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(A) For the reaction: $A_{(g)} \rightleftharpoons B_{(g)} + \frac{1}{2} C_{(g)}$Initial moles: $1, 0, 0$At equilibrium: $(1 - \alpha), \alpha, \frac{\

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) For the reaction: $A_{(g)} \rightleftharpoons B_{(g)} + \frac{1}{2} C_{(g)}$Initial moles: $1, 0, 0$At equilibrium: $(1 - \alpha), \alpha, \frac{\alpha}{2}$Total moles at equilibrium = $(1 - \alpha) + \alpha + \frac{\alpha}{2} = 1 + \frac{\alpha}{2}$Partial pressures are given by $P_i = x_i \cdot p$,where $x_i$ is the mole fraction and $p$ is the total pressure.$P_A = \frac{1 - \alpha}{1 + \frac{\alpha}{2}} p$$P_B = \frac{\alpha}{1 + \frac{\alpha}{2}} p$$P_C = \frac{\alpha/2}{1 + \frac{\alpha}{2}} p$$K_p = \frac{P_B \cdot P_C^{1/2}}{P_A} = \frac{(\frac{\alpha}{1 + \alpha/2} p) \cdot (\frac{\alpha/2}{1 + \alpha/2} p)^{1/2}}{\frac{1 - \alpha}{1 + \alpha/2} p}$Simplifying the expression:$K_p = \frac{\alpha \cdot (\alpha/2)^{1/2} \cdot p^{1/2}}{1 - \alpha} \cdot \frac{(1 + \alpha/2)^{1/2}}{(1 + \alpha/2)^{1/2}} = \frac{\alpha \cdot \alpha^{1/2} \cdot p^{1/2}}{\sqrt{2} \cdot (1 - \alpha) \cdot (\frac{2 + \alpha}{2})^{1/2}} = \frac{\alpha^{3/2} p^{1/2}}{(2 + \alpha)^{1/2}(1 - \alpha)}$

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?

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