A _ (g) ⇌ B _ (g) + C 2 ^ (g) . The correct relationship between K...

$\begin{cases} & {A}_{\left(g\right)} & ⇌ & {B}_{\left(g\right)} & + & \dfrac{1}{2}{C}_{\left(g\right)} \\ att = 0 & 1 & & 0 & & 0 \\ ateq . & 1 - \alpha & & \alpha & & \dfrac{\alpha}{2}\end{cases}$

Total moles at equilibrium: $= 1 - \alpha + \alpha + \dfrac{\alpha}{2} = 1 + \dfrac{\alpha}{2}$

${P}_{B} = \dfrac{\alpha}{1 + \dfrac{\alpha}{2}}p{P}_{c} = \dfrac{\dfrac{\alpha}{2}}{1 + \dfrac{\alpha}{2}}p{P}_{A} = \dfrac{1 - \alpha}{1 + \dfrac{\alpha}{2}}p$

${K}_{p} = \dfrac{{P}_{B} \times {P}_{C}^{\dfrac{1}{2}}}{{P}_{A}} = \dfrac{\left(\dfrac{\alpha}{1 + \dfrac{\alpha}{2}}p\right)\left(\dfrac{\dfrac{\alpha}{2}}{1 + \dfrac{\alpha}{2}}p\right)\dfrac{1}{2}}{\left(\dfrac{1 - \alpha}{1 + \dfrac{\alpha}{2}}p\right)}$

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

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