For a reaction at equilibrium A(g) B(g)+ 1 2 C(g) the relation between dissociation…
Chemistry · JEE Main · NTA Exams — Equilibrium
For a reaction at equilibrium
\(A(g) \rightleftharpoons B(g)+\frac{1}{2} C(g)\)
the relation between dissociation constant (K), degree of dissociation (\(\alpha\)) and equilibrium pressure (p) is given by: (JEE Main 2022)
\(A(g) \rightleftharpoons B(g)+\frac{1}{2} C(g)\)
the relation between dissociation constant (K), degree of dissociation (\(\alpha\)) and equilibrium pressure (p) is given by: (JEE Main 2022)
- \[K=\frac{\alpha^{\frac{1}{2}} p^{\frac{3}{2}}}{\left(1+\frac{3}{2} \alpha\right)^{\frac{1}{2}}(1-\alpha)}\]
- \(K=\frac{\alpha^{\frac{3}{2}} p^{\frac{1}{2}}}{(2+\alpha)^{\frac{1}{2}}(1-\alpha)}\)
- \[K=\frac{(\alpha p)^{\frac{3}{2}}}{\left(1+\frac{3}{2} \alpha\right)^{\frac{1}{2}}(1-\alpha)}\]
- \(K=\frac{(\alpha p)^{\frac{3}{2}}}{(1+\alpha)(1-\alpha)^{\frac{1}{2}}}\)
Answer
(B) K= ^ 3 2 p^ 1 2 (2+ )^ 1 2 (1- )
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