The decomposition of ammonium carbonate at 30oC is represented as- NH 2 COO NH 4 s- 2 NH 3 g- CO 2 gThe equilibrium constant Kp is 2-9- 10 -5 atm 3- What is the total pressure of gases in equilibrium with NH 2 COO NH 4 s at 30oC -

#160; NH _ 2 COO NH _ 4 (s)⇌ 2 NH _ 3 (g)+ CO _ 2 (g) (p_NH_3) = 2x (p_CO_2) = x K _ p = ( p _ NH _ 3 ) ^ 2 × p _ CO _ 2 2.9× 10 ^ ...

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Byju's Answer

Standard XII

Chemistry

Characteristics of Equilibrium Constant

The decomposi...

Question

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 $K_{p}$ is $2.9 \times 10^{- 5} {a t m}^{3}$ . What is the total pressure of gases in equilibrium with ${N H}_{2} C O O {N H}_{4} (s)$ at $30^{o} C$ ?

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Solution

${N H}_{2} C O O {N H}_{4} (s) ⇌ 2 {N H}_{3} (g) + {C O}_{2} (g)$
$(p_{N H_{3}}) = 2 x$
$(p_{C O_{2}}) = x$

$K_{p} = {(p_{{N H}_{3}})}^{2} \times p_{{C O}_{2}}$

$2.9 \times 10^{- 5} = {(2 x)}^{2} \times x = 4 x^{3}$

$x = {(\frac{2.9 \times 10^{- 5}}{4})}^{1 / 3} = 0.0194 a t m$
Total pressure $= 2 x + x = 3 x = 3 \times 0.0194 = 0.058 a t m$

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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 m o l

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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The decomposition of ammonium carbonate at 30oC is represented as:NH2COONH4(s)⇌2NH3(g)+CO2(g)The equilibrium constant Kp is 2.9×10−5 atm3. What is the total pressure of gases in equilibrium with NH2COONH4(s) at 30oC ?

NH2COONH4(s)⇌2NH3(g)+CO2(g)(pNH3)=2x(pCO2)=xKp=(pNH3)2×pCO22.9×10−5=(2x)2×x=4x3x=(2.9×10−54)1/3=0.0194atmTotal pressure =2x+x=3x=3×0.0194=0.058 atm