For the decomposition of the compound represented as NH2COONH4s- 2NH3g-CO2g the Kp-2-9-10-5atm3IF the reaction is started with 1 mol of the compound- the total pressure at equilibrium would be-

The correct option is A 1.94×10^ 2atm NH _ 2 COO NH _ 4( S ) #160; ⇌ 2 NH _ 3( g ) #160; + CO _ 2( g ) #160; K _ P =2.9× 10 ^ 5 atm If P ...

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Byju's Answer

Standard XII

Chemistry

Derivation of Kp and Kc

For the decom...

Question

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:

1.94 \times 10^{- 2} a t m

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5.82 \times 10^{- 2} a t m

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7.66 \times 10^{- 2} a t m

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38.8 \times 10^{- 2} a t m

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Solution

The correct option is A $1.94 \times 10^{- 2} a t m$
${N H}_{2} C O O {N H}_{4 (S)} ⇌ 2 {N H}_{3 (g)} + {C O}_{2 (g)}$
$K_{P} = 2.9 \times 10^{- 5} a t m$
If P is the total pressure at equilibrium
$K_{P} = {(\frac{2 P}{3})}^{2} (\frac{P}{3}) \Rightarrow P^{3} = \frac{27 \times 2.9 \times 10^{- 5}}{4}$
$P = 5.82 \times 10^{- 2} a t m$

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For the decomposition of the compound represented asNH2COONH4(s)⇌2NH3(g)+CO2(g) the Kp=2.9×10−5atm3IF the reaction is started with 1 mol of the compound, the total pressure at equilibrium would be:

The correct option is A 1.94×10−2atmNH2COONH4(S)⇌2NH3(g)+CO2(g)KP=2.9×10−5atmIf P is the total pressure at equilibrium KP=(2P3)2(P3)⇒P3=27×2.9×10−54P=5.82×10−2atm

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: