For the reaction BaO_2(s) leftharpoons BaO(s)+O_2(g) ; Δ H=+ve . In...

For the reaction $2 \mathrm{BaO}_2(\mathrm{~s}) \rightleftharpoons \mathrm{BaO}(\mathrm{s})+\mathrm{O}_2(\mathrm{~g}) ; \Delta \mathrm{H}= +\mathrm{ve}$ In equillibrium $\mathrm{K}_P=\mathrm{PO}_2$ Hence, the value of equilibrium constant depends only upon partial pressure of $\mathrm{O}_2$. Further on increasing temperature formation of $\mathrm{O}_2$ increases as this is an endothermic reaction. Hence, pressure of $\mathrm{O}_2$ is dependent on temperature.