A 10 litre box contains O3 and O2 at equilibrium at 2000 K- K p -4- 10 14 atm for 2 O 3 g- 3 O 2 g- Assume that P O 2 - P O 3 and if total pressure is 8 atm- then partial pressure of O3 will be-

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Standard XII

Chemistry

Le Chatelier's Principle for Del N Equals to 0

A 10 litre ...

Question

A $10$ litre box contains $O_{3}$ and $O_{2}$ at equilibrium at $2000 K$ . $K_{p} = 4 \times 10^{14} a t m$ for $2 O_{3} (g) ⇌ 3 O_{2} (g)$ . Assume that $P_{O_{2}} >> P_{O_{3}}$ and if total pressure is $8 a t m$ , then partial pressure of $O_{3}$ will be:

8 \times 10^{- 5} a t m

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11.3 \times 10^{- 7} a t m

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9.71 \times 10^{- 6} a t m

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9.71 \times 10^{- 2} a t m

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Solution

The correct option is B $11.3 \times 10^{- 7} a t m$
According to question:
The total pressure is $8 a t m$ .
$\Rightarrow P_{O_{2}} + P_{O_{3}} = 8 a t m$
But it is given that $P_{O_{2}} >> P_{O_{3}}$
$\Rightarrow P_{O_{2}} = 8 a t m$ since pressure of $O_{3}$ is negligible compared to $O_{2}$

Now, $k_{p} = \frac{{(P_{O_{2}})}^{3}}{{(P_{O_{3}})}^{2}}$

$4 \times 10^{14} = \frac{{(8)}^{3}}{{(P_{O_{3}})}^{2}}$

$P_{O_{3}} = \sqrt{\frac{8 \times 8 \times 8}{4 \times 10^{14}}}$

$P_{O_{3}} = 11.313 \times 10^{- 7}$
Hence option B is correct.

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Similar questions

A 10 litre box contains $O_{3}$ and $O_{2}$ at equilibrium at 2000K. $K_{P} = 4 \times 10^{14}$ atm for $2 O_{3} (g) ⇌ 3 O_{2} (g)$ Assume that $p_{O_{2}} >> p_{O_{3}}$ and if total pressure is 8 atm, then moles of $O_{3}$ at equilibrium will be:

Take $\sqrt{5} = 2.2$ , $\sqrt{3} = 1.7$ , $\sqrt{2} = 1.4$

Q. A

10 l i t r e

box contains

O_{3}

and

O_{2}

at equilibrium at 2000 K.

K_{p} = 4 \times 10^{14}

atm for

2 O_{3} (g) ⇌ 3 O_{2} (g)

. Assume that

P_{O_{2}} >> P_{O_{3}}

and if total pressure is 8 atm, then patial pressure of

O_{3}

will be:

Q. A

20 L

box contains

O_{3}

and

O_{2}

at equilibrium at

202 K

K_{p} = 2 \times 10^{14}

for

2 O_{3} ⇌ 3 O_{2}

. Assume that

p_{O_{3}} >> p_{O_{2}}

and total pressure is

8 a t m

, then the partial pressure of

O_{3}

is:

A 20 litre box contains $O_{3}$ and $O_{2}$ in equilibrium at 500K. $K_{p} = 4 \times 10^{14}$ atm for $2 O_{3} (g) ⇌ 3 O_{2} (g)$ . Assume that $p_{O_{2}} >> p_{O_{3}}$ and if the total pressure is 8 atm and the moles of $O_{3}$ at equilibrium are $x \times 10^{- 7}$ , x will be

(Take $\sqrt{5} = 2.2$ , $\sqrt{3} = 1.7$ , $\sqrt{2} = 1.4$ , R= 0.08 $a t m L m o l^{- 1} K^{- 1}$ )

Q. 10

l t

box contain

O_{3}

and

O_{2}

at equilibrium at 2000 K. The

Δ G^{o} = - 534.52 k J

at 8 atm equilibrium pressure. The following equilibrium is present in the container,

2 O_{3} (g) ⇌ 3 O_{2} (g)

. The partial pressure of O

_{3}

will be___________.

(l n 10 = 2.3, R = 8.3 J m o l e^{- 1} K^{- 1}

)

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A 10 litre box contains O3 and O2 at equilibrium at 2000 K. Kp=4×1014 atm for 2O3(g)⇌3O2(g). Assume that PO2>>PO3 and if total pressure is 8 atm, then partial pressure of O3 will be:

The correct option is B 11.3×10−7 atmAccording to question:The total pressure is 8atm.⇒PO2+PO3=8atmBut it is given that PO2>>PO3⇒PO2=8atm since pressure of O3 is negligible compared to O2Now, kp=(PO2)3(PO3)2 4×1014=(8)3(PO3)2PO3=√8×8×84×1014PO3=11.313×10−7Hence option B is correct.

A $10$ litre box contains $O_{3}$ and $O_{2}$ at equilibrium at $2000 K$ . $K_{p} = 4 \times 10^{14} a t m$ for $2 O_{3} (g) ⇌ 3 O_{2} (g)$ . Assume that $P_{O_{2}} >> P_{O_{3}}$ and if total pressure is $8 a t m$ , then partial pressure of $O_{3}$ will be: