JEE Main 2022 (24 Jun, Shift 1) Chemistry Q64: Equilibrium

Question

For a reaction at equilibrium
$\mathrm{A(g)} \rightleftharpoons \mathrm{B(g)}+\frac{1}{2}\mathrm{C(g)}$
the relation between dissociation constant (K), degree of dissociation ($\alpha$) and equilibrium pressure (p) is given by :

Accepted Answer

Option 2: $K=\frac{\alpha^{\frac{3}{2}}p^{\frac{1}{2}}}{(2+\alpha)^{\frac{1}{2}}(1-\alpha)}$

Answer

$K=\frac{\alpha^{\frac{1}{2}}p^{\frac{3}{2}}}{\left(1+\frac{3}{2}\alpha\right)^{\frac{1}{2}}(1-\alpha)}$

Answer

$K=\frac{\alpha^{\frac{3}{2}}p^{\frac{1}{2}}}{(2+\alpha)^{\frac{1}{2}}(1-\alpha)}$

Answer

$K=\frac{(\alpha p)^{\frac{3}{2}}}{\left(1+\frac{3}{2}\alpha\right)^{\frac{1}{2}}(1-\alpha)}$

Answer

$K=\frac{(\alpha p)^{\frac{3}{2}}}{(1+\alpha)(1-\alpha)^{\frac{1}{2}}}$

JEE Main 24 June 2022, Shift 1, Chemistry Q64

Same idea in other shifts