One mole of SO 3 was placed in a two litre vessel at a certain temperature. The following equilibrium was established in the vessel SO 3( g ) leftharpoons 2 SO 2( g )+ O 2( g ) At equilibrium, the vessel was found to contain 0.5 mol of SO 3. The value of K would be

Question

One mole of SO 3 was placed in a two litre vessel at a certain temperature. The following equilibrium was established in the vessel SO 3( g ) leftharpoons 2 SO 2( g )+ O 2( g ) At equilibrium, the vessel was found to contain 0.5 mol of SO 3. The value of K would be (A) 0.25 (B) 0.125 (C) 0.5 (D) 1

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<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/e904a36bcebb815418ee8e0d302ea7c0-.png" /> K=([ SO 2]2[ O 2]/[ SO 3]2) =((0.5 / 2)2(0.25 / 2)/(0.5 / 2)2)=0.125

Q. One mole of $S O_{3}$ was placed in a two litre vessel at a certain temperature. The following equilibrium was established in the vessel
$S O_{3} (g) ⇌ 2 S O_{2} (g) + O_{2} (g)$
At equilibrium, the vessel was found to contain $0.5 m o l$ of $S O_{3}$ . The value of $K$ would be

0.25

0.125

0.5

1

$K = \frac{[ S O _{2} ] ^{2} [ O _{2} ]}{[ S O _{3} ] ^{2}}$

$= \frac{( 0.5/2 ) ^{2} ( 0.25/2 )}{( 0.5/2 ) ^{2}} = 0.125$

Q. One mole of SO3​ was placed in a two litre vessel at a certain temperature. The following equilibrium was established in the vessel SO3​(g)⇌2SO2​(g)+O2​(g) At equilibrium, the vessel was found to contain 0.5mol of SO3​. The value of K would be

0.25

0.125

0.5

1

K=[SO3​]2[SO2​]2[O2​]​ =(0.5/2)2(0.5/2)2(0.25/2)​=0.125

Q. One mole of $S O_{3}$ was placed in a two litre vessel at a certain temperature. The following equilibrium was established in the vessel
$S O_{3} (g) ⇌ 2 S O_{2} (g) + O_{2} (g)$
At equilibrium, the vessel was found to contain $0.5 m o l$ of $S O_{3}$ . The value of $K$ would be

$K = \frac{[ S O _{2} ] ^{2} [ O _{2} ]}{[ S O _{3} ] ^{2}}$

$= \frac{( 0.5/2 ) ^{2} ( 0.25/2 )}{( 0.5/2 ) ^{2}} = 0.125$