A 10 L box contains O 3 and O 2 at equilibrium at 2000 K. KP=4.17 × 1014 atm for 2 O 3 leftharpoons 3 O 2 . Assume that P O 2 P O 3 and if total pressure is 7.33 atm, then partial pressure of O 3 will be

Question

A 10 L box contains O 3 and O 2 at equilibrium at 2000 K. KP=4.17 × 1014 atm for 2 O 3 leftharpoons 3 O 2 . Assume that P O 2 > > P O 3 and if total pressure is 7.33 atm, then partial pressure of O 3 will be (A) 9.71 × 10-5 atm (B) 9.71 × 10-7 atm (C) 9.71 × 10-6 atm (D) 9.71 × 10-2 atm

Tardigrade Admin · Accepted Answer

Kp=4.17 × 1014=((P O 2)3/(P O 3)2) ∴ P O 2 > > P O 3 ∴ P O 2+P O 3=7.33 or P O 2=7.33 ∴ P O 3=√((7.33)3/4.17 × 1014) =9.71 × 10-7 atm

Q. A $10 L$ box contains $O_{3}$ and $O_{2}$ at equilibrium at $2000 K$ .
$K_{P} = 4.17 \times 1 0^{14}$ atm for $2 O_{3} ⇌ 3 O_{2} .$ Assume that $P_{O_{2}} >> P_{O_{3}}$ and if total pressure is $7.33 a t m$ , then partial pressure of $O_{3}$ will be

$9.71 \times 1 0^{- 5}$ atm

$9.71 \times 1 0^{- 7}$ atm

$9.71 \times 1 0^{- 6}$ atm

$9.71 \times 1 0^{- 2}$ atm

$K_{p} = 4.17 \times 1 0^{14} = \frac{( P _{O_{2}} ) ^{3}}{( P _{O_{3}} ) ^{2}}$
$∴ P_{O_{2}} >> P_{O_{3}}$
$∴ P_{O_{2}} + P_{O_{3}} = 7.33$
or $P_{O_{2}} = 7.33$
$∴ P_{O_{3}} = \frac{( 7.33 ) ^{3}}{4.17 \times 1 0 ^{14}}$
$= 9.71 \times 1 0^{- 7}$ atm

Q. A 10L box contains O3​ and O2​ at equilibrium at 2000K. KP​=4.17×1014 atm for 2O3​⇌3O2​. Assume that PO2​​>>PO3​​ and if total pressure is 7.33atm, then partial pressure of O3​ will be

9.71×10−5 atm

9.71×10−7 atm

9.71×10−6 atm

9.71×10−2 atm