for the reaction: NH2COONH4 --- 2NH3 + CO2/ the equilibrium constant kp= 2 9*10^ -5 atm^3 the - Chemistry - Equilibrium

Question

for the reaction: NH2COONH4 ---> 2NH3 + CO2/
the equilibrium constant kp= 2 9*10^ -5 atm^3 the total pressure of
gases at equilibrium when 1 0 mole of reactant was heated will be

Geetha · Accepted Answer

NH2COONH4 →2NH3 + CO2AT equilibrium, partial pressure of CO2 = pNH3 = 2pKP = pNH32 × pCO2 = 2p2 × p = 4p34p3 = 2
9 × 10-5p = 1
9×10-2Total pressure = 2p +p = 3p     = 3 × 1
9 × 10-2    = 5
81 × 10-2   = 0 0581 atm

for the reaction: NH2COONH4 ---> 2NH3 + CO2/ the equilibrium constant kp= 2.9*10^ -5 atm^3. the total pressure of gases at equilibrium when 1.0 mole of reactant was heated will be

for the reaction: NH2COONH4 ---> 2NH3 + CO2/
the equilibrium constant kp= 2.9*10^ -5 atm^3. the total pressure of gases at equilibrium when 1.0 mole of reactant was heated will be