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 :

0.0194 atm

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0.0388 atm

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0.0582 atm

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0.0776 atm

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Solution

The correct option is A 0.0582 atm

$N H_{2} C O O N H_{4} \to$ $2 N H_{3}$ $+ C O_{2}$
Initial moles
1
0
0
Moles at equilibrium
1-x
2x
x
Mole fraction

0.66
0.34
Partial pressure

0.66P
0.34P

The expression for the equilibrium constant is

$K_{p} = P_{N H_{3}}^{2} P_{C O_{2}}$
Substitute values in the above expression
$2 \times 10^{- 5} = (0.66 P)^{2} \times 0.34 P$
Hence, $P = 0.0582 a t m$

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Similar questions

Q. The decomposition of ammonium carbonate at

30^{o} C

is represented as:

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

The equilibrium constant

K_{p}

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

. What is the total pressure of gases in equilibrium with

{N H}_{2} C O O {N H}_{4} (s)

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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

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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. 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)

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. If the reaction is started with 1 mol of the compounds, the total pressure at equilibrium woud 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.

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	$N H_{2} C O O N H_{4} \to$	$2 N H_{3}$	$+ C O_{2}$
Initial moles	1	0	0
Moles at equilibrium	1-x	2x	x
Mole fraction		0.66	0.34
Partial pressure		0.66P	0.34P

For the decomposition, NH2COONH4(s)⇌2NH3(g)+CO2(g); Kp=2.9×10−5atm3 The total pressure of gases at equilibrium when 1.0 mol of NH2COONH4(s) was taken to start with, would be :

The correct option is A 0.0582 atmNH2COONH4→2NH3+CO2Initial moles100Moles at equilibrium1-x2xxMole fraction0.660.34Partial pressure0.66P0.34PThe expression for the equilibrium constant is Kp=P2NH3PCO2Substitute values in the above expression2×10−5=(0.66P)2×0.34PHence, P=0.0582atm

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 :