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Standard XII

Chemistry

Characteristics of Chemical Equilibrium

When sulphur ...

Question

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 :

0.2

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0.512

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1.51

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2

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Solution

The correct option is C $0.512$
$S_{1} g$ $⇌$ $a S_{2}$
at $t = 0$ $1$
at $t = e q$ $1 - 0.2$ $4 x 0.2$
$0.1$ $0.8$
$k_{p} = \frac{p (s_{2})}{p S_{g}} = \frac{(0.8)^{4}}{0.8} = (0.8)^{3} = 0.512$

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Similar questions

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

K_{p}

for the following 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

900

S_{8}

is converted to

S_{2}

. What will be the equilibrium constant for the reaction if initial pressure of

1

atm falls by

25 %

at equilibrium?

Q. When sulphur in the form of

S_{8}

is heated at 900

K

, the initial pressure of 1

atm

falls by

20 %

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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	$S_{1} g$	$⇌$	$a S_{2}$
at $t = 0$	$1$
at $t = e q$	$1 - 0.2$		$4 x 0.2$
	$0.1$		$0.8$

When sulphur is heated at 800 K, the initial pressure of 1 atm decreases by 20 at equilibrium. Kp of the reaction S8(s)⇌4S2(g) is :

The correct option is C 0.512S1g⇌aS2at t=01at t=eq1−0.24x0.20.10.8kp=p(s2)pSg=(0.8)40.8=(0.8)3=0.512

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 :

The correct option is C $0.512$
$S_{1} g$ $⇌$ $a S_{2}$
at $t = 0$ $1$
at $t = e q$ $1 - 0.2$ $4 x 0.2$
$0.1$ $0.8$
$k_{p} = \frac{p (s_{2})}{p S_{g}} = \frac{(0.8)^{4}}{0.8} = (0.8)^{3} = 0.512$