When sulphur -in the form of S-8- is heated temperature T- 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 this reaction-0-96None of these6-14204-8

Question

Toppr Admin · Accepted Answer

S8-x21CC-4S2-g-Initially - 1-xA0- -xA0- -xA0- -xA0- -xA0- 0At eqn 1-x2212-x-xA0- -xA0- -xA0-4x-xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0-xA0-x21D2-0-71-2x leads falls by 30-x-03KP-4x-41-x2212-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.
0.96
None of these
6.14
204.8

$S_{8} ⇌ 4 S_{2} (g)$
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.

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.0.96None of these6.14204.8

S8⇌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.
0.96
None of these
6.14
204.8

$S_{8} ⇌ 4 S_{2} (g)$
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.