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Standard XII

Chemistry

Derivation of Kp and Kc

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 atm, because $S_{8 (g)}$ is partially converted into $S_{2 (g)}$ . Find the value of $K_{p}$ for this reaction.

0.96

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6.14

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204.8

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None of these

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Solution

The correct option is B None of these
$S_{8} ⇌ 4 S_{2} (g)$
Initially : $1$ $0$
At eqn $1 - x$ $4 x$ $\Rightarrow 0.7 1.2$
$x$ leads falls by $30$ %
$x = 03$
$K_{P} = \frac{{(4 x)}^{4}}{1 - x} = \frac{{(1.2)}^{4}}{0.7} = 2.96 {a t m}^{3}$ is the correct answer.

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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 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 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_{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.

Q. When sulphur in the form of

S_{8}

is heated at

900 K

, the initial pressure of one

a t m

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 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.

The correct option is B None of theseS8⇌4S2(g)Initially : 1 0At eqn 1−x 4x ⇒0.71.2x leads falls by 30%x=03KP=(4x)41−x=(1.2)40.7=2.96atm3 is the correct answer.

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.

The correct option is B None of these
$S_{8} ⇌ 4 S_{2} (g)$
Initially : $1$ $0$
At eqn $1 - x$ $4 x$ $\Rightarrow 0.7 1.2$
$x$ leads falls by $30$ %
$x = 03$
$K_{P} = \frac{{(4 x)}^{4}}{1 - x} = \frac{{(1.2)}^{4}}{0.7} = 2.96 {a t m}^{3}$ is the correct answer.