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Ammonium carbamate decomposes as NH(2)CO...

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

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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) (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

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

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