Ammonium carbamate decomposes as :
`NH_(2)COONH_(4) (s) rarr 2NH_(3)(g) + CO_(2)(g)`
For the reaction, `K_(P) = 2.9 xx 10^(-5) atm^(3)` If we start with 1 mole of the compound, the total pressure at equilibrium would be

Ammonium carbamate decomposes as : NH_(2)COONH_(4) rarr 2NH_(3)(g) + CO_(2)(g) For the reaction, K_(P) = 2.9 xx 10^(-5) atm^(3) If we start with 1 mole of the compound, the total pressure at equilibrium would be

Ammonium carbamate decomposes as NH_(2)COONH_(4)(s) hArr 2NH_(3)(g) +CO_(2)(g) . The value of K_(p) for the reaction is 2.9 xx 10^(-5) atm^(3) . If we start the reaction with 1 mole of the compound, the total pressure at equilibrium would be

For the decomposition of the compound, represented as NH_(2)COONH_(4)(s)hArr 2NH_(3)(g)+CO_(2)(g) the _K(p)=2.9xx10^(-5)" atm"^(3) . If the reaction is started with 1 mol of the compound, the total pressure equilibrium would be:

NH_(4)COONH_(4)(s) hArr 2NH_(3)(g)+CO_(2)(g) . for the above reaction, K_(p) is 4, K_c will be

Unit of K_p for NH_4COONH_(2(s)) harr 2NH_(3(g))+ CO_(2(g)) is

For the reaction NH_(2)COONH_(4)(g)hArr 2NH_(3)(g)+CO_(2)(g) the equilibrium constant K_(p)=2.92xx10^(-5)atm^(3) . The total pressure of the gaseous products when 1 mole of reactant is heated, will be

For the reaction NH_(2)COONH_(4)(g)hArr 2NH_(3)(g)+CO_(2)(g) the equilibrium constant K_(p)=2.92xx10^(-5)atm^(3) . The total pressure of the gaseous products when 1 mole of reactant is heated, will be

For the decomposition reaction NH_2COONH_4(s) hArr 2NH_3(g) +CO_2(g) the K_p =2.9 xx 10^(-5) "atm"^3 . The total pressure of gases at equilibrium when 1 mole of NH_2COONH_4 (s) was taken to start with would be