Chemistry - Chemical Equilibrium Question with Solution | TestHub

$\begin{array}{cccccc}& {\mathrm{A}}_{\left(\mathrm{g}\right)}& \rightleftharpoons & {\mathrm{B}}_{\left(\mathrm{g}\right)}& +& \frac{1}{2}{\mathrm{C}}_{\left(\mathrm{g}\right)}\\ \mathrm{at} \mathrm{t}=0& 1& & 0& & 0\\ \mathrm{at} \mathrm{eq}.& 1-\mathrm{\alpha }& & \mathrm{\alpha }& & \frac{\mathrm{\alpha }}{2}\end{array}$

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

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

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

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

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