For the equilibrium,
$N H_{2} C O O N H_{4} (s) ⇌ 2 N H_{3} (g) + C O_{2} (g)$ ,
the partial pressure of carbon-di-oxide $(p_{C O_{2}}) = 1 atm$ at $100^{\circ} C .$
The value of equilibrium constant (in atm) is:

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Solution

For the calculation of equilibrium constant, only gaseous reactants are considered.
$N H_{2} C O O N H_{4} (s) ⇌ 2 N H_{3} (g) + C O_{2} (g)$
$t = 0, C 00$
$t = t_{e q m} C - x 2 x x$

Given, $p_{C O_{2}} = 1 a t m$
$∴$ $p_{C O_{2}} = \frac{x}{3 x} P_{t o t a l} = \frac{P_{Total}}{3} = 1 atm$
Again, $P_{N H_{3}} = \frac{2 x}{3 x} P_{t o t a l} = \frac{2 P_{Total}}{3} = 2 atm$
$∴$ $K_{p} = P_{N H_{3}}^{2} . P_{C O_{2}} = 4 {atm}^{3}$

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Q. For the equilibrium,

N H_{2} C O O N H_{4} (s) ⇌ 2 N H_{3} (g) + C O_{2} (g)

,
the partial pressure of carbon-di-oxide

(p_{C O_{2}}) = 1 atm

100^{\circ} C .

The value of equilibrium constant (in atm) is:

N H_{2} C O O N H_{4} (s) ⇌ 2 N H_{3} (g) + C O_{2} (g) .

If equilibrium pressure is 6 atm for the above reaction;

K_{p}

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N H_{2} C O O N H_{4} (s) ⇌ 2 N H_{3} (g) + C O_{2} (g)

. If equilibrium pressure is

3

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will be:

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$N H_{2} C O O N H_{4} (s) ⇋ 2 N H_{3} (g) + C O_{2} (g)$ 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. Solid Ammonium carbamate dissociates as:

N H_{2} C O O N H_{4} (s) ⇌ 2 N H_{3} (g) + C O_{2} (g)

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For the equilibrium, NH2COONH4(s)⇌2NH3(g)+CO2(g), the partial pressure of carbon-di-oxide (pCO2)=1 atm at 100 ∘C. The value of equilibrium constant (in atm) is:

For the calculation of equilibrium constant, only gaseous reactants are considered. NH2COONH4(s)⇌2NH3(g)+CO2(g) t=0, C 0 0 t=teqm C−x 2x x Given, pCO2=1 atm ∴ pCO2=x3xPtotal=PTotal3=1 atm Again, PNH3=2x3xPtotal=2 PTotal3=2 atm ∴ Kp=P2NH3.PCO2=4 atm3

For the equilibrium,
$N H_{2} C O O N H_{4} (s) ⇌ 2 N H_{3} (g) + C O_{2} (g)$ ,
the partial pressure of carbon-di-oxide $(p_{C O_{2}}) = 1 atm$ at $100^{\circ} C .$
The value of equilibrium constant (in atm) is: