For the reaction $\mathrm{BaO}_2(s) \rightleftharpoons \mathrm{BaO}(s)+\mathrm{O}_2(g) ; \Delta…

For the reaction
$\mathrm{BaO}_2(s) \rightleftharpoons \mathrm{BaO}(s)+\mathrm{O}_2(g) ; \Delta \mathrm{H}=+\mathrm{ve} \text {. }$
In an equilibrium condition, pressure of $\mathrm{O}_2$ depends on:
  1. increased mass of $\mathrm{BaO}_2$
  2. increased mass of $\mathrm{BaO}$
  3. increased temperature at equilibrium
  4. increased mass of both $\mathrm{BaO}_2$ and $\mathrm{BaO}$

Solution

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.

Asked in: NEET 2002

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