You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Factors Affecting Rate of Reaction

For the decom...

Question

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.0388 atm

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.0582 atm

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

0.0776 atm

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

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, $P = 0.0582 a t m$

Suggest Corrections

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)

30^{o} C

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:

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)

; 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:

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.

Join BYJU'S Learning Program

	$N H_{2} C O O N H_{4} \to$	$2 N H_{3}$	$+ C O_{2}$
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 :