Ammonium carbonate decomposes as NH 2 COONH 4s - 2NH 3g - CO 2g- For the reaction- K p - 2-9- 10 -5 atm 3- If we start with1 mole of the compound - the total pressure at equilibrium would be-

The correct option is B 0.0582atmsolution: NH_2COONH_4(s)⇌ 2NH_3(g)+CO_2(g) #160;If partial pressure of CO_2 at equilibrium is p #160;Then ...

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

Byju's Answer

Standard XII

Chemistry

Percentage Composition

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, $K_{p} = {2.9 \times 10}^{- 5} {a t m}^{3}$ . If we start with1 mole of the compound , the total pressure at equilibrium would be?

0.0766atm

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

0.0582atm

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

0.0388atm

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

0.0194atm

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

Open in App

Solution

The correct option is B 0.0582atm
solution:
$N H_{2} C O O N H_{4} (s) ⇌ 2 N H_{3} (g) + C O_{2} (g)$
If partial pressure of $C O_{2}$ at equilibrium is $p$
Then partial pressure of $N H_{3}$ is 2p
$K_{p} = (p_{N H 3})^{2} \times (p_{C O 2}) = (2 p)^{2} (p) = 4 p^{3}$ Now $4 p^{3} = 2.9 \times 10^{- 5}$
or $p = 1.935 \times 10^{- 2}$
$= 5.82 \times 10^{- 2} a t m$
hence the correct option:B

Suggest Corrections

Similar questions

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

mol 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

Join BYJU'S Learning Program

Ammonium carbonate decomposes as NH2COONH4(s)⇌2NH3(g)+CO2(g). For the reaction, Kp=2.9×10−5atm3. If we start with1 mole of the compound , the total pressure at equilibrium would be?

The correct option is B 0.0582atmsolution:NH2COONH4(s)⇌2NH3(g)+CO2(g) If partial pressure of CO2 at equilibrium is p Then partial pressure of NH3 is 2p Kp=(pNH3)2×(pCO2)=(2p)2(p)=4p3 Now 4p3=2.9×10−5or p=1.935×10−2 =5.82×10−2atmhence the correct option:B

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?