For first order reaction,

$\mathrm{X}\rightleftharpoons \mathrm{Y}$

Given, ${\mathrm{K}}_{\mathrm{f}}=2 {\mathrm{s}}^{-1}$

${\mathrm{K}}_{\mathrm{b}}=1 {\mathrm{s}}^{-1}$

$\therefore {\mathrm{K}}_{\mathrm{eq}}=\frac{{\mathrm{K}}_{\mathrm{f}}}{{\mathrm{K}}_{\mathrm{b}}}=\frac{\left[\mathrm{Y}\right]}{\left[\mathrm{X}\right]}$

${\mathrm{K}}_{\mathrm{eq}}=\frac{2}{1}=\frac{\left[\mathrm{Y}\right]}{\left[\mathrm{X}\right]}$

$\Rightarrow 2\left[\mathrm{X}\right]=\left[\mathrm{Y}\right]$

or $2{\left[\mathrm{X}\right]}_{\mathrm{eq}}={\left[\mathrm{Y}\right]}_{\mathrm{eq}}$

$\therefore \left[\mathrm{Y}\right]>\left[\mathrm{X}\right]$

Thus, as the rate of reaction increases, there will be increased in the concentration ${\mathrm{Y}}_{\mathrm{eq}},$ but it will be less than initial concentration of $\mathrm{X}\left({\mathrm{X}}_0\right),$ whereas the concentration of ${\mathrm{X}}_{\mathrm{eq}}$ decreases.