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Standard XII

Chemistry

Law of Mass Action

A 20 litre bo...

Question

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 $a t m L m o l^{- 1} K^{- 1}$ )

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Solution

$P V = n R T$

$8 \times 20 = n \times 0.08 \times 500$

$n \approx 4 = \sum n$

$k_{p} = \frac{(n O_{2})^{3}}{(n O_{3})^{2}} \times \frac{P}{\sum n}$

$P \propto n$

$p O_{2} >> p O_{3} \to n O_{2} >> n O_{3}$

$∴ \sum n = n O_{2}$

$4 \times 10^{14} = \frac{\sum n^{2}}{(n O_{3})^{2}} \times P$

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

$n O_{3} = 4 \times \sqrt{2} \times 10^{- 7} = 5.6 \times 10^{- 7}$

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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

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:

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}

)

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:

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A 20 litre box contains O3 and O2 in equilibrium at 500K. Kp=4×1014 atm for 2O3(g)⇌3O2(g). Assume that pO2>>pO3 and if the total pressure is 8 atm and the moles of O3 at equilibrium are x×10−7, x will be (Take √5=2.2, √3=1.7, √2=1.4, R= 0.08 atm L mol−1 K−1)

PV=nRT 8×20=n×0.08×500 n≈4=∑n kp=(nO2)3(nO3)2×P∑n P∝n pO2>>pO3 → nO2>>nO3 ∴∑n=nO2 4×1014=∑n2(nO3)2×P (nO3)2=42×84×1014 nO3=4×√2×10−7=5.6×10−7