The degree of dissociation of $I_{2}$ molecule at $1000^{o}$ C and under atmospheric pressure is $40 %$ by volume. The total pressure on the gas at equilibrium so that dissociation is reduced to $20 %$ at the same temperature, will be.

4.57

atm

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

2.83

atm

No worries! We‘ve got your back. Try BYJU‘S free classes today!

5.33

atm

No worries! We‘ve got your back. Try BYJU‘S free classes today!

7.57

atm

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $4.57$ atm
$I_{2}$ dissociates as
$I_{2} ⇌ 2 I$

If $x$ is the degree of dissociation at $1000^{o} C$ under atmospheric pressure, then

Initial concentration: $[I_{2}] = 1, I = 0$

At equilibrium $I_{2} = 1 - x, I = 2 x$

The total number of moles at equilibrium $= 1 - x + 2 x = 1 + x$

Partial pressure of $I_{2}$ and $I$ will be

$P_{I_{2}} = (1 - x) p / (1 + x)$

$P_{I} = (2 x) p / (1 + x)$

$K_{p} = P_{I}^{2} / P_{I_{2}} = \frac{(2 x)^{2} p^{2} (1 + x)}{(1 - x) p (1 + x)^{2}} = \frac{(2 x)^{2} p}{(1 - x^{2})}$

If $p = 1$ atm and $x = 0.4$

$K_{p} = \frac{4 \times 0.16}{(0.6) (1.4)} = 0.76$

At $x = 0.2$ let pressure $p^{'}$

$K_{p} = \frac{(2 x)^{2} p^{'}}{(1 - x^{2})} 0.76 = \frac{(4 \times 0.04) p^{'}}{(0.8) (1.2)}$
$\Rightarrow p^{'} = 4.57 a t m$

Suggest Corrections

Similar questions

The degree of dissociation of I₂ molecule at 1000⁰C and under atmospheric pressure is 40% by volume. The total pressure on the gas at equilibrium so that the dissociation is reduced to 20% at the same temperature, will be

Q. The degree of dissociation of

P C l_{5}

at a certain temperature and under atmospheric pressure is 0.2. Calculate the pressure at which it will be half dissociated at the same temperature:

Q. Ammonia gas at 15 atm is introduced in a rigid vessel at 300 K. At equilibrium the total pressure of the vessel is found to be 40.11 atm at 300 C. The degree of dissociation of

N H_{3}

will be :

Q. Ammonia gas at

15

atm is introduced in a rigid vessel at

300

K. At equilibrium, the total pressure of the vessel is found to be

40.11

atm at

300^{0}

C. The degree of dissociation of

N H_{3}

will be :

Q. At some temperature and under a pressure of 4 atm,

P C l_{5}

is 10% dissociated. Calculated the pressure at which

P C l_{5}

will be 20% dissociated temperature remaining same.

Join BYJU'S Learning Program

The degree of dissociation of I2 molecule at 1000oC and under atmospheric pressure is 40% by volume. The total pressure on the gas at equilibrium so that dissociation is reduced to 20% at the same temperature, will be.

The degree of dissociation of $I_{2}$ molecule at $1000^{o}$ C and under atmospheric pressure is $40 %$ by volume. The total pressure on the gas at equilibrium so that dissociation is reduced to $20 %$ at the same temperature, will be.