The equilibrium constant ( K p ) for the decomposition of water H 2 O ( g ) arrow H 2( g )+(1/2) O 2( g ) Is given by Where α is the degree of dissociation and p is the pressure of the reaction mixture at equilibrium.

Question

The equilibrium constant ( K p ) for the decomposition of water H 2 O ( g ) arrow H 2( g )+(1/2) O 2( g ) Is given by Where α is the degree of dissociation and p is the pressure of the reaction mixture at equilibrium. (A) K P =(α3 P (1/2)/√2) (B) K P =(α3 P (3/2)/(1-α)(2+α)(1)2) (C) K P =(α3 P (1/2)/(1+α)(2-α)(1)2) (D) K P =(α3 / 2 P 1 / 2/(1-α)(2+α)1 / 2)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/1eca89d40075c3437877e0ba054b25fe-.png" /> K p =((α)(α / 2)1 / 2/(1-α))(( p /1+(α)2))1 / 2 =(α3 / 2 P 1 / 2 √2/√2(1-α)(2+α)1 / 2) =(α3 / 2 P 1 / 2/(1-α)(2+α)1 / 2)

Q. The equilibrium constant $(K_{p})$ for the decomposition of water $H_{2} O (g) \to H_{2} (g) + \frac{1}{2} O_{2} (g)$ Is given by
Where $α$ is the degree of dissociation and $p$ is the pressure of the reaction mixture at equilibrium.

$K_{P} = \frac{α ^{3} P ^{\frac{1}{2}}}{2}$

$K_{P} = \frac{α ^{3} P ^{\frac{3}{2}}}{( 1 - α ) ( 2 + α ) ^{\frac{1}{2}}}$

$K_{P} = \frac{α ^{3} P ^{\frac{1}{2}}}{( 1 + α ) ( 2 - α ) ^{\frac{1}{2}}}$

$K_{P} = \frac{α ^{3/2} P ^{1/2}}{( 1 - α ) ( 2 + α ) ^{1/2}}$

$K_{p} = \frac{( α ) ( α /2 ) ^{1/2}}{( 1 - α )} (\frac{p}{1 + \frac{α}{2}})^{1/2}$
$= \frac{α ^{3/2} P ^{1/2} 2}{2 ( 1 - α ) ( 2 + α ) ^{1/2}}$
$= \frac{α ^{3/2} P ^{1/2}}{( 1 - α ) ( 2 + α ) ^{1/2}}$

Q. The equilibrium constant (Kp​) for the decomposition of water H2​O(g)→H2​(g)+21​O2​(g) Is given by Where α is the degree of dissociation and p is the pressure of the reaction mixture at equilibrium.

KP​=2​α3P21​​

KP​=(1−α)(2+α)21​α3P23​​

KP​=(1+α)(2−α)21​α3P21​​

KP​=(1−α)(2+α)1/2α3/2P1/2​