The degree of dissociation of I2 molecule at 10000C and under atmospheric pressure is 40

L_2(g) ⇌2l(g) nbsp; At nbsp;t=0 1 0 At equilibrium 1 x 2 x nbsp;x=0.40, p_1=1 atm in case I 1 y 2y y=0.20, p_2=? nbsp; atm in case II ...

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Standard XII

Chemistry

Order of Reaction

The degree of...

Question

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

4.57 atm

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2.83 atm

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5.33 atm

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7.57 atm

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Solution

The correct option is C
5.33 atm

$l_{2} (g) ⇌ 2 l (g)$

At $t = 010$

At equilibrium $1 - x 2 x x = 0.40, p_{1} = 1 a t m in case I$

$1 - y 2 y y = 0.20, p_{2} = ? a t m in case I I$

$K_{p} = \frac{p_{1}^{2}}{p_{1_{2}}} = \frac{4 x^{2} p_{1}}{(1 - x)} = \frac{4 y^{2} p_{2}}{(1 - y)}$

$∴ \frac{4 \times 0.16 \times 1}{(1 - 0.4)} = \frac{4 \times 0.04 \times p_{2}}{(1 - 0.2)}$
$\Rightarrow p_{2} = 5.33 a t m$

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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

The correct option is C 5.33 atml2(g)⇌2l(g) At t=010 At equilibrium 1−x2x x=0.40,p1=1 atm in caseI 1−y2y y=0.20,p2=? atm in caseII Kp=p21p12=4x2p1(1−x)=4y2p2(1−y) ∴4×0.16×1(1−0.4)=4×0.04×p2(1−0.2) ⇒p2=5.33 atm

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