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

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2 atm

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Solution

The correct option is C

$N H_{2} C O O N H_{4 (s)}$ $⇋$ $2 N H_{3 (g)}$ + $C O_{2 (g)}$ active mass of $N H_{2} C O O N H_{4 (s)}$ is unity as it is solid,equilibrium pressures are independent of amount of $N H_{2} C O O N H_{4 (s)}$ . Let $P_{C O_{2}}$ = 'x' then $P_{N H_{2}}$ = '2x' $K_{p}$ = ${P_{N H_{2}}}^{2}$ $\times$ $P_{C O_{2}}$ ; 2.9 $\times$ $10^{- 5}$ = ${(2 x)}^{2}$ $\times$ x = 4 $x^{3}$ x = $\sqrt[3]{\frac{2.9 \times 10^{- 5}}{4}}$ atm

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Similar questions

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:

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 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:

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For the decomposition reaction NH2COONH4(s)⇋ 2NH3(g) + CO2(g). The Kp= 2.9 × 10−5atm2. The partial pressure of CO2 at equilibrium when 2 mole of NH2COONH4(s) was taken to start with would be

The correct option is C NH2COONH4(s)⇋ 2NH3(g) + CO2(g) active mass of NH2COONH4(s) is unity as it is solid,equilibrium pressures are independent of amount of NH2COONH4(s). Let PCO2 = 'x' then PNH2 = '2x' Kp = PNH22 ×PCO2; 2.9 × 10−5 = (2x)2 ×x = 4 x3 x = 3√2.9×10−54atm

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