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Byju's Answer

Standard XII

Physics

Velocity Displacement Relationship

When Sulphur ...

Question

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 $K_{p}$ for the following reaction:
$S_{8} (g) ⇌ 4 S_{2} (g)$

1.16 {a t m}^{3}

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0.71 {a t m}^{3}

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2.55 {a t m}^{3}

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5.1 {a t m}^{3}

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Solution

The correct option is C $2.55 {a t m}^{3}$
$S_{8} (q) ⇌ 4 S_{2} (g)$
Initial (atm) $1$ $0$
Change (atm) $- 0.29$ $4 \times 0.29$
(29% decremas of initial)
Final (atm) $0.71$ $1.16$
(as $e q u^{m})$
$k_{p} = \frac{[^{p} S_{2}]^{4}}{[^{p} S_{8}]} = \frac{(1.16)^{4}}{0.71}$
$= 2.55 a t m^{3}$

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Similar questions

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_{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)

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)

Q. When sulphur is heated at

800 K

, the initial pressure of

1

atm decreases by

20

at equilibrium.

K_{p}

of the reaction

S_{8} (s) ⇌ 4 S_{2} (g)

is :

Q. When sulphur in the form of

S_{8}

is heated at 900K, the initial pressure of one atm of

S_{(} 8)

falls by 29% at equilibrium. This is because of conversion of some

S_{8}

S_{2}

. Find the value of equilibrium constant for this reaction.

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When Sulphur in the form of Sa is heated at 900K, the initial pressure of one atm falls by 29% at equilibrium. Find out Kp for the following reaction:S8(g)⇌4S2(g)

The correct option is C 2.55atm3 S8(q)⇌4S2(g)Initial (atm) 1 0Change (atm) −0.29 4×0.29 (29% decremas of initial)Final (atm) 0.71 1.16(as equm)kp=[pS2]4[pS8]=(1.16)40.71=2.55atm3

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 $K_{p}$ for the following reaction:
$S_{8} (g) ⇌ 4 S_{2} (g)$

The correct option is C $2.55 {a t m}^{3}$
$S_{8} (q) ⇌ 4 S_{2} (g)$
Initial (atm) $1$ $0$
Change (atm) $- 0.29$ $4 \times 0.29$
(29% decremas of initial)
Final (atm) $0.71$ $1.16$
(as $e q u^{m})$
$k_{p} = \frac{[^{p} S_{2}]^{4}}{[^{p} S_{8}]} = \frac{(1.16)^{4}}{0.71}$
$= 2.55 a t m^{3}$