For the rxn NH2COONH4-s- - 2 NH3-g- - CO2-g- the equi constant Kp- 5 - 10-4 atm3 the tottal pressure of the gaseous product when 1 mole of reactant is heated will be - 0-15 atm -

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Standard XII

Chemistry

Lattice Energy

for the rxn N...

Question

for the rxn NH2COONH4(s) = 2 NH3(g) + CO2(g) the equi constant Kp= 5 * 10^-⁴ atm3 the tottal pressure of the gaseous product when 1 mole of reactant is heated will be :( 0.15 atm )

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Solution

Make the ICE table for reaction:
NH2COONH4(s) 2 NH3(g) CO2(g)
Initial pressure conc. of 1 mole 0 0
Final pressure 1-x 2x x
Let us write the expression for Kp.As ammonium carbamate is a solid, it does not appear in the expression for Kp
Kp = (pNH3(g)²CO₂(g) = (2x)²(x) = 4x³
Given Kp = 5 x 10^- ⁴atm³
âˆ´5 x 10^- ⁴= 4x³
X = 0.05atm
Total pressure at equilibrium = 2x + x = 2*0.05 + 0.5 = 0.15atm

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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.

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	NH2COONH4(s)	2 NH3(g)	CO2(g)
Initial pressure	conc. of 1 mole	0	0
Final pressure	1-x	2x	x

for the rxn NH2COONH4(s) = 2 NH3(g) + CO2(g) the equi constant Kp= 5 * 10-4 atm3 the tottal pressure of the gaseous product when 1 mole of reactant is heated will be :( 0.15 atm )

for the rxn NH2COONH4(s) = 2 NH3(g) + CO2(g) the equi constant Kp= 5 * 10^-⁴ atm3 the tottal pressure of the gaseous product when 1 mole of reactant is heated will be :( 0.15 atm )