Chemistry - Chemical Equilibrium Question with Solution | TestHub

a moles of $A(g)$ taken initially and at time Now moles fraction of $\mathrm{A}(\mathrm{g}),\mathrm{B}(\mathrm{g})$ and $\mathrm{C}(\mathrm{g})$ are
$\begin{array}{rl}& X_A=\frac{a-a\alpha }{a+a\alpha }=\frac{1-\alpha }{1+\alpha }\\ & X_B=\frac{a\alpha }{a+a\alpha }=\frac{\alpha }{1+\alpha }\\ & X_C=\frac{a\alpha }{a+a\alpha }=\frac{\alpha }{1+\alpha }\end{array}$
Now if $P$ is total pressure then partial pressure of $\mathrm{A}(\mathrm{g}),\mathrm{B}(\mathrm{g})$ and $\mathrm{C}(\mathrm{g})$ are
$\begin{array}{rl}& {\mathrm{P}}_{\mathrm{A}}=\left(\frac{1-\alpha }{1+\alpha }\right)\mathrm{P}\\ & {\mathrm{P}}_{\mathrm{B}}=\left(\frac{\alpha }{1+\alpha }\right)\mathrm{P}\\ & {\mathrm{P}}_{\mathrm{C}}=\left(\frac{\alpha }{1+\alpha }\right)\mathrm{P}\\ & {\mathrm{K}}_{\mathrm{P}}=\frac{\left(\frac{\alpha }{1+\alpha }\right)\mathrm{P}\left(\frac{\alpha }{1+\alpha }\right)\mathrm{P}}{\left(\frac{1-\alpha }{1+\alpha }\right)\mathrm{P}}\\ & {\mathrm{K}}_{\mathrm{P}}=\frac{{\alpha }^2\mathrm{P}}{1-{\alpha }^2}\end{array}$
As $K_P$ is only function of temperature.
So as P $\uparrow \alpha \downarrow$