Ammonium carbamate decomposes as, NH 2 COONH 4(s) leftharpoons 2 NH 3(g)+ CO 2(g) For the reaction, Kp=2.9 × 10-5 atm 3 If we start with 1 mole of the compound, the total pressure at equilibrium would be

Question

Ammonium carbamate decomposes as, NH 2 COONH 4(s) leftharpoons 2 NH 3(g)+ CO 2(g) For the reaction, Kp=2.9 × 10-5 atm 3 If we start with 1 mole of the compound, the total pressure at equilibrium would be (A) 0.0766 atm (B) 0.0582 atm (C) 0.0388 atm (D) 0.0194 atm

Tardigrade Admin · Accepted Answer

NH 2 COONH 4(s) leftharpoons 2 NH 3(g)+ CO 2( g ) At eqm, if partial pressure of CO 2=p, then that of NH 3=2 p. Kp=p NH 32 × p CO 2=(2 p)2 × p=4 p3 2.9 × 10-5=4 p3 or p3=7.25 × 10-6 ⇒ p=1.935 × 10-2 Hence, total pressure =3 p=5.81 × 10-2=0.0581 atm

Q. Ammonium carbamate decomposes as,
$N H_{2} COON H_{4 (s)} ⇌ 2 N H_{3 (g)} + C O_{2 (g)}$
For the reaction, $K_{p} = 2.9 \times 1 0^{- 5} a t m^{3}$ If we start with $1$ mole of the compound, the total pressure at equilibrium would be

$0.0766 a t m$

$0.0582 a t m$

$0.0388 a t m$

$0.0194 a t m$

Q. Ammonium carbamate decomposes as, NH2​COONH4(s)​⇌2NH3(g)​+CO2(g)​ For the reaction, Kp​=2.9×10−5atm3 If we start with 1 mole of the compound, the total pressure at equilibrium would be

0.0766atm

0.0582atm

0.0388atm

0.0194atm

NH2​COONH4(s)​⇌2NH3(g)​+CO2(g)​ At eqm, if partial pressure of CO2​=p, then that of NH3​=2p. Kp​=pNH3​2​×pCO2​​=(2p)2×p=4p3 2.9×10−5=4p3 or p3=7.25×10−6 ⇒p=1.935×10−2 Hence, total pressure =3p=5.81×10−2=0.0581atm

Q. Ammonium carbamate decomposes as,
$N H_{2} COON H_{4 (s)} ⇌ 2 N H_{3 (g)} + C O_{2 (g)}$
For the reaction, $K_{p} = 2.9 \times 1 0^{- 5} a t m^{3}$ If we start with $1$ mole of the compound, the total pressure at equilibrium would be

$0.0766 a t m$

$0.0582 a t m$

$0.0388 a t m$

$0.0194 a t m$