Question

When sulphur is heated at 900 K, si converted to $\mathrm { S } _ { 2 }$ what will be the equilibrium constant for the reaction if initial pressure of 1 atm falls by 25% at equilibrium?

• A. $0.75 \mathrm {~atm} ^ { 3 }$
• B. $2.55 \mathrm {~atm} ^ { 3 }$
• C.
• D. $1.33 \mathrm {~atm} ^ { 3 }$

Answer 1

Answer: (D)

$1.33 \mathrm {~atm} ^ { 3 }$

Answer 2

: $S _ { 8 ( \mathrm {~g} ) }$ $4 S _ { 2 ( \mathrm {~g} ) }$

Initial pressure 1 0

At equilibrium $1 - \frac { 25 } { 100 }$ $4 \times \frac { 25 } { 100 }$

$= 0.75$ = 1

$\mathrm { K } _ { \mathrm { p } } = \frac { \left( \mathrm { P } _ { \mathrm { S } _ { 2 } } \right) ^ { 4 } } { \mathrm { P } _ { \mathrm { S } _ { 8 } } } = \frac { ( 1 ) ^ { 4 } } { 0.75 } = 1.33 \mathrm {~atm} ^ { 3 }$