Reaction  ${\mathrm{BaO}}_2\left(\mathrm{s}\right) \rightleftharpoons \mathrm{BaO}\left(\mathrm{s}\right) + {\mathrm{O}}_2\left(\mathrm{g}\right);$  $∆\mathrm{H}=+\mathrm{ve}$. In equilibrium condition, pressure of ${\mathrm{O}}_2$ depends on

${\mathrm{BaO}}_2\left(\mathrm{s}\right) \rightleftharpoons \mathrm{BaO}\left(\mathrm{s}\right) + {\mathrm{O}}_2\left(\mathrm{g}\right);$    $∆\mathrm{H}=+\mathrm{ve}$

According to the law of mass action, the rate of forward reaction = ${\mathrm{r}}_1$

${\mathrm{r}}_1 \propto \left[{\mathrm{BaO}}_2\right] \mathrm{or} {\mathrm{r}}_1 ={\mathrm{k}}_1 \left[\mathrm{BaO}\right]$

But concentration of solid = 1, then ${\mathrm{r}}_1 ={\mathrm{k}}_1$

Similarly the rate of backward reaction = $r_2$

${\mathrm{r}}_2 \propto \left[\mathrm{BaO}\right] \left[{\mathrm{O}}_2\right] \mathrm{or} {\mathrm{r}}_2 ={\mathrm{k}}_2 \left[\mathrm{BaO}\right] \left[{\mathrm{O}}_2\right]$

$\because \mathrm{conc}. \mathrm{of} \left[\mathrm{BaO}\right]=1$

or    ${\mathrm{r}}_2 ={\mathrm{k}}_2 \left[{\mathrm{O}}_2\right]$

At equilibrium, ${\mathrm{r}}_1$ = ${\mathrm{r}}_2$

${\mathrm{k}}_1 ={\mathrm{k}}_2 \left[{\mathrm{O}}_2\right]$   or    ${\mathrm{k}}_1 ={\mathrm{k}}_2. {\mathrm{p}}_{{\mathrm{O}}_2}$

where, $p_{O_2}$ = partial pressure of ${\mathrm{O}}_2$

or   $\frac{{\mathrm{k}}_1}{{\mathrm{k}}_2} = {\mathrm{p}}_{{\mathrm{O}}_2}$ (equilibrium constant)

$\because \frac{{\mathrm{k}}_1}{{\mathrm{k}}_2} = \mathrm{k} \mathrm{or} {\mathrm{p}}_{{\mathrm{O}}_2}$

So, from the above it is clear that pressure of ${\mathrm{O}}_2$ does not depend upon the concentration of reactants. The given equation is an endothermic reaction. It the temperature of such reaction is increased, then dissociation of ${\mathrm{BaO}}_2$ would increase, and more ${\mathrm{O}}_2$ is produced.