In the reaction $N H_{2} C O O N H_{4} (s) ⇌ 2 N H_{3} (g) + C O_{2} (g)$
The equilibrium pressure was $3 a t m$ at $1000 K$ . The $K_{p}$ of the reaction:

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Solution

The correct option is B $4$

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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

atm for the given reaction,

K_{p}

will be:

Q. Solid ammonium carbamate dissociated according to the given reaction

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

Total pressure of the gases in equilibrium is

5

atm. Hence

K_{p}

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

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.

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}

will be:

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In the reaction NH2COONH4(s)⇌2NH3(g)+CO2(g) The equilibrium pressure was 3 atm at 1000K. The Kp of the reaction:

The correct option is B 4

In the reaction $N H_{2} C O O N H_{4} (s) ⇌ 2 N H_{3} (g) + C O_{2} (g)$
The equilibrium pressure was $3 a t m$ at $1000 K$ . The $K_{p}$ of the reaction:

The correct option is B $4$