For the decomposition reaction NH2COONH4 (s) leftharpoons 2NH3(g) + CO2(g) The Kp = 2.9 × 10-5 atm3. The total pressure of gases at equilibrium when 1 mole of NH2COONH4 (s) was taken to start with would be

Question

For the decomposition reaction NH2COONH4 (s) leftharpoons 2NH3(g) + CO2(g) The Kp = 2.9 × 10-5 atm3. The total pressure of gases at equilibrium when 1 mole of NH2COONH4 (s) was taken to start with would be (A) 0.0194 atm (B) 0.0388 atm (C) 0.0582 atm (D) 0.0766 atm

Tardigrade Admin · Accepted Answer

underset12COOH4(s) leftharpoons underset22HH3 + underset1CO2(g) Kp = 2.9 × 10-5atm3 If the P is the total pressure at equilibrium Kp =( (2p/3))2 ((P/3)) ∴ P3 =(27 ×2.9×10-5/4) =4.69575 P=√[3]1.6575 =0.058

Q. For the decomposition reaction
$N H_{2} COON H_{4} (s) ⇌ 2 N H_{3} (g) + C O_{2} (g)$ The $K p = 2.9 \times 1 0^{- 5} a t m^{3}$ . The total pressure of gases at equilibrium when $1$ mole of $N H_{2} COON H_{4} (s)$ was taken to start with would be

$0.0194 a t m$

$0.0388 a t m$

$0.0582 a t m$

$0.0766 a t m$

$1 2 COO H_{4} (s) ⇌ 2 2 H H_{3} + 1 C O_{2} (g)$
$K_{p} = 2.9 \times 1 0^{- 5} a t m^{3}$
If the P is the total pressure at equilibrium
$K_{p} = (\frac{2 p}{3})^{2} (\frac{P}{3})$
$∴ P^{3} = \frac{27 \times 2.9 \times 1 0 ^{- 5}}{4} = 4.69575$
$P = 3 1.6575 = 0.058$

Q. For the decomposition reaction NH2​COONH4​(s)⇌2NH3​(g)+CO2​(g) The Kp=2.9×10−5atm3. The total pressure of gases at equilibrium when 1 mole of NH2​COONH4​(s) was taken to start with would be

0.0194atm

0.0388atm

0.0582atm

0.0766atm