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Standard XII

Chemistry

Types of Equillibrium

When sulphur ...

Question

When sulphur (in the form of $S_{8}$ ) is heated at temperature $T$ , at equilibrium, the pressure of $S_{8}$ falls by $30 %$ from $1.0 a t m$ , because $S_{8} (g)$ is partially converted into $S_{2} (g) .$
Find the value of $K_{p}$ for this reaction $S_{8} (g) ⇌ 4 S_{2} (g)$ .

2.96

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6.14

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4.28

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None of these

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Solution

The correct option is A $2.96$
$S_{8} (g) ⇌ 4 S_{2} (g)$
$Initial pressure 1 a t m 0$
$At Eqm (1 - 0.3) (4 \times 0.3)$
$= 0.7 a t m = 1.2 a t m$

$K_{p} = \frac{P_{S_{2}}^{4} (g)}{P_{S_{8} (g)}} K_{p} = \frac{(1.2)^{4}}{0.70} = 2.96$

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Similar questions

Q. When sulphur (in the form of

S_{8}

) is heated at temperature T, at equilibrium, the pressure of

S_{8}

falls by

30 %

from 1.0 atm, because

S_{8 (g)}

is partially converted into

S_{2 (g)}

. Find the value of

K_{p}

for this reaction.

Q. When sulphur ( in the form of

S_{8}

) is heated to temperature T, at equilibrium, the pressure of

S_{8}

falls by 30% from 1.0 atm, because

S_{8}

(g) is partially converted into

S_{2}

(g). Find the value of

K_{p}

for this reaction.

Q. When sulphur in the form of

S_{2} (g)

is heated at

900 K

, the initial partial pressure of

S_{8} (g)

which was

1 a t m

falls by

29 %

at equilibrium. This is because of conversion of some

S_{8} (g)

S_{2} (g)

. Find the

K_{p}

for reaction,

S_{8} (g) ⇌ 4 S_{2} (g)

Q. When Sulphur in the form of

S_{a}

is heated at

900 K

, the initial pressure of one atm falls by

29

% at equilibrium. Find out

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for the following reaction:

S_{8} (g) ⇌ 4 S_{2} (g)

Q. When S in the form of

S_{8}

is heated at 900K, the initial pressure of 1 atmosphere falls by 29% at equilibrium. This is because of conversion of some

S_{8} t o S_{2}

. Find the

K_{p}

for reaction :

S_{8} (g) \Leftrightarrow 4 S_{2} (g)

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Types of Equillibrium

Standard XII Chemistry

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When sulphur (in the form of S8) is heated at temperature T, at equilibrium, the pressure of S8 falls by 30% from 1.0 atm, because S8(g) is partially converted into S2(g). Find the value of Kp for this reaction S8(g)⇌4S2(g).

The correct option is A 2.96 S8(g)⇌4S2(g) Initial pressure 1 atm 0 At Eqm (1−0.3) (4×0.3) =0.7 atm =1.2 atm Kp=P4S2(g)PS8(g)Kp=(1.2)40.70=2.96

When sulphur (in the form of $S_{8}$ ) is heated at temperature $T$ , at equilibrium, the pressure of $S_{8}$ falls by $30 %$ from $1.0 a t m$ , because $S_{8} (g)$ is partially converted into $S_{2} (g) .$
Find the value of $K_{p}$ for this reaction $S_{8} (g) ⇌ 4 S_{2} (g)$ .

The correct option is A $2.96$
$S_{8} (g) ⇌ 4 S_{2} (g)$
$Initial pressure 1 a t m 0$
$At Eqm (1 - 0.3) (4 \times 0.3)$
$= 0.7 a t m = 1.2 a t m$

$K_{p} = \frac{P_{S_{2}}^{4} (g)}{P_{S_{8} (g)}} K_{p} = \frac{(1.2)^{4}}{0.70} = 2.96$