibrium woud be:

For the decomposition reaction $N H_{2} C O O N H_{4 (s)}$ $⇋$ $2 N H_{3 (g)}$ + $C O_{2 (g)}$ . The $K_{p}$ = 2.9 $\times$ $10^{- 5}$ ${a t m}^{2}$ . The partial pressure of $C O_{2}$ at equilibrium when 2 mole of $N H_{2} C O O N H_{4 (s)}$ was taken to start with would be

Q. Solid ammonium carbamate dissociated according to the given reaction

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

Total pressure of the gases in equilibrium is

5

atm. Hence

K_{p}

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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 atmCorrect 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.