For a reaction NH4COONH2(s) leftharpoons 2NH3(g) +CO2(g), the equilibrium pressure is 3 atm. Kp for the reaction will be

Question

For a reaction NH4COONH2(s) leftharpoons 2NH3(g) +CO2(g), the equilibrium pressure is 3 atm. Kp for the reaction will be (A) 27 (B) 4 (C) 3 (D) 9

Tardigrade Admin · Accepted Answer

NH4COONH2(s) leftharpoons underset text2p2NH3(g)+ underset textpCO2(g) When volume and temperature are constant, the number of moles of a gas is proportional to its partial pressure. So, 2p + p = 3 3p=3, ∴ p=1 atm Kp=(2p)2 × p=4p3 =4 × (1)3=4 atm3

Q. For a reaction $N H_{4} COON H_{2 (s)} ⇌ 2 N H_{3 (g)} + C O_{2 (g)}$ , the equilibrium pressure is $3 a t m$ . $K_{p}$ for the reaction will be

$27$

$4$

$3$

$9$

$N H_{4} COON H_{2 (s)} ⇌ 2p 2 N H_{3 (g)} + p C O_{2 (g)}$
When volume and temperature are constant, the number of moles of a gas is proportional to its partial pressure.
So, $2 p + p = 3$
$3 p = 3$ ,
$∴ p = 1 a t m$
$K_{p} = (2 p)^{2} \times p = 4 p^{3}$
$= 4 \times (1)^{3} = 4 a t m^{3}$

Q. For a reaction NH4​COONH2(s)​⇌2NH3(g)​+CO2(g)​, the equilibrium pressure is 3atm. Kp​ for the reaction will be

27

4

3

9

NH4​COONH2(s)​⇌2p2NH3(g)​​+pCO2(g)​​ When volume and temperature are constant, the number of moles of a gas is proportional to its partial pressure. So, 2p+p=3 3p=3, ∴p=1atm Kp​=(2p)2×p=4p3 =4×(1)3=4atm3

Q. For a reaction $N H_{4} COON H_{2 (s)} ⇌ 2 N H_{3 (g)} + C O_{2 (g)}$ , the equilibrium pressure is $3 a t m$ . $K_{p}$ for the reaction will be

$27$

$4$

$3$

$9$

$N H_{4} COON H_{2 (s)} ⇌ 2p 2 N H_{3 (g)} + p C O_{2 (g)}$
When volume and temperature are constant, the number of moles of a gas is proportional to its partial pressure.
So, $2 p + p = 3$
$3 p = 3$ ,
$∴ p = 1 a t m$
$K_{p} = (2 p)^{2} \times p = 4 p^{3}$
$= 4 \times (1)^{3} = 4 a t m^{3}$